Non-Overlapping Rings
Do you know of a function or series of functions that I can use to create non-overlapping rings around a table of points? Non-overlapping rings are circular buffers around points. Areas of overlap are removed where rings cross.
The link below is a non-overlapping rings example:
http://www.mappinganalytics.com/map/images/radius-map.jpg
If you do not have a function, what would be the logic for creating such a function? Any guidance would be welcome.
Thanks
Edited by: user11121688 on Jun 24, 2010 8:47 AM
Hi,
Sorry I misunderstood your non-overlapping ring definition . The boundaries of the rings are straight lines from the intersection points.
the following is my suggestion :
First figure out which buffers intersect. Using the sdo_distance between two points and the buffer distance (radius)
should do (distance < radius means intersecting!). Each buffer can have a set of intersecting buffers.
Then for each intersecting pair (buffer_i, buffer_j), find out the pairwise non-overlapping ring for each buffer.
Assume that buffer_i has the center c_i and , buffer_j has the center, c_j
If a buffer has no intersecting buffer,
its pairwise non-overlapping ring is itself.
else
use the exterior rings (as linestring) of each buffer and find out their intersections points, i_1 and i_2.
You might want to take care of some exception cases like buffers are touching (1 intersection pt cases)
or the same (same point cases).
Construct a 3-pt polygon t_j (c_j,i_1,i_2).
The pairwise non-overlapping ring for buffer_i w.r.t. buffer_j is sdo_difference(buffer_i,t_j, tolernace).
Find out all pairwise non-overlapping rings for buffer_i with all other intersecting buffers.
We have a set of pair-wise non-overlapping ring for buffer_i.
Find out the intersection for all such buffers and we have the aggregate non-overlapping ring for buffer_i.
For a buffer without any pairwise non-overlapping ring, its aggregate non-overlapping ring is itself.
Follow the same approach for all other buffers to find out all aggregate non-overlapping rings.
This approach should be applicable to both geodetic and projected coordinate systems.
hope it helps!
jack
Similar Messages
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Pure SQL to partition date-time occurrences into non-overlapping windows?
i've a question that so far, i've never been able to solve via a pure SQL option.
it's hard to explain in words, but it's something like this:
given a set of date-time, i would like to partition the data into non-overlapping windows of 30 minutes each.
the data is supposed to be partitioned into windows of 30 minutes, meaning when the data is within 30 minutes of the first occurrence, only the first occurrence will be returned. in the next second after the 30th minute, the record will be considered as the start of a new window and is also returned. so those data that occurs within the window period are suppressed. the first occurrence is not necessarily occurring on the 00th minute, so the window start will never be constant.
run the below query to look at the dummy data.
SELECT 'A' AS ID
, TRUNC(SYSDATE) + 7 / 24 + 1 *(ROWNUM - 1) / 1440 AS datetime
FROM DUAL
CONNECT BY ROWNUM <= 50
UNION ALL
SELECT 'A' AS ID
, TRUNC(SYSDATE) + 9 / 24 + 8 / 1440 + 1 *(ROWNUM - 1) / 1440 AS datetime
FROM DUAL
CONNECT BY ROWNUM <= 35
UNION ALL
SELECT 'B' AS ID
, TRUNC(SYSDATE) + 7 / 24 + 5 *(ROWNUM - 1) / 1440 AS datetime
FROM DUAL
CONNECT BY ROWNUM <= 15this is supposed to be the output.
ID DATETIME
A 5/19/2010 07:00:00
A 5/19/2010 07:30:00
A 5/19/2010 09:08:00
A 5/19/2010 09:38:00
B 5/19/2010 07:00:00
B 5/19/2010 07:30:00
B 5/19/2010 08:00:00so far, i'm using a PL/SQL to pipe the records. but i would like to know if this is achievable via SQL or not.
i've tried looking at analytics, width_bucket, ntile and alll options i can think of, but i just can't solve this at all.hey Bob,
your answer is most definitely correct and does what i want. i've verified it again my data set and it returns the results as required!
you've definitely proven me wrong. i was always under the impression that this wasn't possible. thanks!
just a small note:
i need the windows to be binned by seconds, so have changed the numtodsinterval to raw numbers.
