Elapsed time to run subvi
hi....
first of all ..sorry if the subject title didn't match with my problem. I have no idea what is the suitable title should i put there.
anyway my question is..how should i do to set every one hour i can run a subvi?for example i have a subvi that use to send a msg to the gsm card to delete all the read msg inside the simcard. so means that I want to delete all the messages every one hour. if i have delete the msg i will stop the subvi and wait for next hour to delete the message.
anyone can help?..pls..hope that's all clear..
thanks..
Or you could use the Elapsed Time function wired to a case statement with you subVI inside that.
Similar Messages
-
Check Scan for NI435X and elapsed time doesn't run in loop
Dear all Labview Experts
i have 2 questions
(1) i written program with Check Scan for NI435X (input to no of scan to NI 435X read ) and one with only NI435X read( to read in voltage) with constant input to no of scan,
the data i obtained is attached in 2 doucments; With Check Scan and without check scan ( the left column is temperature read in ( a formula is used to convert volt to temperature, right column is time)
strangely the program with check scan shows values in negative ==> Voltage read in is 0, strange, why is this so??? and why is there a diff?? Hope u understand my question
(2) For elapsed time, i want to find the total elapsed time for the whole program as well as individual cases, there is no "running" elapsed time or individual running time,
what happen is that when the each case end, the time will jump (e.g. from 0 at start, it will change to 5.35 at end of case) , is there any where to make the time run, i need it because some case required
c
ondition such as case will move to next after 5 seconds, but if above case happen, the vi will only stop if i put >= 5 rather than just place = 5
hope u understand,
does it help if i placed the condition as another loop , will it mean the same things
Attach are vi and some doucments, however as my vi contains PID tool kit as will as ni435x ni DC power which cannot be opened if you do
not have these kits.......however mayb u can help to look at the portion of timing as this is the main problem i faced now.........
Hope it is not too confusing for you to understand me...........
i can only attach max 3, the next post i will attach 2 vis...........
Attachments:
Picture of check scan and without.rtf 15 KB
With Check Scan.txt 4 KB
Without Check Scan.txt 1 KBThks Brian
Attached are 3 VIs
1.the first vi is the one with the check scan
2. the next two vi the only diff is the placement of the where i put the elapsed time and time of each case.........(without check scan)--the first
one is within the big loop, the other outside big loop -- the purpose i did this is because when i run the vi with the timing in the big loop, the timing does not run continuously but "skip". for e.g. Denature time: from 0 . it jump to 5.15s without showing the running.............
i will attached another 2 vi s in my next attachment
Hope this clarifies a bit
your help is really appreciated
thks
Attachments:
PCR Control_State Machine_4 channel_Timing within Loop.vi 438 KB
PCR Control_State Machine_4 channel_with_check scan.vi 443 KB
PCR Control_State Machine_4 channel_Timing outside of loop.vi 442 KB -
I've hours per run wired to elapsed time in my VI. The output elapsed time is display out to a cluster display.
How do I get the VI to stop once it has react hours per run elasped time?
And how would I wire to show that time on (hours) once the VI has stopped?
Please let me know what I'm doing wrong.
thanks
Attachments:
on-off.JPG 35 KBWell, I 'll give you one trick that I use alot when I ever wanted to know:
When will a certain event occur in the future ?
In your case: the event is "Time elapsed for all the events or runs".
It may help you, and may not help you, Good luckMessage Edited by Ayman_Metwally on 02-28-2005 10:36 PM
Ayman Mohammad Metwally
Automation Engineer
Egypt - Cairo
Attachments:
current.GIF 18 KB -
How to stop the "Elapsed Time" counting without stop the program running?
