topic Re: sum in internal table in Application Development and Automation Discussions

sum in internal table

Former Member — Thu, 01 Feb 2007 15:25:51 GMT

Hallow I have a internal table with employee number and number of hours that employee done in jobs the problem is that I have for one employee 3 kind of hour.

Pernr hour

123 10

123 10.5

123 15

456 4

456 9

How can I do that I sum the hour ex.

Emp num 123 35.5

Emp num 456 13

Because its internal table I cant do a select count

thankes

Re: sum in internal table

Former Member — Thu, 01 Feb 2007 15:27:09 GMT

try control break event like......

at end of pernr.

sum.

endat.

regds,

kiran

Re: sum in internal table

Former Member — Thu, 01 Feb 2007 15:29:02 GMT

Or use a <b>collect</b> statement !

( collect instead of append, it'll do the job )

Hope this helps,

Erwan

Re: sum in internal table

Former Member — Thu, 01 Feb 2007 15:30:55 GMT

Hi,

COLLECT statement will give u the desired output.

regards,

madhumitha

Re: sum in internal table

Former Member — Thu, 01 Feb 2007 15:38:34 GMT

Hi Antonio,

Please try this by creating new internal table (ITAB2) with the same structure (ITAB).

...

SORT ITAB.

LOOP ITAB.
  ITAB2-PERNR = ITAB-PERNR.
  ITAB2-HOUR = ITAB-HOUR.
  COLLECT ITAB2.
ENDLOOP.

...

Now, your ITAB2 has the total hours by PERNR.

Regards,

Ferry Lianto

Re: sum in internal table

Former Member — Thu, 01 Feb 2007 15:40:14 GMT

Hi Antonio,

Use control-break statements to achieve your requirement.

DATA:

BEGIN OF ITAB OCCURS 0,

PERNR TYPE PERNR.

HOUR TYPE I,

END OF ITAB.

DATA ITAB_NEW TYPE ITAB OCCURS 0 WITH HEADER LINE.

DATA V_COUNT TYPE I.

DATA V_PERNR TYPE PERNR.

SORT ITAB BY PERNR.

LOOP AT ITAB.

V_PERNR = ITAB-PERNR.

AT NEW PERNR.

CLEAR V_COUNT.

ENDAT.

V_COUNT = V_COUNT + ITAB-HOUR.

AT END OF PERNR.

ITAB_NEW-PERNR = V_PERNR.

ITAB_NEW-HOUR = V_C0UNT.

APPEND ITAB_NEW.

ENDAT.

ENDLOOP.

Thanks,

Vinay