PWA Timesheet - How to limit hours per day on timesheet submission?
I've set "Maximum Hours per day" limiation to 24 hours on "Timesheet settings and defaults" and it works when a user click "Save" on "Timesheet" view.
But when the user click "Send" without saving first, it bypasses this limitation and save the timesheet reporting with any amount of hours that the user entered.
Is there any way to solve this and limit the user so he won't be able to submit more than 24 hours per day?
Thanks
Under configure timesheet settings and defaults
In the Hourly Reporting Limits section, specify the maximum and minimum hours allowed in a timesheet and the maximum number of hours allowed to be reported in a day. If team members report time beyond these limits, errors appear on their timesheets when
they submit them.
make it 8 hours then check.
kirtesh
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Looping through date column - Summing hours per day
All,
Running the query below against my timesheet tables I get the following results:
-----SQL CODE------
SELECT
ts.ts_date Date,
ts.user_name Name,
tc.account Account,
ts.no_of_hrs Hours,
SUM(ts.no_of_hrs) OVER(PARTITION BY ts.ts_date) Daily_Total
FROM
eba_time_timesheet ts,
eba_time_timecodes tc
WHERE
ts.timecode_id = tc.id AND
ts.user_name LIKE 'JohnD'
ORDER BY
1
-----RESULTS-------
Date Name Account Hours Daily Total
1-Dec-09 JOHND 489310 1.5 8
1-Dec-09 JOHND 486830 1.5 8
1-Dec-09 JOHND 481710 3 8
1-Dec-09 JOHND 481210 0.5 8
1-Dec-09 JOHND 486840 0.5 8
1-Dec-09 JOHND 485710 0.5 8
1-Dec-09 JOHND 481010 0.5 8
2-Dec-09 JOHND 481710 1 8
2-Dec-09 JOHND 485710 7 8
3-Dec-09 JOHND 481710 6 8
3-Dec-09 JOHND 488810 1.5 8
3-Dec-09 JOHND 481310 0.5 8
4-Dec-09 JOHND 489710 8 8
7-Dec-09 JOHND 481110 0.5 8
7-Dec-09 JOHND 489710 7 8
7-Dec-09 JOHND 481210 0.5 8
However, I would prefer the Daily Total column be a row in the results instead of a column. Here is an example of how I would prefer the results. This statement will then be sent to a calendar for each user to see there time for each account and total time per day.
Date Name Account Hours
1-Dec-09 JOHND 489310 1.5
1-Dec-09 JOHND 486830 1.5
1-Dec-09 JOHND 481710 3
1-Dec-09 JOHND 481210 0.5
1-Dec-09 JOHND 486840 0.5
1-Dec-09 JOHND 485710 0.5
1-Dec-09 JOHND 481010 0.5
*1-Dec-09 JOHND Daily Total 8*
2-Dec-09 JOHND 481710 1
2-Dec-09 JOHND 485710 7
*2-Dec-09 JOHND Daily Total 8*
3-Dec-09 JOHND 481710 6
3-Dec-09 JOHND 488810 1.5
3-Dec-09 JOHND 481310 0.5
*3-Dec-09 JOHND Daily Total 8*
4-Dec-09 JOHND 489710 8
*4-Dec-09 JOHND Daily Total 8*
7-Dec-09 JOHND 481110 0.5
7-Dec-09 JOHND 489710 7
7-Dec-09 JOHND 481210 0.5
*7-Dec-09 JOHND Daily Total 8*
Any help would be greatly appreciated.
This is my 1st post so if I've left something out or you need additional info please let me know.