我使用Labview 8.2,配合USB DAQ來做測試治具。目的:
當檢測到某個電壓(Analog Input)到達某個電壓值(5V)之後,開始啟動計時。若是在15秒之內,另一個電壓值降為0之後為正常,此時停止計時(但是數值不能歸零);若是超過15秒還沒有降到0伏,則表示錯誤,此時亮起紅色 indicator警示,但是時間仍然繼續計時,直到電壓降到0為止才停(但是數值不能歸零)。
我使用 Elapsed Time 模組來使用,可是問題是如何讓該模組停止計時同時不能將時間歸零?因為此模組只有 "Reset" 控制功能,一但將條件結果輸入到 Reset input 之後,電壓達到零之後就會將此 Elapsed Time 歸零了。有沒有解決辦法?Elapsed Time 確實會將時間reset,如果你想要保留時間的話其實有很多做法
例如使用 tick count,他會顯示出相對時間數值,先放一個tick count在迴圈開始前
記得用sequence以確保他會在回區尚未開始運作前先記錄時間
然後再放一個tick count在迴圈內,再用case或select讓電壓等於零時記錄下tick count的值
再將兩者相減即可
Chris -
Elapsed Time in my V$SQL
Hi,
Can someone answer this please?. I'm bit perplexed by the elapsed time in my v$sql
UPDATE ods_router_tables SET last_data_received = :b1, last_data_received_dst = :b3 WHERE
table_name = :b2 AND NVL(last_data_received,TO_DATE('01-JAN-1980','DD-MON-YYYY')) < :b1
Executions: 2786327
Disk reads: 0
Buffer gets: 6023381
Rows processed: 423713
Optimizer cost: 1
Sorts: 0
CPU Time: 898500000
Elapsed Time: 1.84467440681166E19
This table is quite small, only 44 records and if i run the statement in Sql plus, it runs in a
flash. But why this enormous Elapsed time in V$SQL. Is my conversion of this time to
seconds/execution correct?
Elapsed time in seconds/execution: 1.84467440681166E19/1000000/2786327=6620451 seconds!
I guess there is something seriously wrong here. I'm running on a 4 CPU server. DB version is 9.2.
I see similar figures in some of the innocent looking Insert statements as well.
thanks in advance,
SunilELAPSED_TIME in v$sql also includes all wait times for that sql. Is it possible that update is waiting for locks some time? See the following case.
SQL> create table test2 as select * from dual;
Table created.
SQL> update test2 set dummy='A';
1 row updated.From another session try to update the same row.
SQL> update test2 set dummy='B';Wait for some time. Then rollback the first session.
SQL> rollback;
Rollback complete.Now look at v$sql.
SQL> select sql_text,executions,elapsed_time from v$sql where sql_text like 'update%test2%';
SQL_TEXT
EXECUTIONS ELAPSED_TIME
update test2 set dummy='B'
1 29413712
update test2 set dummy='A'
1 2506The first update's elapsed time is very small. The second update's elapsed time is high because of the wait for the lock. -
ORA-1555 ORA-3136 errors:: elapsed time vs Query Duration
Dear all,
- My Database version is 11.2.0.2, Solaris.
- We have been having a problem in the production database where the front end nodes start going up and down for couple of hours sometimes. ; When node flapping is going on we get connection timed out alerts.
WARNING: inbound connection timed out (ORA-3136) opiodr aborting
process unknown ospid (4342) as a result of ORA-609 opiodr aborting
process unknown ospid (4532) as a result of ORA-609 opiodr aborting
process unknown ospid (4534) as a result of ORA-609 opiodr aborting....
Since this week node flapping is happening every day. Since past 2 days after or during node flapping we are getting ORA-1555 error.