WITH t AS
(SELECT 'A' AS ID
, TRUNC(SYSDATE) +(6.75 / 24) AS datetime
FROM DUAL
UNION ALL
SELECT 'A' AS ID
, TRUNC(SYSDATE) +(6.75 / 24) AS datetime
FROM DUAL
UNION ALL
SELECT 'A' AS ID
, TRUNC(SYSDATE) + 7 / 24 + 1 *(ROWNUM - 1) / 1440 AS datetime
FROM DUAL
CONNECT BY ROWNUM <= 50
UNION ALL
SELECT 'A' AS ID
, TRUNC(SYSDATE) + 9 / 24 + 8 / 1440 + 1 *(ROWNUM - 1) / 1440 AS datetime
FROM DUAL
CONNECT BY ROWNUM <= 35
UNION ALL
SELECT 'B' AS ID
, TRUNC(SYSDATE) + 7 / 24 + 5 *(ROWNUM - 1) / 1440 AS datetime
FROM DUAL
CONNECT BY ROWNUM <= 15)
,a AS
(SELECT ID
,datetime
,LAG(datetime) OVER(PARTITION BY ID ORDER BY datetime) AS prevtime
,LAST_VALUE(datetime) OVER(PARTITION BY ID ORDER BY datetime RANGE BETWEEN CURRENT ROW AND 30 / 1440 + 1 / 86400 FOLLOWING) AS interval_end
FROM t)
,b AS
(SELECT ID
,datetime
,LEAD(datetime) OVER(PARTITION BY ID ORDER BY datetime) AS nexttime
FROM t)
,ab AS
(SELECT a.ID
,a.datetime
,a.prevtime
,a.interval_end
,b.datetime as b_datetime
,b.nexttime
FROM a JOIN b ON(a.ID = b.ID
AND a.interval_end = b.datetime)
SELECT ID
,datetime
FROM ab
START WITH prevtime IS NULL
CONNECT BY ID = PRIOR ID
AND datetime = PRIOR nexttime
ORDER BY ID
,datetime;this most definitely proves that i'm still not sure of how to use hierarchy queries.
Edited by: casey on May 20, 2010 11:20 AM -
Groups having non-overlapping intervals
Hi there,
I need help on some mind boggling overlap problem, or perhaps non-overlap problem.
Having the sample data below, I need to put together subcodes in groups having non-overlapping
intervals [MIN_VAL;MAX_VAL[
I have this table:
SQL>create table t (product varchar2(1) not null
2 ,subcode number(1) not null, constraint subcode_chk check (subcode >= 0)
3 ,min_val number(2) not null, constraint min_val_chk check (min_val >= 0)
4 ,max_val number(2) not null, constraint max_val_chk check (max_val >= 0)
5 ,constraint t_pk primary key (product, subcode)
6 ,constraint t_val_chk check (min_val < max_val));
Table created.
SQL>
SQL>insert into t (product, subcode, min_val, max_val) values ('A', 0, 0, 99);
1 row created.
SQL>insert into t (product, subcode, min_val, max_val) values ('A', 1, 0, 99);
1 row created.
SQL>insert into t (product, subcode, min_val, max_val) values ('A', 2, 18, 67);
1 row created.
SQL>insert into t (product, subcode, min_val, max_val) values ('A', 3, 20, 65);
1 row created.
SQL>insert into t (product, subcode, min_val, max_val) values ('A', 4, 0, 99);
1 row created.
SQL>insert into t (product, subcode, min_val, max_val) values ('A', 5, 45, 60);
1 row created.
SQL>insert into t (product, subcode, min_val, max_val) values ('A', 6, 20, 40);
1 row created.
SQL>insert into t (product, subcode, min_val, max_val) values ('A', 7, 0, 99);
1 row created.
SQL>insert into t (product, subcode, min_val, max_val) values ('A', 8, 60, 65);
1 row created.
SQL>insert into t (product, subcode, min_val, max_val) values ('A', 9, 0, 99);
1 row created.
SQL>commit
2 /
Commit complete.
SQL>
SQL> select product, subcode, min_val, max_val
2 from t
3 where product = 'A'
4 order by product, subcode;
PRODUCT SUBCODE MIN_VAL MAX_VAL
A 0 0 99
A 1 0 99
A 2 18 67
A 3 20 65
A 4 0 99
A 5 45 60
A 6 20 40
A 7 0 99
A 8 60 65
A 9 0 99
10 rows selected.
SQL>The records of interest are subcodes 5,6,8 since they in certain cases can be considered as one subcode.
It is OK, that MAX_VAL of one record = MIN_VAL of other record. The main thing is that the subcodes
within a group are mutual disclosing on MIN_VAL, MAX_VAL.
