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author's profile photo Richa Katiyar
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Richa Katiyar

Check if a non-key column of internal table holds same value for all records

Posted on Aug 18, 2016 at 01:59 PM 2.1k Views

Hello folks,

What is the best way to compare records of an internal table based on a single column values?

Say, I have an internal table with values as below:

A B C 1001 10 1001 20 X 1001 30

My requirement is to check if the column C contains identical values or not. In case of blank, all records should hold blank value in column C; in case 'X, all records should hold 'X'.

Kindly suggest the best approach to achieve this.

Thanks,

Rics

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6 Answers

  • Best Answer
    author's profile photo Bartosz Ziólkowski
    Bartosz Ziólkowski
    Posted on Aug 18, 2016 at 09:45 PM
    2

    And what should happen if you have 2 blank fields and 2 with value 'X'?

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  • author's profile photo Former Member
    Former Member
    Posted on Aug 19, 2016 at 04:00 AM
    1

    Hi Rics,

    1. Copy the internal table to another internal table it_tab2[] = it_tab1[].

    2. Read the first record of it_tab1[].

    Read table it_tab1 into wa_tab1 index 1.

    3. Then, delete it_tab2[] with following condition,

    Delete it_tab2 where c = wa_tab1-c.

    4. If all the records of it_tab2 is deleted, then C field has identical values. If it_tab2 is not completed deleted then C field has values which are not identical

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  • author's profile photo Horst Keller
    Horst Keller
    Posted on Aug 19, 2016 at 05:57 AM
    1

    To check if there is only one given value in a column of an internal table, you can do the following:

    READ TABLE itab WITH KEY c = 'X' TRANSPORTING NO FIELDS.

    DATA(subrc1) = sy-subrc.

    LOOP AT itab TRANSPORTING NO FIELDS WHERE c <> 'X'.

    EXIT.

    ENDLOOP.

    DATA(subrc2) = sy-subrc.

    IF subrc1 = 0 AND subrc2 <> 0.

    "only X

    ENDIF.

    If you can assert that you have exactly two values as in your scenario above, you can do it as follows:

    READ TABLE itab WITH KEY c = 'X' TRANSPORTING NO FIELDS.

    DATA(subrc1) = sy-subrc.

    READ TABLE itab WITH KEY c = ' ' TRANSPORTING NO FIELDS.

    DATA(subrc2) = sy-subrc.

    IF subrc1 = 0 AND subrc2 <> 0.

    "only X

    ELSEIF subrc1 <> 0 AND subrc2 = 0.

    "only space

    ELSE.

    "both

    ENDIF.


    Or in a modern way (elegant, but in fact too much read accesses ...):


    IF line_exists( itab[ c = 'X' ] ) AND

    NOT line_exists( itab[ c = ' ' ).

    "only X

    ELSEIF NOT line_exists( itab[ c = 'X' ] ) AND

    line_exists( itab[ c = ' ' ] ).

    "only space

    ELSE.

    "both

    ENDIF.


    In case that there are more values possible than 'X' and ' ', see my answer above.

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    • Glen Simpson
      Aug 19, 2016 at 07:00 AM

      Assuming you're on release 7.40 or higher, you can use IF NOT ( <expression1> EQUIV <expression2> ) to implement an "exclusive or" statement.

      So to refine Horst's example a little :

      TYPES: BEGIN OF ts_itab, c TYPE c, END OF ts_itab.

      TYPES tt_itab TYPE STANDARD TABLE OF ts_itab WITH DEFAULT KEY.

      DATA(itab) = VALUE tt_itab( ( c = 'X' ) ( c = 'X' ) ( c = 'X' ) ).

      IF NOT ( line_exists( itab[ c = 'X' ] ) EQUIV line_exists( itab[ c = ' ' ] ) ).

      "table contains only X or only ' '

      ENDIF.

  • author's profile photo Chris Bosler
    Chris Bosler
    Posted on Aug 18, 2016 at 08:37 PM
    0

    Probably just a SORT using COMPARE BY.

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  • author's profile photo Richa Katiyar
    Richa Katiyar
    Posted on Aug 19, 2016 at 07:21 AM
    0

    Hi All,

    Thank you for your quick responses, which were all very helpful.

    My problem has been solved with @Horst Keller solution. Thank you for your time and patience. 😊

    Regards,

    Rics

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  • author's profile photo Former Member
    Former Member
    Posted on Aug 22, 2016 at 08:23 AM
    0

    itab2[] = itab1[].

    delete itab2 where c = itab1[1]-c.

    if itab2[] is not initial and

    itab1[1]-c is initial.

    itab1-c = 'X'.

    else

    itab1-c = space.

    endif.

    modify itab1 transforting c where a = itab1[1]-a.

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