Extract from alert log error:
ORA-01555 caused by SQL statement below (SQL ID: g8804k5pkmtyt, Query Duration=19443 sec, SCN: 0x0001.07bd90ed):
SELECT d.devId, d.vendor, d.model, d.productClass, d.oui, d.parentDeviceId, d.created, d.lastModified AS devLastMod, d.customerId, d.userKey1, d.userKey2, d.userKey4, d
.userKey5, d.firmwareFamily, d.softwareVer, d.serialNum, d.ip, d.mac, d.userKey3, d.userKey6, d.provisioningId, d.status, d.classification, d.population, d.name, d.ipRe
solver, d.ipExpirationTime, d.geoLocationId,contact.firstContactTime, ifaces.id, ifaces.type AS ifaceType, ifaces.lastModified AS ifaceLastMod, ifaces.timeoutname, ifac
es.username1, ifaces.password1, ifaces.username2, ifaces.password2, ifaces.connReqUrl, ifaces.connReqScheme, ifaces.srvNonce, ifaces.deviceNonce, ifaces.phoneNumber,ifa
ces.bootstrapSecMethod, ifaces.srvAuthentication, ifaces.deviceAuthentication, ifaces.userPIN, ifaces.networkID, ifaces.omaSessionID, ifaces.portNum, ifaces.mgtIp, ifac
es.cmtsIp, ifaces.mgtReadCommunity, ifaces.mgtWriteCommunity, ifaces.cmtsReadCommunity, ifaces.cmtsWriteCommunity, devto.name AS devtoName, devto.rebootTimeout, devto.sessionInitiationI run Statspack report from the whole day duration, and looking into the elapsed time in seconds no more than 3739.61 sec (too lower than run duration in the alert log file of 19443 sec); So I would like to know if there is any co-relations between the ORA-3136 errors and the ORA-1555 errors?
CPU CPU per Elapsd Old
Time (s) Executions Exec (s) %Total Time (s) Buffer Gets Hash Value
tTime <= :3 ) AND (endTime IS NULL OR endTime >= :4 )
2773.77 7,787,914 0.00 3.4 3739.61 112,671,645 1909376826
Module: JDBC Thin Client
SELECT d.devId, d.vendor, d.model, d.productClass, d.oui, d.pare
ntDeviceId, d.created, d.lastModified AS devLastMod, d.customerI
d, d.userKey1, d.userKey2, d.userKey4, d.userKey5, d.firmwareFam
ily, d.softwareVer, d.serialNum, d.ip, d.mac, d.userKey3, d.user
SQL> show parameter UNDO_MANAGEMENT
NAME TYPE VALUE
undo_management string AUTO
SQL> show parameter UNDO_RETENTION
NAME TYPE VALUE
undo_retention integer 10800BR,
DiegoThank you. Please let me know if it is enough or you need more information;
SQL ordered by Gets DB/Inst: DB01/db01 Snaps: 14835-14846
-> End Buffer Gets Threshold: 100000 Total Buffer Gets: 677,689,568
-> Captured SQL accounts for 73.6% of Total Buffer Gets
-> SQL reported below exceeded 1.0% of Total Buffer Gets
CPU Elapsd Old
Buffer Gets Executions Gets per Exec %Total Time (s) Time (s) Hash Value
21,286,248 2,632,793 8.1 3.4 666.73 666.76 3610154549
Module: JDBC Thin Client
SELECT d.devId, d.vendor, d.model, d.productClass, d.oui, d.pare
ntDeviceId, d.created, d.lastModified AS devLastMod, d.customerI
d, d.userKey1, d.userKey2, d.userKey4, d.userKey5, d.firmwareFam
ily, d.softwareVer, d.serialNum, d.ip, d.mac, d.userKey3, d.user
17,029,561 1,176,849 14.5 2.7 417.32 416.73 1909376826
Module: JDBC Thin Client
SELECT d.devId, d.vendor, d.model, d.productClass, d.oui, d.pare
ntDeviceId, d.created, d.lastModified AS devLastMod, d.customerI
d, d.userKey1, d.userKey2, d.userKey4, d.userKey5, d.firmwareFam
ily, d.softwareVer, d.serialNum, d.ip, d.mac, d.userKey3, d.user
17,006,795 37 459,643.1 2.7 367.61 368.95 4045552861
Module: JDBC Thin Client
SELECT d.devId, d.vendor, d.model, d.productClass, d.oui, d.pare
ntDeviceId, d.created, d.lastModified AS devLastMod, d.customerI
d, d.userKey1, d.userKey2, d.userKey4, d.userKey5, d.firmwareFam
ily, d.softwareVer, d.serialNum, d.ip, d.mac, d.userKey3, d.userAnother Statspack report for the whole day shows;
SQL ordered by CPU DB/Inst: DB01/db01 Snaps: 14822-14847
-> Total DB CPU (s): 82,134
-> Captured SQL accounts for 40.9% of Total DB CPU
-> SQL reported below exceeded 1.0% of Total DB CPU