I’m using Oracle 10guser9160575 wrote:
Thanks for all the input! I ended up using the GROUP BY ROLLUP and adding a CASE statement for inserting "DAILY TOTAL" in the account column.If you are on at least 10g, model solution could be simpler:
with t as (
select to_date('1-Dec-09','dd-mon-yy') dt,'JOHND' name,489310 account,1.5 hours,8 daily_total from dual union all
select to_date('1-Dec-09','dd-mon-yy'),'JOHND',486830,1.5,8 from dual union all
select to_date('1-Dec-09','dd-mon-yy'),'JOHND',481710,3,8 from dual union all
select to_date('1-Dec-09','dd-mon-yy'),'JOHND',481210,0.5,8 from dual union all
select to_date('1-Dec-09','dd-mon-yy'),'JOHND',486840,0.5,8 from dual union all
select to_date('1-Dec-09','dd-mon-yy'),'JOHND',485710,0.5,8 from dual union all
select to_date('1-Dec-09','dd-mon-yy'),'JOHND',481010,0.5,8 from dual union all
select to_date('2-Dec-09','dd-mon-yy'),'JOHND',481710,1,8 from dual union all
select to_date('2-Dec-09','dd-mon-yy'),'JOHND',485710,7,8 from dual union all
select to_date('3-Dec-09','dd-mon-yy'),'JOHND',481710,6,8 from dual union all
select to_date('3-Dec-09','dd-mon-yy'),'JOHND',488810,1.5,8 from dual union all
select to_date('3-Dec-09','dd-mon-yy'),'JOHND',481310,0.5,8 from dual union all
select to_date('4-Dec-09','dd-mon-yy'),'JOHND',489710,8,8 from dual union all
select to_date('7-Dec-09','dd-mon-yy'),'JOHND',481110,0.5,8 from dual union all
select to_date('7-Dec-09','dd-mon-yy'),'JOHND',489710,7,8 from dual union all
select to_date('7-Dec-09','dd-mon-yy'),'JOHND',481210,0.5,8 from dual
select dt "Date",
name "Name",
account "Account",
hours "Hours"
from t
model
dimension by(dt,name,to_char(account) account)
measures(hours,daily_total,0 seq)
rules upsert all(
hours[any,any,'Daily Total'] = max(daily_total)[cv(dt),cv(name),any],
seq[any,any,'Daily Total'] = 1
order by dt,
name,
seq,
account
Date Name Account Hours
01-DEC-09 JOHND 481010 .5
01-DEC-09 JOHND 481210 .5
01-DEC-09 JOHND 481710 3
01-DEC-09 JOHND 485710 .5
01-DEC-09 JOHND 486830 1.5
01-DEC-09 JOHND 486840 .5
01-DEC-09 JOHND 489310 1.5
01-DEC-09 JOHND Daily Total 8
02-DEC-09 JOHND 481710 1
02-DEC-09 JOHND 485710 7
02-DEC-09 JOHND Daily Total 8
Date Name Account Hours
03-DEC-09 JOHND 481310 .5
03-DEC-09 JOHND 481710 6
03-DEC-09 JOHND 488810 1.5
03-DEC-09 JOHND Daily Total 8
04-DEC-09 JOHND 489710 8
04-DEC-09 JOHND Daily Total 8
07-DEC-09 JOHND 481110 .5
07-DEC-09 JOHND 481210 .5
07-DEC-09 JOHND 489710 7
07-DEC-09 JOHND Daily Total 8
21 rows selected.
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How to add extra per day records based on maximum occurances in the week.
Hello,
I need to add extra records based on maximum occurances on a day in a week.
For example, the result of the query shows that maximum occuances in a week 0 is 3 on so I need to add empty records in each day in that week.