SQL> select product, subcode, min_val, max_val
2 from t
3 where product = 'A'
4 and subcode in (5,6,8)
5 order by min_val;
PRODUCT SUBCODE MIN_VAL MAX_VAL
A 6 20 40
A 5 45 60
A 8 60 65
SQL>I have started out by trying to solve it using lag/lead analytical functions, but without luck.
Next, I've come up with this hierarchial query, but I don't quite understand it myself, and that bothers me!
SQL> select distinct -- This, distinct, bothers me!
2 product
3 ,subcode
4 ,min_val
5 ,max_val
6 -- ,connect_by_isleaf
7 -- ,connect_by_root subcode
8 from t
9 where connect_by_isleaf = 1 -- Why does this help me?
10 start with -- This, start with, seems "clumpsy"
11 min_val in ( select min_val
12 from t
13 group by product, subcode, min_val)
14 connect by nocycle -- This, nocycle, really bothers me!
15 min_val > prior min_val
16 and max_val <= prior max_val
17 and product = prior product
18 and subcode <> prior subcode
19 order by product
20 ,subcode
21 ,min_val
22 ,max_val;
PRODUCT SUBCODE MIN_VAL MAX_VAL
A 5 45 60
A 6 20 40
A 8 60 65
SQL>Currently I'm struggling with just identifying the three subcodes. In the perfect world this would be better output
PRODUCT SUBCODE MIN_VAL MAX_VAL GROUP_FLAG
A 0 0 99
A 1 0 99
A 2 18 67
A 3 20 65
A 4 0 99
A 5 45 60 1
A 6 20 40 1
A 7 0 99
A 8 60 65 1
A 9 0 99Or even better, if using herarchial query:
PRODUCT SUBCODE MIN_VAL MAX_VAL ROOT_SUBCODE
A 0 0 99 0
A 1 0 99 1
A 2 18 67 2
A 3 20 65 3
A 4 0 99 4
A 5 45 60 6
A 6 20 40 6
A 7 0 99 7
A 8 60 65 6
A 9 0 99 9Any help and inspiration would be much appreciated. But please don't get offended if I don't follow up the next 12-14 hrs.
Regards
Peter
BANNER
Oracle Database 10g Enterprise Edition Release 10.2.0.4.0 - 64biEdited by: Peter on May 19, 2009 12:52 PM
- Changed line 15 in hierarchial queryBrillant as always, Frank. Ten points and mange tak for you.
(1) What is your concept of "group" in this problem? In what sense do subcodes 5, 6 and 8 form a group? Is it that they share the quality of not overlapping with some other row with the same product? That is, will there be at most two groups per product: rows that overlap with every other row, and rows that don't?By group I mean, that when isolated from other subcodes, a group does not overlap. For product A, I have 10 different subcodes. In certain situations I'm allowed to consider 5,6 and 8 as one, since they don't overlap. In general the data presented results in 8 groups, 7 groups having 1 subcode, 1 group having 3 subcodes.
(2) What you mean by "mutually disclosing"? Is it that the min_val to max_val ranges do not overlap?Yes. They are however allowed to be equal.
As to your query. Seems you're right, it might actually be that "simple". I changed slightly, allowing >= and <=
SQL> SELECT m.*,
2 CASE
3 WHEN
4 EXISTS (
5 SELECT NULL
6 FROM t
7 WHERE product = m.product
8 AND ( min_val >= m.max_val
9 OR max_val <= m.min_val
10 )
11 )
12 THEN 1 END group_flag
13 FROM t m;
PRODUCT SUBCODE MIN_VAL MAX_VAL GROUP_FLAG
A 0 0 99
A 1 0 99
A 2 18 67
A 3 20 65
A 4 0 99
A 5 45 60 1
A 6 20 40 1
A 7 0 99
A 8 60 65 1
A 9 0 99
10 rows selected.
SQL>
This assumes that min_val <= max_val on each row.Your assumption is perfectly correct, as expessed in check constraint on table t.
>
The rest of this message concerns the CONNECT BY query you posted, in case you want to understand it better.
When I run the CONNECT BY query you posted, I get these results:
P SUBCODE MIN_VAL MAX_VAL
A 0 0 99
A 1 0 99
A 4 0 99
A 5 45 60
A 6 20 40
A 7 0 99
A 8 60 65
A 9 0 99
Ouch, my bad. Somehow I posted the wrong query. I have edited line 15 from
15 min_val >= prior min_valinto
15 min_val > prior min_val
The START WITH clause includes all rows: you might as well not have a START WITH clause:
10 start with -- This, start with, seems "clumpsy"
11 min_val in ( select min_val
12 from t
13 group by product, subcode, min_val)
That's great, somehow I was misled to believe that wew should always have a START WITH.