CPU CPU per Elapsd Old
Time (s) Executions Exec (s) %Total Time (s) Buffer Gets Hash Value
tTime <= :3 ) AND (endTime IS NULL OR endTime >= :4 )
2773.77 7,787,914 0.00 3.4 3739.61 112,671,645 1909376826
Module: JDBC Thin Client
SELECT d.devId, d.vendor, d.model, d.productClass, d.oui, d.pare
ntDeviceId, d.created, d.lastModified AS devLastMod, d.customerI
d, d.userKey1, d.userKey2, d.userKey4, d.userKey5, d.firmwareFam
ily, d.softwareVer, d.serialNum, d.ip, d.mac, d.userKey3, d.user
SQL ordered by Gets DB/Inst: DB01/db01 Snaps: 14822-14847
-> End Buffer Gets Threshold: 100000 Total Buffer Gets: 1,416,456,340
-> Captured SQL accounts for 55.8% of Total Buffer Gets
-> SQL reported below exceeded 1.0% of Total Buffer Gets
CPU Elapsd Old
Buffer Gets Executions Gets per Exec %Total Time (s) Time (s) Hash Value
86,354,963 7,834,326 11.0 6.3 2557.34 2604.08 906944860
Module: JDBC Thin Client
SELECT d.devId, d.vendor, d.model, d.productClass, d.oui, d.pare
ntDeviceId, d.created, d.lastModified AS devLastMod, d.customerI
d, d.userKey1, d.userKey2, d.userKey4, d.userKey5, d.firmwareFam
ily, d.softwareVer, d.serialNum, d.ip, d.mac, d.userKey3, d.user
.....BR,
Diego
Edited by: 899660 on 27-ene-2012 7:43
Edited by: 899660 on 27-ene-2012 7:45 -
How can I get the elapse time for execution of a Query for a session
Hi ,
How can I get the elapse time for execution of a Query for a session?
Example - I have a report based on the procedure ,when the user execute that it takes say 3 min. to return rows.
Is there any possible way to capture this session info. for this particular execution of query along with it's execution elapse time?
Thanks in advance.Hi
You can use the dbms_utility.get_time tool (gives binary_integer type value).
1/ Initialize you time and date of beginning :
v_beginTime := dbms_utility.get_time ;
2/ Run you procedure...
3/ Get end-time with :
v_endTime := dbms_utility.get_time ;
4/ Thus, calculate elapsed time by difference :
v_elapsTime := v_endTime - v_beginTime ;
This will give you time elapsed in of 100th of seconds...
Then you can format you result to give correct print time.
Hope it will help you.
AL -
How to use elapsed time function with state machine in Lab VIEW
Hello
I've been trying to use state machine with elapsed time function in order to sequentially start and stop my code. The arrangement is to start the code for 1 minute then stop for 5 minutes. I've attached the code, the problem is when I place the elapsed time function out of the while loop it doesn't work, on the other hand when I place it inside the loop it does work but it doesn't give the true signal to move to the next state.
Could you please have a look to my code and help me to solve this issue.
Regards
Rajab
Solved!
Go to Solution.
Attachments:
daq assistance thermocouple(sate machine raj).vi 436 KBRajab84 wrote:
Thanks apok for your help
even with pressing start it keeps running on wait case
could you please explain the code for me, the use of Boolean crossing, increment , and equal functions
Best Regards
Rajab
OK..I modded the example to stop after 2 cycles. Also recommend taking the free online LabVIEW tutorials.
run vi. case statement goes to "initialize", shift registers are initialized to their constants. goto "wait"
"start"= false, stay in current state. If true, transition to "1 min" case
reset elapsed timer with True from shift register(counter starts at zero)."time has elapsed"=false, stay in current state(1 min). If true, goto "5min" case
reset elapsed timer with True from shift register of previous case(counter starts at zero)."time has elapsed"=false, stay in current state(5 min). If true, goto "1min" case. Also, bool crossing is looking for "true-false" from "5 min" compare function to add cycle count.