SQL QUERY:
with numbers as (
SELECT ROWNUM N FROM dual
CONNECT BY LEVEL <=35
),months as (
SELECT ADD_MONTHS('01-JUN-10',n-1) AS current_month,
TO_CHAR(ADD_MONTHS('01-JUN-10',n-1),'fmMonth-YYYY') AS current_month_label
FROM numbers
WHERE ADD_MONTHS('01-JUN-10',n-1) <= NVL(add_months('01-JUN-10', :monthcount-1),:p_month)
ORDER BY n ASC
) , dates as (
select trunc(trunc(m.current_month, 'MM'), 'D') + n - 1 calendar_date,
trunc((n - 1) / 7) AS week_num,
m.current_month as current_month
from numbers, months m
where trunc(dom.schedule.calendar_begin(m.current_month) + n - 1, 'd') <= last_day(m.current_month)
Order By week_num, calendar_date
select dates.week_num, edate, name from (
--week 0
select '30-MAY-10' as edate, 'JJ' as name from dual union
select '31-MAY-10' as edate, 'MK' as name from dual union
select '01-JUN-10' as edate, 'MK' as name from dual union
select '01-JUN-10' as edate, 'BK' as name from dual union
select '01-JUN-10' as edate, 'JJ' as name from dual union
select '04-JUN-10' as edate, 'KK' as name from dual union
select '04-JUN-10' as edate, 'LA' as name from dual union
-- week 1
select '06-JUN-10' as edate, 'BK' as name from dual union
select '06-JUN-10' as edate, 'JJ' as name from dual union
select '07-JUN-10' as edate, 'KK' as name from dual union
select '11-JUN-10' as edate, 'LA' as name from dual
) a join dates on dates.calendar_date = a.edate
order by week_num, to_date(edate);CURRENT RESULT :
WEEK_NUM EDATE NAME
0 30-MAY-10 JJ
0 31-MAY-10 MK
0 01-JUN-10 MK
0 01-JUN-10 JJ
0 01-JUN-10 BK
0 04-JUN-10 KK
0 04-JUN-10 LA
1 06-JUN-10 BK
1 06-JUN-10 JJ
1 07-JUN-10 KK
1 11-JUN-10 LAREQUIRED RESULT:
WEEK_NUM EDATE NAME
0 30-MAY-10 JJ
0 30-MAY-10 null
0 30-MAY-10 null
0 31-MAY-10 MK
0 31-MAY-10 null
0 31-MAY-10 null
0 01-JUN-10 BK
0 01-JUN-10 JJ
0 01-JUN-10 MK
0 02-JUN-10 null
0 02-JUN-10 null
0 02-JUN-10 null
0 03-JUN-10 null
0 03-JUN-10 null
0 03-JUN-10 null
0 04-JUN-10 KK
0 04-JUN-10 LA
0 04-JUN-10 null
0 05-JUN-10 null
0 05-JUN-10 null
0 05-JUN-10 null
--Number of Week 1 records should be 14 = (max occurances on 06-JUN-10) * 7
1 06-JUN-10 BK
1 06-JUN-10 JJ
1 07-JUN-10 KK
1 07-JUN-10 null
1 08-JUN-10 null
1 08-JUN-10 null
1 09-JUN-10 null
1 09-JUN-10 null
1 10-JUN-10 null
1 10-JUN-10 null
1 11-JUN-10 LA
1 11-JUN-10 null
1 12-JUN-10 null
1 12-JUN-10 nullI would appreciate it if you could help me to find efficient way to achieve this?
Thanks,
GM
Edited by: user12068331 on Sep 28, 2010 2:17 PM
Edited by: user12068331 on Sep 28, 2010 2:48 PMOk,
Few things to clarify first :
1 - How do you define week number ?
Oracle has 2 different way for week number : normal week number (WW) and isoweek (IW).
is your weeknumber one of those, or is it something else ?
2 - Depending on the country the week doesn't start on the same day : in France (as most of Europe I think) first day is monday, where in the USA (as far as I know) first day is sunday.