Thank you for your remainding comments on hierarchial query. I'll definitely have to play around with it to fully understand. With the edit mentioned, and
without a START WITH it now looks as,
SQL> select distinct -- This, distinct, bothers me!
2 product
3 ,subcode
4 ,min_val
5 ,max_val
6 from t
7 where connect_by_isleaf = 1 -- Why does this help me?
8 connect by nocycle -- This, nocycle, really bothers me!
9 min_val > prior min_val
10 and max_val <= prior max_val
11 and product = prior product
12 and subcode <> prior subcode -- Not necessary
13 order by product
14 ,subcode
15 ,min_val
16 ,max_val;
PRODUCT SUBCODE MIN_VAL MAX_VAL
A 5 45 60
A 6 20 40
A 8 60 65
3 rows selected.
SQL>One final question. In your solution with EXISTS, you use < and >. In my rewritten example I use <= and >=. It seems that it still works, even without a condition on subcode != m.subcode in the subquery.
Maybe I'm too tired by now, but why is that? - just don't get it
Regards
Peter -
Non-Overlapping channels bleed over
I have multiple AP's in an arena enviornment. Now without violating the rule where an AP is assigned a non-overlapping channel, what is best practice if there are neighboring AP's within close proximity of each other and they are assigned the same channel?
For example, in our press level I have 2 AP's both on channel 11, the suite level below has a few more ap's where the signal bleeds over to the press level and even if some were on channel 1, 6, & 11 there would still be conflicting channels. I hope this questions makes sense.Thanks for the points Rob! Things have been pretty busy the last couple months.
Planning RF fields is a three-dimensional thing, more or less dependent on the construction materials laterally and vertically.
Given that there are only the three channels (for 802.11g), you have no option other than stacking and staggering, like:
level3: 1--6--11--1--6--11--1--6--11
Level2: 6--11--1--6--11--1--6--11--1
Level1: 11--1--6--11--1--6--11--1--6
With the only real variation being the type of antennas you choose, the topology of the spaces, and the power levels of the APs. There's really no other way to do it (using only 802.11g).
NOW, what I'd recommend that you recommend is to add the additional capability for 802.11a. Many/most computer systems have dual band capability, most (Cisco) APs are rigged for dual band.
So light up that 802.11a channel and encourage the folks that can to move up there. It's an easy sell: more bandwidth ("sorta" ...but we're marketing here so outright lies are generally permitted), less interference, and better security (from the aspect of fewer hackers work this band because there's not enough "interesting" stuff on it, etc.).
The benefit to you is that you would have 40 non-overlapping channels ... much easier to plan.
I understand that you have some equipment and users that must remain on 802.11g (try to forbit B if you can, it's making the system slower and reducing your availability), but usually if you offer / encourage those with the capability to move to the better band, they'll take it.
Otherwise, start looking for places where you could say " You know, if I had an antenna with *this* kind of pattern, I could ..." because there probably is an antenna, somewhere, with that radiation pattern.
What you can't get in pattern, you can make up for with barriers, reflectors, etc.
Good Luck
Scott -
Non-overlapping channels - 802.11a/n - need clarification
There seems to conflicting or inconclusive information the topics below. I have just been readying various docs using google searches. I am hoping to get some solid answers.
1) What channels for 802.11a are really non-overlapping? 8 or 12? I have read in some sources that it is 12 but the last 4 are for outdoor use(?).
2) in regards to 802.11a, channels 149 - 165 are considered for outdoor use so indoor availability is only 8 channels, correct?
3) What would be the non-overlapping channels for 802.11n? Indoor use only.
4) What happens when an 802.11n/a device and a traditional 802.11a device connect to the same network?
Thank you,Hi Jacob,
Following diagram should helps you (Ref source )
Referencing this & if you look at your 4 queries you can find the proper answer. Here is my response
1) What channels for 802.11a are really non-overlapping? 8 or 12? I have read in some sources that it is 12 but the last 4 are for outdoor use(?).
As you can see, excluding UNII-2 extended there are 12 non overlapping channels (13 including CH165), depends on the different country regulations you can use those for indoor/outdoor deployments.
2) in regards to 802.11a, channels 149 - 165 are considered for outdoor use so indoor availability is only 8 channels, correct?