Once cycle count reaches 2, stop while loop....
Attachments:
Untitled%202[1].vi 42 KB -
Direct Path Read waits are not showing in Elapsed time
Hi,
I'm having a question regarding interpretation of a SQL trace file. I'm on Oracle 11.2.0.1 HP/UX 64 bit.
Following is only the overall result of the trace (it is quite big).
My question is about the Direct Path Read waits which are totallizing 268s of wait but are not showing in the fetch elapsed time (49.58s) and are not showing anywhere in the trace except in the overall result.
I do not understand why it is not part of the Elapsed time...
For info, the trace is for the specific session that was performing all the required queries to display an online report. The database is accessed by the Java application using Hybernate.
The trace was obtained by the following SQL:
exec sys.dbms_monitor.serv_mod_act_trace_enable(service_name=>'SYS$USERS',waits=>true,binds=>true);Then I query the sessions to find the one created by the application.
OVERALL TOTALS FOR ALL NON-RECURSIVE STATEMENTS
call count cpu elapsed disk query current rows
Parse 36 0.43 0.51 0 5 0 0
Execute 62 0.01 0.01 0 0 0 0
Fetch 579 4.01 49.06 3027 153553 0 5516
total 677 4.45 49.58 3027 153558 0 5516
Misses in library cache during parse: 29
Misses in library cache during execute: 2
Elapsed times include waiting on following events:
Event waited on Times Max. Wait Total Waited
---------------------------------------- Waited ---------- ------------
SQL*Net message to client 32754 0.00 0.03
SQL*Net message from client 32753 2.33 232.01
Disk file operations I/O 179 0.00 0.02
db file sequential read 2979 0.54 45.72
SQL*Net more data to client 133563 0.04 5.30
direct path read 34840 0.94 268.21
SQL*Net more data from client 1075 0.00 0.02
db file scattered read 6 0.03 0.11
asynch descriptor resize 52 0.00 0.00
OVERALL TOTALS FOR ALL RECURSIVE STATEMENTS
call count cpu elapsed disk query current rows
Parse 25 0.00 0.02 0 0 0 0
Execute 58 0.05 0.04 0 0 0 0
Fetch 126 0.00 0.04 4 161 0 123
total 209 0.05 0.11 4 161 0 123
Misses in library cache during parse: 3
Misses in library cache during execute: 3
Elapsed times include waiting on following events:
Event waited on Times Max. Wait Total Waited
---------------------------------------- Waited ---------- ------------
Disk file operations I/O 1 0.00 0.00
db file sequential read 4 0.01 0.03
asynch descriptor resize 1 0.00 0.00
37 user SQL statements in session.
57 internal SQL statements in session.
94 SQL statements in session.
Trace file: oxd1ta00_ora_16542.trc
Trace file compatibility: 11.1.0.7
Sort options: default
1 session in tracefile.
37 user SQL statements in trace file.
57 internal SQL statements in trace file.
94 SQL statements in trace file.
57 unique SQL statements in trace file.
241517 lines in trace file.
568 elapsed seconds in trace file.Thanks
ChristopheChristophe Lize wrote:
Closing this thread even if it's not answered...Sorry, I don't have time to test this myself now, but you shouldn't mark this thread as answered if it is not, because other people might find it and think they find an answer if they have a similar question.
I suggest you try the following to narrow down things:
1. Open the RAW trace file and check the cursor numbers of the "direct path reads" - check if you can find any references for those cursor numbers manually. The cursor numbers are those numbers behind the WAIT #<xx>, and you can check if you find any other entry unequal to WAIT #<xx> with the same #<xx>, for example EXEC #<xx> or FETCH #<xx>
A short primer on how to interpret the raw trace file can also be found in MOS document 39817.1
2. Run the RAW trace file through alternative free trace file analyzers like SQLDeveloper (yes it can process raw trace files), OraSRP or Christian Antognini's TVD$XTAT. If you have My Oracle Support access you can also try Oracle's own extended Trace Analyzer (TRCA / TRCANLZR). See MOS Note 224270.1
Check if these tools tell you more about your specific wait event and oddities with the trace file in general.