the to_char builtin function can display weeknumber :with a as (
--week 0
select 0 as weeknum, to_date('30-MAY-2010','DD-MON-YYYY') as edate, 'JJ' as name from dual union all
select 0 as weeknum, to_date('31-MAY-2010','DD-MON-YYYY') as edate, 'MK' as name from dual union all
select 0 as weeknum, to_date('01-JUN-2010','DD-MON-YYYY') as edate, 'MK' as name from dual union all
select 0 as weeknum, to_date('01-JUN-2010','DD-MON-YYYY') as edate, 'BK' as name from dual union all
select 0 as weeknum, to_date('01-JUN-2010','DD-MON-YYYY') as edate, 'JJ' as name from dual union all
select 0 as weeknum, to_date('04-JUN-2010','DD-MON-YYYY') as edate, 'KK' as name from dual union all
select 0 as weeknum, to_date('04-JUN-2010','DD-MON-YYYY') as edate, 'LA' as name from dual union all
-- week 1
select 1 as weeknum, to_date('06-JUN-2010','DD-MON-YYYY') as edate, 'BK' as name from dual union all
select 1 as weeknum, to_date('06-JUN-2010','DD-MON-YYYY') as edate, 'JJ' as name from dual union all
select 1 as weeknum, to_date('07-JUN-2010','DD-MON-YYYY') as edate, 'KK' as name from dual union all
select 1 as weeknum, to_date('11-JUN-2010','DD-MON-YYYY') as edate, 'LA' as name from dual
select to_char(a.edate,'IW') isoweek, to_char(a.edate,'WW') numweek, a.*
from a;
IS NU WEEKNUM EDATE NA
21 22 0 Sunday 30/05/2010 00:00:00 JJ
22 22 0 Monday 31/05/2010 00:00:00 MK
22 22 0 Tuesday 01/06/2010 00:00:00 MK
22 22 0 Tuesday 01/06/2010 00:00:00 BK
22 22 0 Tuesday 01/06/2010 00:00:00 JJ
22 23 0 Friday 04/06/2010 00:00:00 KK
22 23 0 Friday 04/06/2010 00:00:00 LA
22 23 1 Sunday 06/06/2010 00:00:00 BK
22 23 1 Sunday 06/06/2010 00:00:00 JJ
23 23 1 Monday 07/06/2010 00:00:00 KK
23 24 1 Friday 11/06/2010 00:00:00 LA
11 rows selected. -
How many iPhones stolen per day? Prevent shutdown.
Dear friends.
Since I had my iPhone stolen I'm searching for a way (without jailbreaking it) to prevent that my new iPhone gets lost.
"Find my iPhone" is only useful when the thief is dum.b as hel.l and when nobody has found it (yet).
Does Apple see comercial advantages on this situation? How many iPhones are stolen a day?!
I mean, why in the world wouldn't you develop security apps and security features so we don't lose our beloved phones?
This is serious, and it seems Apple doesn't care. I don't want to jailbreak my iPhone, but you are not helping, Apple!
I would like to see a Restriction setting that would block the shutdown of my iPhone when it's locked. It would require a passcode to shut it down. It's so simple!
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The same works for Airplane mode and Wi-Fi. A passcode should be required to turn it off (if I enable this feature).
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SECURITY!
So far nothing was done in this area. I believe Apple should pay more attention to this.
Best regards,
WillWill Segatto wrote:
Dear friends.
Since I had my iPhone stolen I'm searching for a way (without jailbreaking it) to prevent that my new iPhone gets lost.
"Find my iPhone" is only useful when the thief is dum.b as hel.l and when nobody has found it (yet).
Does Apple see comercial advantages on this situation? How many iPhones are stolen a day?!
I mean, why in the world wouldn't you develop security apps and security features so we don't lose our beloved phones?
This is serious, and it seems Apple doesn't care. I don't want to jailbreak my iPhone, but you are not helping, Apple!
I would like to see a Restriction setting that would block the shutdown of my iPhone when it's locked. It would require a passcode to shut it down. It's so simple!
This way thieves wouldn't be able to shut it down!
The same works for Airplane mode and Wi-Fi. A passcode should be required to turn it off (if I enable this feature).
Advanced considerations: DFU mode and recovery mode should "expire", the iPhone should reset itself after some time (e.g. 10 minutes) so the iPhone would come back to life and allow tracking.
SECURITY!
So far nothing was done in this area. I believe Apple should pay more attention to this.
Best regards,
Will
Sorry requiring a passcode to shut down the phone isn't a good idea.
do you know how many users forget their 4 digit passcode to unlock their phones - use the search function at the top right and you'll see all the threads.
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