It depend on the courntry regulation. I think in US you can use that band(UNII-3) for both indoor/outdoor deployments. So in that case 12 CH availabel excluding UNII-2 extend band.
3) What would be the non-overlapping channels for 802.11n? Indoor use only.
802.11n in this band, you can aggregate two 20MHz channel, (effective 40MHz bandwidth). You can see 6 non-overlapping channels available with 40MHz width in those bands (80MHz & 160MHz option shown in the above diagram applicable to 802.11ac standard)
4) What happens when an 802.11n/a device and a traditional 802.11a device connect to the same network
802.11n device will communicate with 40MHz (if configured on you wireless network & client support channel bonding) & normal 802.11a client will connect using 20MHz channel without any CH bonding or aggregation
If that answers to your query pls mark this thread as "Answered"
HTH
Rasika
**** Pls rate all useful responses **** -
Overlapping v's Non-overlapping channels...
I know it is bad practise to have more than one AP within the same cell area using the same channel, but wondered what the effects of this were. For example, if I had an AP in channel 6 and another AP ten yards from it in channel 6 also, what would be the repercussions?
Thinking about it logically, there would be radio collisions. Each AP could potentially try to transmit at the same time causing collisions. Therefore, you would see slow throughput and high error counts on the AP? Also, if a WLAN client transmitted a frame to a user on the wired network, each AP would receive it (regardless of association) and forward the frame on - resulting in the wired client receiving two copies of the same frame (one from each AP)???
Can anyone, explain why it is not good practise to have over-lapping channels (or any useful web links/documents)?
I will continue to use non-overlapping channels, but it would be nice to know why.
ThanksHi Darren,
That is a great question. Here are some links that have really good answers;
Channel Deployment Issues for 2.4-GHz 802.11 WLANs
From this doc;
http://www.cisco.com/en/US/products/hw/wireless/ps430/prod_technical_reference09186a00802846a2.html#wp134263
Assigning 802.11b Access Point Channels
An important concept to note regarding channel assignments is that the channel actually represents the center frequency that the transceiver within the radio and access point uses (e.g., 2.412 GHz for channel 1 and 2.417 GHz for channel 2). There is only 5 MHz separation between the center frequencies, and an 802.11b signal occupies approximately 30 MHz of the frequency spectrum. The signal falls within about 15 MHz of each side of the center frequency. As a result, an 802.11b signal overlaps with several adjacent channel frequencies. This leaves you with only three channels (channels 1, 6, and 11 for the U.S.) that you can use without causing interference between access points.
From this link;
http://www.wi-fiplanet.com/tutorials/article.php/972261
Channels and international compatibility
Although the statement that channels 1, 6, and 11 are "non-overlapping" is incomplete, the 1, 6, 11 guideline has merit. If transmitters are closer together than channels 1, 6, and 11 (e.g. 1, 4, 7, and 10), overlap between the channels will probably cause unacceptable degradation of signal quality and throughput.
http://en.wikipedia.org/wiki/IEEE_802.11
Hope this helps!
Rob
Please remember to rate helpful posts......... -
If customer sites are in a single VPN and they dont need inter-VPN communication.. is it necessary to configure RT ?
Hello Mukarram,
as Shivlu has explained route targets are used to decide what vpnv4 prefixes to import in a specific VRF.
without route targets you just get a collection of isolated VRF sites not a working MPLS VPN.
Advanced usage of RTs allows for extranets or hub and spoke topologies.
Being a BGP extended community like standard community multiple RTs can be associated to a single vpnv4 route (for example for implementing extranets) but at least one has to be present.
any-to-any connectivity between VRF sites require only a single RT.
It's also important to understand differences between RT and RD:
RD is prepended to the IPv4 prefix to build the vpnv4 prefix sent in address-family vpnv4.
RD is there to allow for address overlapping in different or same PE node: with distinct RDs we can distinguish between two same IP subnets in two different VRFs.
RTs are associated as extended BGP communities to the vpnv4 routes.
RTs are used to build the IP routing tables of each VRFs and provide IP connectivity between VRF sites.
Often simple examples use the same value for RD and RT because both are 64 bits and they can use the same notation but they are different.
Also RDs can be different in each VRF site without causing problems.
Actually in some complex topologies with multi PE connections the usage of a different RD is mandatory to achieve the desired effects in terms of fault tolerance and load balancing.
Hope to help
Giuseppe -
SCVMM Unable to assign the subnet because it overlaps an existing one
Hello,
We are using SCVMM to manage an Hyper-V cluster (2012 R2).