Regards,
Randolf
Oracle related stuff blog:
http://oracle-randolf.blogspot.com/
Co-author of the "OakTable Expert Oracle Practices" book:
http://www.apress.com/book/view/1430226684
http://www.amazon.com/Expert-Oracle-Practices-Database-Administration/dp/1430226684 -
Same sqlID with different execution plan and Elapsed Time (s), Executions time
Hello All,
The AWR reports for two days with same sqlID with different execution plan and Elapsed Time (s), Executions time please help me to find out what is reason for this change.
Please find the below detail 17th day my process are very slow as compare to 18th
17th Oct 18th Oct
221,808,602
21
2tc2d3u52rppt
213,170,100
72,495,618
9c8wqzz7kyf37
209,239,059
71,477,888
9c8wqzz7kyf37
139,331,777
1
7b0kzmf0pfpzn
144,813,295
1
0cqc3bxxd1yqy
102,045,818
1
8vp1ap3af0ma5
128,892,787
16,673,829
84cqfur5na6fg
89,485,065
1
5kk8nd3uzkw13
127,467,250
16,642,939
1uz87xssm312g
67,520,695
8,058,820
a9n705a9gfb71
104,490,582
12,443,376
a9n705a9gfb71
62,627,205
1
ctwjy8cs6vng2
101,677,382
15,147,771
3p8q3q0scmr2k
57,965,892
268,353
akp7vwtyfmuas
98,000,414
1
0ybdwg85v9v6m
57,519,802
53
1kn9bv63xvjtc
87,293,909
1
5kk8nd3uzkw13
52,690,398
0
9btkg0axsk114
77,786,274
74
1kn9bv63xvjtc
34,767,882
1,003
bdgma0tn8ajz9
Not only queries are different but also the number of blocks read by top 10 queries are much higher on 17th than 18th.
The other big difference is the average read time on two days
Tablespace IO Stats
17th Oct
Tablespace
Reads
Av Reads/s
Av Rd(ms)
Av Blks/Rd
Writes
Av Writes/s
Buffer Waits
Av Buf Wt(ms)
INDUS_TRN_DATA01
947,766
59
4.24
4.86
185,084
11
2,887
6.42
UNDOTBS2
517,609
32
4.27
1.00
112,070
7
108
11.85
INDUS_MST_DATA01
288,994
18
8.63
8.38
52,541
3
23,490
7.45
INDUS_TRN_INDX01
223,581
14
11.50
2.03
59,882
4
533
4.26
TEMP
198,936
12
2.77
17.88
11,179
1
732
2.13
INDUS_LOG_DATA01
45,838
3
4.81
14.36
348
0
1
0.00
INDUS_TMP_DATA01
44,020
3
4.41
16.55
244
0
1,587
4.79
SYSAUX
19,373
1
19.81
1.05
14,489
1
0
0.00
INDUS_LOG_INDX01
17,559
1
4.75
1.96
2,837
0
2
0.00
SYSTEM
7,881
0
12.15
1.04
1,361
0
109
7.71
INDUS_TMP_INDX01
1,873
0
11.48
13.62
231
0
0
0.00
INDUS_MST_INDX01
256
0
13.09
1.04
194
0
2
10.00
UNDOTBS1
70
0
1.86
1.00
60
0
0
0.00
STG_DATA01
63
0
1.27
1.00
60
0
0
0.00
USERS
63
0
0.32
1.00
60
0
0
0.00
INDUS_LOB_DATA01
62
0
0.32
1.00
60
0
0
0.00
TS_AUDIT
62
0
0.48
1.00
60
0
0
0.00
18th Oct
Tablespace
Reads
Av Reads/s
Av Rd(ms)
Av Blks/Rd
Writes
Av Writes/s
Buffer Waits
Av Buf Wt(ms)
INDUS_TRN_DATA01
980,283
91
1.40
4.74The AWR reports for two days with same sqlID with different execution plan and Elapsed Time (s), Executions time please help me to find out what is reason for this change.
