Bookmark to a section with a derived value

WEBI Report:
I have a section that is based on a derived value (=MonthNumberOfYear([Closed Date])) called [Month of Period].
I want to Bookmark that section, but when I view the Map screen, "#FORMAT" is displayed for each record. Clicking on any of them does access the appropriate record, but there is no way of telling what that is by the "#FORMAT" label.
Any suggestions?
Edited by: jandon on Aug 6, 2010 10:12 PM

i dont see any problem with that
you just create a new variable Month of Period
=MonthNumberOfYear([Closed Date])
and section based on this variable
Regards
Amr

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SSRS Report with mutli select values in parameters

Hi All,
I am creating a SSRS report and displaying the result in tablix. I want to make the report to filter by multi select values from the parameters.
I have 3 parameters. For all the three parameters, I have set "Default Values" and "Available Values". Also, modified my result dataset to get the values from parameters using "WHERE ID = @Parameter1" (example) and so
on..
3 Parameters types: 1. Yes or No
2. Date 3. Values with NULL in the DB
Problems:
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Any suggestions for the above three problems for SSRS with multi select values as filter will be helpful.
Maruthu | My Blog

Hi Maruthi,
Regarding the three things:
1.) I believe your first parameter is mutliselect , please convert it to single select.
Steps:
a.) Select the parameter, right click and go to parameter properties
b.) Under General Tab in DataType section remove the checkbox for Allow Multiple Value.
2.) Instead of returning as the date please return string. For example :
SELECT CONVERT(VARCHAR, GETDATE(),101)
Here is the list of conversion : http://www.sqlusa.com/bestpractices/datetimeconversion/
3.) As such there is no concept of null in SSRS. (Its a relational concept) . We do have nothing , which means empty. In Order to use null please select the Allow null option in the DataType section of the Parameter, This will add an checkbox for null in
the report preview. Its developer job to handle the null values in there query.
Regards Harsh

I had my screen sectioned with an application running permanently in it and have lost it, please can someone tell me how to get it to do that again? Thank you.

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Compare the current value with the previous value in the same column

Hi all,
I have to include a statement in a query which allows to compare the current value of column A with the previous value of column A (same column). from there, I need to add a condition in order to have the expected result.
Let's take an example to illustrate what I want to achieve:
I have the following columns in table called 'Charges':
Ship_id batch_nr Order_nr Price
SID1111 9997 MD5551 50
SID1111 9998 MD5552 50
SID1111 9999 MD5553 50
SID2222 8887 MD6661 80
SID2222 8887 MD6662 80
SID2222 8887 MD6662 80
SID3333 6666 MD7771 90
I want to check if the ship_id of row 2,3 (and more if available) is equal to the ship_id of row 1.
If it is the case, then value 'together with the first batch_nr' in row 2 and 3 under Price column. If not, then keep the original value of Price column
PLease see below the expected result:
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SID1111 9997 MD5551 50
SID1111 9998 MD5552 together with 9997
SID1111 9999 MD5553 together with 9997
SID2222 8887 MD6661 80
SID2222 8887 MD6662 together with 8887
SID2222 8887 MD6663 together with 8887
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Imco20030

Hi,
user11961002 wrote:
Hi,
Here is the query that I use:
[ select
sl.ship_id,
o.ordnum,
o.reffld_5 "BatchNR",
sum(tc1.chrg_amt) "FreightPRC",
sum(tc2.chrg_amt) "FuelPRC",
sum (tc1.chrg_amt + tc2.chrg_amt + tc3.chrg_amt) "Total Price"
from ord_line ol
join ord o on (ol.ordnum = o.ordnum and ol.client_id = o.client_id)
join shipment_line sl on (ol.ordnum = sl.ordnum and ol.client_id = sl.client_id and ol.ordlin = sl.ordlin)
join adrmst a2 on (o.rt_adr_id = a2.adr_id)
left join tm_chrg tc1 on (tc1.chargetype = 'FREIGHT' and tc1.chrg_role = 'PRICE' and tc1.ship_id = sl.ship_id)
left join tm_chrg tc2 on (tc2.chargetype = 'FUELSURCHARGE'and tc2.chrg_role = 'PRICE' and tc2.ship_id = sl.ship_id)
where sl.ship_id = 'SID0132408'
group by o.client_id, o.ordnum, o.reffld_2, sl.ship_id, a2.adrnam, a2.adrln1, a2.adrpsz, a2.adrcty, a2.ctry_name,
o.reffld_5, ol.early_shpdte
order by ship_id
]That looks like the query you were using before you started this thread.
Modify it, using the analytic fucntions FIRST_VALUE and LAG, like I showed you.
I see that you did simplify the problem quite a bit, and it's good that you did that.
It doesn't matter that your real problem involves joins or GROUP BY. Analytic functions are calculated on the results after all joins and GROUPS BYs are done. Just substitute your real expressions for the simplified ones.
For example, in your simplified problem, there was a column called order_nr, but I see now that's it's really called o.ordnum. Where the solution I posted earlier says "ORDER BY order_nr", you should say "ORDER BY o.ordnum".
Here's a less obvious example: in your simplifed problem, there was a column called price, but I see now that it's really SUM (tc1.chrg_amt + tc2.chrg_amt + tc3.chrg_amt). Where the solution I posted earlier says "TO_CHAR (price)", you should say "TO_CHAR (SUM (tc1.chrg_amt + tc2.chrg_amt + tc3.chrg_amt))". (You can't use an alias, like "Total Price", in the same SELECT clasue where it is defined.)
I removed some columns from the select as they are not relevant for the wanted action like 'adress details or other references'.
Now here is the result:
Shipment ID     Order Number     WMS Batch     Freight      Fuel Price Order Total Price
SID0132408     MDK-000014-05602649     04641401     110     10 120
SID0132408     MDK-000014-05602651     04641402     110     10 120
SID0132408     MDK-000014-05602652     04641363     110     10 120
as you can see, the 3 orders have the same shipment ID.
The expected result should be shown under column 'Total Price' as follows:
Shipment ID     Order Number     WMS Batch     Freight      Fuel Price Order Total Price
SID0132408     MDK-000014-05602649     04641401     110     10 120
SID0132408     MDK-000014-05602651     04641402     110     10 tog with 04641401
SID0132408     MDK-000014-05602652     04641363     110     10 tog with 04641401Okay, so those are the correct results that I asked for, plus the incorrect results you're getting now. Thanks; extra information doesn't hurt.
But where is the raw data that you're starting with?
It looks like you tried to format the code (but not the results) by typing this 1 character:
before the formatted section and this different character
after the formatted section. To post formatted text on this site, type these 6 characters
before the formatted section, and the exact same 6 characters again after the formatted section.

With Out Derivation in Char relationship

Hi,
I am trying to understand the concept of char relationship with derivation and without derivation. I would request you to please tell me the basic difference between with and with out derivation?
we can say that, with derivation is used to derive certain values based on a relationship. So what exactly with out derivation does in input ready query and automatic planning scenarios?
Please give me an example to use with out derivation. I have read the help document, but it is not clear.
Regards
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So according to your explanation: the conclusion is : both with derivation and without derivation checks valid combinations.
with derivation - apart from checking combination - it also derives the value. In this case,
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With out derivation - it can't derive any char. it only checks the valid combination. both chars need to be in the aggregation level.
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Thanks

LGGeiger wrote:
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Bookmark to a section with a derived value

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