When we tried to create a new VLAN we got the following error:
Error (13638)
Unable to assign the subnet 172.22.172.128/25 because it overlaps with an existing subnet.
Recommended Action
Specify a non-overlapping subnet, and then try the operation again.
After revising the rest of subnets we haven't found any overlaps.
We have another VLAN defined with 172.22.172.0/25
I've checked with a CIDR calculator and the IP Range is not overlapped.
How can we define both VLANs with their corresponding Subnets?
RegardsHi Pep,
Have you tried a different range just to check there is not something overlapping that's been missed?
I have a single C class range I then subnet up further in our SCVMM deployment depending type of VM with no issues. With associated IP Pools sat on top. In fact I have just added your two ranges with two different VLANs with no issues.
The 255.255.255.128 or /25 subnet breaks your IP range up into 0-127 and 128 to 255 as expected. There is no difference to this and further sub netting a B or C class range for a normal enterprise deployment. Granted its not a default class and
will give you two different subnets but that's correct behaviour and shouldn't be an issue in SCVMM assigning each as a separate VLAN.
Kind Regards
Michael Coutanche
Blog:
Twitter: LinkedIn:
Note: Posts are provided “AS IS” without warranty of any kind, either expressed or implied, including but not limited to the implied warranties of merchantability and/or fitness for a particular purpose. -
Airport Express as access point non-Apple router's network?
I have a first generation Airport Express and I'd like to use it as an access point on a friend's network with another brand's router. Here's what I have in mind, hopefully others here can confirm/deny that this will work:
In the Internet tab of the Airport Express I will choose manual IP addressing and pick an address on the same subnet as the main router that's outside the range the router distributes.
In the Wireless tab I will choose Create a Wireless Network, name it identically to the existing network with identical security, and choose a non-overlapping channel.
The Airport Express will be connected by ethernet cable to a Lan port of the main router.
This should work, right? Thanks.I forgot to say, but it will not detect the Slingbox on my network when I'm connected through the Airport Express. It detects if fine if I plug in my laptop to the router.
-
Adding/merging non intersectng elements.
I've got a question that I have for a few months.
I know this was possible in CS3 but since CS4 I think it has been deleted.
I'm Struggling with the follwing:
- Let's say I have 5 elements that are NOT overlapping. I know that it was in possible in CS3 to "add" these together so that you'd have 1 element that was completely editable. You could for instance put an object on top of the elements > use the pathfinder to delete a part, and all previous 5 elements would be edited (because now it was 1 element).
- If I try this by doing this the same way as I used to in CS5, this doesn't work anymore. No matter how hard I try to join the non overlapping elements it doesn't work in CS5. If I do the exact same thing as described above I'll be left with the same 5 elements and the deletion will only happen in 1 element.
Super frustrating...
Was this something that was altered in CS4?
And if so, how do I do it now?
Thanks in advance!I believe you are desiring to have a live pathfinder effect. This chnaged to be the opposite of CS3 in CS4. In CS4 hold down option before making a clicking on pathfinder button.
If you hover the cursor over a pathifnder button, a yellow post it help note will explain this. -
Trouble with creating interlocking rings (olympic style)
I'm trying to make interlocking rings (olympic style), but every single time I go to the pathfinder and click "divide", so that I filll in the overlapping areas of the rings with different colors, it ttells me "the filter produced no results, please select one or more filled paths" even though I already selected the three overlapping rings. Im in adobe CS5.
Please help!Select the art and then grab the Shape Builder tool...
Then clikc-drag across sections you want to unify...
The image below is an animated gif. It will not animate on the forum by default. Right-click/Control-Click the image and choose "Open Link in New Tab/Window" to view the animation. -
I need help.
lynus,
The error is the message.
I guess it is one of the Pathfinder options, such as Intersect shape areas.
If you have one path or two non overlapping paths, make suree you change it.
If you have two non locked overlapping paths, something else is wrong. You may try things on the following list.
The following is a general list of things you may try when the issue is not in a specific file (you may have tried/done some of them already); 1) and 2) are the easy ones for temporary strangenesses, and 3) and 4) are specifically aimed at possibly corrupt preferences); 5) is a list in itself, and 6) is the last resort.