Please find the below detail 17th day my process are very slow as compare to 18th
You wrote with different execution plan, I think, you saw plans. It is very difficult, you get old plan.
I think Execution plans is not changed in different days, if you not added index or ...
What say ADDM report about this script?
As you know, It is normally, different Elapsed Time for same statement in different day.
It is depend your database workload.
It think you must use SQL Access and SQl Tuning advisor for this script.
You can get solution for slow running problem.
Regards
Mahir M. Quluzade -
Write elapsed time to a spreadsheet in hours:minutes:seconds format
Hi everyone,
I've been trying to write an elapsed time to a spreadsheet file in an hours:minutes:seconds format, but the time is displayed in a floating point value of seconds..
how can I write to a spreadsheet in an hours:minutes:seconds format.
Thank you,
James-I often use a subVI that converts Seconds to Hours, Minutes and Seconds. Use the Quotient and Remainder function to divide your elapsed time by 3600, 60 and 1. You can then convert those values to a modified string and use the Write to Spreadsheet File.
As Dennis said, newer versions of LabVIEW's Write to Spreadsheet File.VI can handle arrays of Double, Integer or String automatically, and in older versions, the Write to Spreadsheet File.VI can be modified and copied to handle strings.
Hope this helps.
(Written in 8.5)
Message Edited by LabViewGuruWannabe on 01-18-2008 09:28 PM
Attachments:
TimeToSpreadsheet.vi 26 KB
SecondstoHMS.png 32 KB -
Can't get from time elapsed/time left to volume control
On my iPod Classic 80 GB, "Now Playing" defaults to the "time elapsed/time left" view. When I try to get to "volume control," I stroke the clickwheel, but it often doesn't respond. I have the iPod in a Belkin rubbery protective skin that covers the clickwheel, but it seems responsive enough when I'm already at the control I want to use.
Is there a way to explicitly go from "time elapsed/time left" to "volume control" in a way that doesn't require the stroking action on the clickwheel? Alternatively is this a sign that the sensitivity of the clickwheel has somehow deteriorated? I've only had the classic for about 6 months and it shows no other signs of malfunction.
TIA!You shouldn't be deleting old Tima Machine backups. When TM runs out of space it automatically deletes the oldest backups to make rioom for the new
http://pondini.org/TM/12.html -
Query Execution/Elapsed Time and Oracle Data Blocks
Hi,
I have created 3 tables with one column only. As an example Table 1 below:
SQL> create table T1 ( x char(2000));
So 3 tables are created in this way i.e. T1,T2 and T3.
T1 = in the default database tablespace of 8k (11g v11.1.0.6.0 - Production) (O.S=Windows).
T2 = I created in a Tablespace with Blocksize 16k.
T3 = I created in a Tablespace with Blocksize 4k. In the same Instance.
Each table has approx. 500 rows (So, table sizes are same in all the cases to test Query execution time ). As these 3 tables are created under different data block sizes so the ALLOCATED no. of data blocks are different in all cases.
T1 = 8k = 256 Blocks = 00:00:04.76 (query execution time/elapsed time)
T2 = 16k=121 Blocks = 00:00:04.64
T3 = 4k = 490 Blocks = 00:00:04.91
Table Access is FULL i.e. I have used select * from table_name; in all 3 cases. No Index nothing.
My Question is why query execution time is nearly the same in all 3 cases because Oracle has to read all the data blocks in each case to fetch the records and there is a much difference in the allocated no. of blocks ???
In 4k block size example, Oracle has to read just 121 blocks and it's taking nearly the same time as it's taking to read 490 blocks???
This is just 1 example of different data blocks. I have around 40 tables in each block size tablespace and the result are nearly the same. It's very strange for me because there is a much difference in the no. of allocated blocks but execution time is almost the same, only difference in milliseconds.
I'll highly appreciate the expert opinions.
Bundle of thanks in advance.
Best Regards,Hi Chris,
No I'm not using separate databases, it's 8k database with non-standard blocksizes of 16k and 4k.
Actually I wanted to test the Elapsed time of these 3 tables, so for that I tried to create the same size
tables.
