如何在内部表中生成新的自己的列表？ 阿巴

Question

List in the internal table at the beginning:

code | name | sum
 22  | Jon  | 234.3
 22  | Jon  | 34.2
 22  | Jon  | 0
 22  | Jon  | 0
 12  | Bob  | 999.4
 12  | Bob  | 0
 45  | Anna | 0
 45  | Anna | 0
 11  | Mike | 0
 11  | Mike | 234.3

To get the output of such a list from the internal table:

code | name | sum
 22  | Jon  | 234.3
 22  | Jon  | 34.2
 12  | Bob  | 999.4
 45  | Anna | 0
 11  | Mike | 234.3

Conditions for the formation of a new (outgoing) list:

1. If the column name with several identical names (for example with four - Jon ) has a value (column sum ) 34.5 ... and 0, then throw away all 0 and print only non-zero.
2. If the column name with several identical names (for example two - Anna ) has a value (column sum ) 0 - then print only one name with the value 0.
3. The list can not be sorted - the output must have a list with the same order as the input.

Answer 1

I'm considering lt_grp1 will contains beginning internal table records. And i've declared lt_grp2 of same structure as lt_grp1 .

  DATA: lv_index TYPE i VALUE 0.

  APPEND LINES OF lt_grp1 TO lt_grp2.
  DELETE ADJACENT DUPLICATES FROM lt_grp2 COMPARING code name.

  LOOP AT lt_grp2 INTO ls_grp1.

    LOOP AT lt_grp1 INTO ls_grp2
          WHERE code = ls_grp1-code
            AND name = ls_grp1-name.
      lv_index = lv_index + 1.
      IF ls_grp2-sum = 0.
        IF lv_index > 1.
          DELETE lt_grp1 INDEX sy-tabix.
        ENDIF.
      ELSE.
        IF lv_index > 1.
          DELETE lt_grp1 WHERE sum = 0
                          AND code = ls_grp1-code.
        ENDIF.
      ENDIF.
      CLEAR: ls_grp2.
    ENDLOOP.
    CLEAR : lv_index.
  ENDLOOP.


  CLEAR :ls_grp1.
  LOOP AT lt_grp1 INTO ls_grp1.
    WRITE: / ls_grp1-code, ls_grp1-name, ls_grp1-sum.
  ENDLOOP.

Hope this helps!

For those who think i didn't tested it.

Here is input table -

Output -

Answer 2

This has O(n log(n)) runtime.

lt_copy = lt_original.
SORT lt_copy BY code name ASCENDING sum DESCENDING. "if there is a non-zero line, it is at the top "
DELETE ADJACENT DUPLICATES FROM lt_copy.

LOOP AT lt_original INTO DATA(ls_original).
  IF ls_original-sum = 0. "only 0 lines need to be checked"
    READ TABLE lt_copy ASSIGNING <fs_max_sum> BINARY SEARCH
        WITH KEY code = <fs_original>-code
                 name = <fs_original>-name.
    IF <fs_max_sum>-sum = 0.
      "there are only zeros for this code and name, we need it"
    ELSE.
      DELETE lt_original.
    ENDIF.
  ENDIF.
ENDLOOP.

Answer 3

I don't have enough points to comment on someone else's post but I see that they are sorting the tables when the spec said that it must not be sorted. You are unfortunately going to have to sort this table (which is actually fortunate thing). You need to ask why it must remain in that specific manner.

I tried an approach with no sorting but I lose Anna when extracting the data.

LOOP At itab.
    READ TABLE itab WITH KEY code = itab-code TRANSPORTING NO FIELDS.
    IF SY-SUBRC EQ 0.
      IF itab-sum NE 0.
        wa-code = itab-code.
        wa-name = itab-name.
        wa-sum = itab-sum.
        APPEND wa.
      ENDIF.
    ENDIF
  ENDLOOP.