1) Close down Illy and open again;
2) Restart the computer (you may do that up to 3 times);
3) Close down Illy and press Ctrl+Alt+Shift/Cmd+Option+Shift during startup (easy but irreversible);
4) Move the folder (follow the link with that name) with Illy closed (more tedious but also more thorough and reversible);
5) Look through and try out the relevant among the Other options (follow the link with that name, Item 7) is a list of usual suspects among other applications that may disturb and confuse Illy, Item 15) applies to CC, CS6, and maybe CS5);
Even more seriously, you may:
6) Uninstall, run the Cleaner Tool (if you have CS3/CS4/CS5/CS6/CC), and reinstall.
http://www.adobe.com/support/contact/cscleanertool.html -
Time overlapping of data record 3 with 4
Hi all
While Running the DTP for loading master data from PSA to INFO OBJECT i am getting this error "Time overlapping of data record 3 with 4 ".
can anyone help me for resolving this
thanks
Preethi.....The datefrom/validfrom and dateto/validto of those two records have some overlapping period. Browse the PSA table to see the date from and dateto values.
Analyze why the extractor is bringing records with overlapping periods. If its a custom datasource make sure the view/functionmodule is generating proper non-overlapping records using RSA3 -
hi all,
we are updating the number range for the fiscal year in FBN1 where we are getting the error message as ENTER INTERVALS WITHOUT OVERLAP
no year from to
20 2008 2008000000 2008999999
20 2009 2009000000 2009999999
year 2011 and 2012 2013 not used
20 2014 2014000000 2014999999 here its showing the error ENTER INTERVALS WITHOUT OVERLAP
kindly help how to solve this problem
thanks and regards,
chethanHi
In table NRIV, give object name as RF_BELEG, Subobject value as company code. Sort entries in ascending order of "From number" and check the overlapping entry.
In FBN1 when you add interval use F4 for "from" field there you will be able to see non-overlapping entries and available options. If this doesn't help you to identity overlapping number range, then select any available number range, save. Then go to added number range try changing it there, when you try to save it there, it will display all overlapping entries.
Regards
Aniket -
Overlapping Dates , Denormaization of the tables
Hi Guys,
I am in a situation that we are denormalising few tables for better performance and reduce the joins. For which I am joining the tables using the logic to check for the dates and add the fields corresponding to those dates to represent the joined data correct at any given point of time . A simple example is :
In the below tables : in Table1, the case_number 0023A is having Status "AC" during the time period 15-jan-08 and 31-dec-9999 ( open-ended or current date), and in Table2 the same case is having center_code "C12" during 22-jun-08 and 31-dec-9999 . When the tables are joined the over-lapped dates will be having common data like *0023A, 22-JUN-2008, 31-DEC-9999, AC, C12*
My [previous posting|http://forums.oracle.com/forums/thread.jspa?forumID=75&threadID=682324] on the forums which has been answered by Frank and was perfectly working for me in getting the results for different data as in the above mentioned thread, but the present data set is throwing me the problem by not selecting the Non-Overlapping Dates
CREATE TABLE TABLE1 (
CASE_NUMBER VARCHAR2(5),
CHANGE_EFF_DATE DATE,
END_EFF_DATE DATE,
STATUS VARCHAR2(2) );
INSERT INTO TABLE1 VALUES
( '0023A'
,TO_DATE('15-JAN-2008','DD-MON-YYYY')
,TO_DATE('31-DEC-9999','DD-MON-YYYY')
,'AC'
INSERT INTO TABLE1 VALUES
( '0023A'
,TO_DATE('07-OCT-2007','DD-MON-YYYY')
,TO_DATE('14-JAN-2008','DD-MON-YYYY')
,'CL'
INSERT INTO TABLE1 VALUES
( '0023A'
,TO_DATE('08-APR-2007','DD-MON-YYYY')
,TO_DATE('06-OCT-2007','DD-MON-YYYY')
,'AC'
INSERT INTO TABLE1 VALUES
( '0023A'
,TO_DATE('13-MAR-2007','DD-MON-YYYY')
,TO_DATE('07-APR-2007','DD-MON-YYYY')
,'RJ'
INSERT INTO TABLE1 VALUES
( '0023A'
,TO_DATE('31-MAY-2005','DD-MON-YYYY')
,TO_DATE('12-MAR-2007','DD-MON-YYYY')
,'AP'
CREATE TABLE TABLE2 (
CASE_NUMBER VARCHAR2(5),
CHANGE_EFF_DATE DATE,
END_EFF_DATE DATE,
CENTER_CODE VARCHAR2(3) );
INSERT INTO TABLE2 VALUES
( '0023A'
,TO_DATE('22-JUN-2007','DD-MON-YYYY')
,TO_DATE('31-DEC-9999','DD-MON-YYYY')
,'C12'
INSERT INTO TABLE2 VALUES
( '0023A'
,TO_DATE('09-MAR-2007','DD-MON-YYYY')
,TO_DATE('21-JUN-2007','DD-MON-YYYY')
,'101'
SQL> SELECT * FROM TABLE1;
CASE_ CHANGE_EF END_EFF_D ST
0023A 15-JAN-08 31-DEC-99 AC
0023A 07-OCT-07 14-JAN-08 CL
0023A 08-APR-07 06-OCT-07 AC
0023A 13-MAR-07 07-APR-07 RJ
0023A 31-MAY-05 12-MAR-07 AP
SQL> SELECT * FROM TABLE2;
CASE_ CHANGE_EF END_EFF_D CEN
0023A 22-JUN-07 31-DEC-99 C12
0023A 09-MAR-07 21-JUN-07 101 -----
Here is the query I am running to get the joined information from both the tables with regards to specific point in time.
