assigning

[Former Member]

Hi!

Suppose itab2 contains lots of values lets say about 20.

How can I assign its values to itab3' s all components.

Hopefully, somebody can give helpful idea.

Regards

itab3-one= itab2-firstvalue

itab3-two= itab2-secondvalue

itab3-three= itab2-thirdvalue

itab3-four= itab2-fourthvalue

append itab3

TYPES: begin of typ1, 
       one(10) type c, 
       two(10) type c, 
       three(10) type c, 
       four(10) type c, 
       end of typ1. 
        
TYPES: begin of typ2, 
       first(10) type c, 
       end of typ2. 


data itab type ALSMEX_TABLINE occurs 0 with header line. 
data itab2 type typ2 occurs 0 with header line . 
data itab3 type typ1 occurs 0 with header line .


LOOP AT ..
...
ENDLOOP.

[Former Member]

Hi,

Use the LOOP AT..

LOOP AT ITAB2.

• MOVE THE VALUES FOR EACH COLUMN.

ITAB3-ONE = ITAB2-FIRST.

ITAB3-..

APPEND ITAB3.

ENDLOOP.

Thanks,

Naren

[Former Member]

Hello Naren,

I think thats wrong. Please note within the LOOP iteration

you will assign always the same values for all itab3' s

components.

According to your suugestion it looks as follows:

LOOP AT ITAB2.

• MOVE THE VALUES FOR EACH COLUMN.

ITAB3-ONE = ITAB2-FIRST.

ITAB3-TWO = ITAB2-FIRST.

ITAB3-THREE = ITAB2-FIRST.

ITAB3-FOUR = ITAB2-FIRST.

APPEND ITAB3.

ENDLOOP.

[Former Member]

Ignore my answer.

Message was edited by:

Venkat Muthyala

[Former Member]

Hi Ilhan,

This should be the best and simple way of doing this, according to me.

<code>

append lines of itab2 to itab3.

</endcode>

Remember to reward points, if you are satisifed with this answer.

Rgds,

Prashanth.

SAP.

[Former Member]

hi Prashanth,

can you please write a snippet to demonstrate it concretely ?

regards

Ilhan

[Former Member]

Hi,

Since ITAB2 is declared with a header line..

And I am looping at ITAB2..For each iteration the header line will be populated..

<b>So according to this logic..ITAB3 will have the values of ITAB2..And not the same values...</b>

LOOP AT <b>ITAB2</b>.

• MOVE THE VALUES FOR EACH COLUMN.

ITAB3-ONE = ITAB2-FIRST.

ITAB3-TWO = ITAB2-FIRST.

ITAB3-THREE = ITAB2-FIRST.

ITAB3-FOUR = ITAB2-FIRST.

APPEND ITAB3.

ENDLOOP.

Thanks,

Naren

[Former Member]

Dear Narendran,

TYPES: begin of typ1,

one(10) type c,

two(10) type c,

three(10) type c,

four(10) type c,

end of typ1.

TYPES: begin of typ2,

first(10) type c,

end of typ2.

data itab2 type typ2 occurs 0 with header line .

data itab3 type typ1 occurs 0 with header line .

itab2-first = '2'.

itab2-first = '5'.

itab2-first = '6'.

itab2-first = '12'.

itab2-first = '32'.

itab2-first = '35'.

itab2-first = '40'.

itab2-first = '12'.

itab2-first = '45'.

itab2-first = '48'.

itab2-first = '58'.

itab2-first = '60'.

append itab2.

LOOP AT ITAB2.

ITAB3-ONE = ITAB2-FIRST.

ITAB3-TWO = ITAB2-FIRST.

ITAB3-THREE = ITAB2-FIRST.

ITAB3-FOUR = ITAB2-FIRST.

APPEND ITAB3.

ENDLOOP.

LOOP AT itab3.

write:/ itab3.

ENDLOOP.