And how I equalize these is like I have created one column table with char(2000).
555 MB is the figure I wanted to use for these 3 tables ( no special figure, just to make it bigger than the
RAM used for my db at the db startup to be sure of not retrieving the records from cache).
so row size with overhead is 2006 * 290,000 rows = 581740000(bytes) / 1024 = 568105KB / 1024 = 555MB.
Through this math calculation I thought It will be the total table size. So I Created the same no. of rows in 3 blocksizes.
If it's wrong then what a mes because I was calculating tables sizes in the same way from the last few months.
Can you please explain a little how you found out the tables sizes in different block sizes.Though I understood how you
calculated size in MB from these 3 block sizes
T8K =97177 BLOCKS=759MB *( 97177*8 = 777416KB / 1024 = 759MB )*
T16K=41639 BLOCKS=650MB
BT4K=293656 BLOCKS=1147MB
For me it's new to calculate the size of a table. Can you please tell me then how many rows I can create in each of
these 3 tables to make them equal in MB to test for elapsed time.
Then I'll again run my test and put the results here. Because If I've wrongly calculated table sizes then there is no need to talk about elapsed time. First I must equalize the table sizes properly.
SQL> select sum(bytes)/1024/1024 "Size in MB" from dba_segments> 2 where segment_name = 'T16K';
Size in MB
655
Is above SQL is correct to calculate the size or is it the correct alternative way to your method of calculating the size??
I created the same table again with everything same and the result is :
SQL> select num_rows,blocks from user_tables where table_name = 'T16K';NUM_ROWS BLOCKS
290000 41703
64 more blocks are allocated this time so may be that's y it's showing total size of 655 instead of 650.
Thanks alot for your help.
Best Regards,
KAm
Edited by: kam555 on Nov 20, 2009 5:57 PM -
Execution time, elapsed time of an sql query
Can you please tell me how to get the execution time, elapsed time of an sql query
user8680248 wrote:
I am running query in the database
I would like to know how long the query take the time to completeWhy? That answer can be totally meaningless as the VERY SAME query on the VERY SAME data on the VERY SAME database in the VERY SAME Oracle session can and will show DIFFERENT execution times.
So why do you want to know a specific query's execution time? What do you expect that to tell you?
If you mean that you want to know how long an existing query being executed is still going to take - that's usually quite difficult to determine. Oracle does provide a view on so-called long operations. However, only certain factors of a query's execution will trigger that this query is a long operation - and only for those specific queries will there be long operation stats that provide an estimated completion time.
If your slow and long running query does not show in long operation, then Oracle does not consider it a long operation - it fails to meet the specific criteria and factors required as a long operation. This is not a bug or an error. Simply that your query does not meet the basic requirements to be viewed as a long operation.
Oracle however provides the developer with the means to create long operations (using PL/SQL). You need to know and do the following:
a) need to know how many units of work to do (e.g. how many fetches/loop iterations/rows your code will process)
b) need to know how many units of work thus far done
c) use the DBMS_APPLICATION_INFO package to create a long operation and continually update the operation with the number of work units thus far done
It is pretty easy to implement this in PL/SQL processing code (assuming requirements a and b can be met) - and provide long operation stats and estimated completion time for the DBA/operators/users of the database, waiting on your process to complete. -
Restarting a elapsed time function in a for loop
Hello,
I am having an issue while using an elasped time function inside of a while loop (I wire the Time has elapsed? boolean to the stop function of the while loop), that is consequently inside of a for loop. The elapsed time function works correctly on its own in a while loop, but when added to a for loop will not reset for each iteration of the for loop. I can post an example if it is needed, but has anyone run into this problem before? I have tried wiring all the reset boolean's but I cannot get the timer to reset untill the for loop has finished.
Any suggestions on how to achieve this another way are welcome.
Thanks,
AndrewI found the issue, I was using 2 timers inside of the for loop on the assumption that they are completely independant of one another....they are not. if time has elapsed on one timer then it uses that time to start off the next timer, kind of a wierd behaviour but I guess I am not really using the timer in a normal manner.
Andrew
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