SELECT T1.CASE_NUMBER
,GREATEST(T1.CHANGE_EFF_DATE,T2.CHANGE_EFF_DATE) CHANGE_EFF_DATE
,LEAST(T1.END_EFF_DATE,T2.END_EFF_DATE) END_EFF_DATE
,T1.STATUS
,T2.CENTER_CODE
FROM
TABLE1 T1
LEFT OUTER JOIN
TABLE2 T2
ON
T1.CASE_NUMBER=T2.CASE_NUMBER AND
T1.CHANGE_EFF_DATE <= T2.END_EFF_DATE AND
T2.CHANGE_EFF_DATE <= T1.END_EFF_DATE
ORDER BY 2;
Here is the result-set I am getting :
CASE_ CHANGE_EF END_EFF_D ST CEN
0023A 09-MAR-07 12-MAR-07 AP 101
0023A 13-MAR-07 07-APR-07 RJ 101
0023A 08-APR-07 21-JUN-07 AC 101
0023A 22-JUN-07 06-OCT-07 AC C12
0023A 07-OCT-07 14-JAN-08 CL C12
0023A 15-JAN-08 31-DEC-99 AC C12But my result-set should include the Non-Overlapping dates as well which should look like this, but I am missing the top record ( First record) in my output :
CASE_ CHANGE_EF END_EFF_D ST CEN
0023A 31-MAY-07 08-MAR-07 AP
0023A 09-MAR-07 12-MAR-07 AP 101
0023A 13-MAR-07 07-APR-07 RJ 101
0023A 08-APR-07 21-JUN-07 AC 101
0023A 22-JUN-07 06-OCT-07 AC C12
0023A 07-OCT-07 14-JAN-08 CL C12
0023A 15-JAN-08 31-DEC-99 AC C12I shall be really thankful if you guys can me help me .
Thanks.
Venu
Edited by: Oracle, Developer on Dec 11, 2008 9:30 PMHi, Venu,
If I understand the problem correctly, you need a row in table2 that covers all dates before the earliest change_eff_date, similar to the way you have a row that covers all dates after the latest change_eff_date. You don't have to store such a row in the table: you can generate one at run time, and use UNION to add it to your real data.
WITH t2 AS
SELECT case_number
, change_eff_date
, end_eff_date
, center_code
FROM table2
UNION
SELECT case_number
, TO_DATE (1, 'J') AS change_eff_date -- Earliest possible date
, MIN (change_eff_date) - 1 AS end_eff_date
, NULL AS center_code
FROM table2
GROUP BY case_number
SELECT T1.CASE_NUMBER
, GREATEST (T1.CHANGE_EFF_DATE, T2.CHANGE_EFF_DATE) CHANGE_EFF_DATE
, LEAST (T1.END_EFF_DATE ,T2.END_EFF_DATE) END_EFF_DATE
, T1.STATUS
, T2.CENTER_CODE
FROM TABLE1 T1
LEFT OUTER JOIN T2
ON T1.CASE_NUMBER = T2.CASE_NUMBER
AND T1.CHANGE_EFF_DATE <= T2.END_EFF_DATE
AND T2.CHANGE_EFF_DATE <= T1.END_EFF_DATE
ORDER BY 1, 2;Note that the main query is exactly what you had before, except for the definition of t2.
Where t2 had been simply table2, now it is the UNION of table2 with one row per case_number, with a change_eff_date in 4712 BCE.
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