*purpose -> RESULT SHOULD LOOK LIKE see below

*2 5 6 12

*32 35 40 12

*45 48 58 60

• this logic doesn't provide desired result

• I get only

• 60 60 60 60

*LOOP ....

• Firsttime of iteration

*ITAB3-ONE = ITAB2-FIRST. 2

*ITAB3-TWO = ITAB2-FIRST. 5

*ITAB3-THREE = ITAB2-FIRST. 6

*ITAB3-FOUR = ITAB2-FIRST. 12

*ENDLOOP.

[Former Member]

Sorry I have to make correction.

instead

• this logic doesn't provide desired result

• I get only

• 60 60 60 60

please consider

• this logic doesn't provide desired result

• I get only

2 2 2 2

5 5 5 5

6 6 6 6

12 12 12 12

32 32 32 32

35 35 35 35

40 40 40 40

12 12 12 12

45 45 45 45

48 48 48 48

58 58 58 58

60 60 60 60

[Former Member]

Hi,

Okay ..I got it..

Try this..

DATA: V_INDEX TYPE SYINDEX.

FIELD-SYMBOLS: <FS>.

V_INDEX = 1.

DO.

READ TABLE ITAB2 INDEX V_INDEX.

IF SY-SUBRC <> 0.

EXIT.

ENDIF.

DO 4 TIMES.

READ TABLE ITAB2 INDEX V_INDEX.

IF SY-SUBRC <> 0.

EXIT.

ENDIF.

ASSIGN SY-INDEX COMPONENT OF STRUCTURE ITAB3 TO <FS>.

<FS> = ITAB2-FIRST.

IF SY-SUBRC <> 0.

EXIT.

ENDIF.

V_INDEX = V_INDEX + 1.

ENDDO.

APPEND ITAB3.

ENDDO.

Thanks,

Naren

[Former Member]

Hi Naren,

I'm working with R/3 4.6 C and I have got the following

syntax error when I did check your addtional code.

-> COMPONENT OF STRUCTURE ITAB3 ist not expected

any idea whats wrong

friendly reagrds

Ilhan

[Former Member]

Ok Naren I got it too.

excuse the oversight you wrote:

ASSIGN SY-INDEX COMPONENT OF STRUCTURE ITAB3 TO <FS>.

instead of

ASSIGN COMPONENT SY-INDEX OF STRUCTURE ITAB3 TO <FS>.

It doesn't greatly matter. I thank you very much for your kindly efforts

to help me. It provides the desired result now.

Is that really the sole possibility to solve that problem?

[Former Member]

Hi,

Sorry for the syntax error..This is one of possibility..You can also write four read statements to do that..Instead of DO..ASSIGN COMPONENT..ENDDO...Then you can insert after 4 reads.

DATA: V_INDEX TYPE SYINDEX.

DATA: V_MOD TYPE SYINDEX.

V_INDEX = 1.

DO.

READ TABLE ITAB2 INDEX V_INDEX.

IF SY-SUBRC <> 0.

EXIT.

ENDIF.

ITAB3-ONE = ITAB2-FIRST.

V_INDEX = V_INDEX + 1.

READ TABLE ITAB2 INDEX V_INDEX.

IF SY-SUBRC <> 0.

EXIT.

ENDIF.

ITAB3-SECOND = ITAB2-FIRST.

V_INDEX = V_INDEX + 1.

READ TABLE ITAB2 INDEX V_INDEX.

IF SY-SUBRC <> 0.

EXIT.

ENDIF.

ITAB3-THIRD = ITAB2-FIRST.

V_INDEX = V_INDEX + 1.

READ TABLE ITAB2 INDEX V_INDEX.

IF SY-SUBRC <> 0.

EXIT.

ENDIF.

ITAB3-FOURTH = ITAB2-FIRST.

V_INDEX = V_INDEX + 1.

APPEND ITAB3.

ENDDO.

Thanks,

Naren