matinal:SAP ABAP 如何处理负号和千分位符号

2023-10-14 15:20:00 浏览数 (1)

写一个功能把

代码语言:javascript复制
*&----------------------------------------------------------------------*
*&      符号(l_temp2)移动,千分位加符号(l_temp4)
*&----------------------------------------------------------------------*
FORM move_dig CHANGING p_temp.
  DATA:  l_temp1(17) TYPE c,
         l_temp2(2)  TYPE c VALUE '-',
         l_temp3(1)  TYPE c,
         l_temp4(1)  TYPE c VALUE ',',
         i_len       TYPE i,
         i_len1      TYPE i,
         i_count     TYPE i,
         i_count1    TYPE i,
         c1(17)      TYPE c,
         c2(17)      TYPE c,
         c3(2)       TYPE c.
  IF p_temp < 0.
    SPLIT p_temp AT '-' INTO l_temp1 l_temp3.
    CONCATENATE l_temp2  l_temp1 INTO p_temp.
    CONDENSE p_temp NO-GAPS.
    i_len = strlen( p_temp ).
    MOVE '-' TO c3.
  ELSE.
    CONDENSE p_temp NO-GAPS.
    i_len = strlen( p_temp ).
    MOVE ' ' TO c3.
  ENDIF.

* When ' '
  IF c3 = ' ' AND i_len > 3.
    i_count  = i_len DIV 3.
    DO i_count TIMES.
      i_len1 = i_len - 3.
      MOVE p_temp i_len1(3) TO c1.
      CONCATENATE l_temp4 c1 c2 INTO c2.
      MOVE i_len1 TO i_len.
      CLEAR i_len1.
    ENDDO.
*  Last Record
    i_count = i_len MOD 3.
    IF i_count > 0.
      MOVE p_temp 0(i_count) TO c1.
      CONCATENATE c1 c2 INTO c2.
    ELSE.
      SHIFT c2.
    ENDIF.
    MOVE c2 TO p_temp.
*    clear p_temp.
    CLEAR i_count.
  ENDIF.
************ 
* When '-'
  IF c3 = '-' AND i_len > 4.
    SHIFT p_temp.
    i_len = strlen( p_temp ).
    i_count = i_len DIV 3.
    DO i_count TIMES.
      i_len1 = i_len - 3.
      MOVE p_temp i_len1(3) TO c1.
      CONCATENATE l_temp4 c1 c2 INTO c2.
      MOVE i_len1 TO i_len.
      CLEAR i_len1.
    ENDDO.
* Last record
    i_count = i_len MOD 3.
    IF i_count > 0.
      MOVE p_temp 0(i_count) TO c1.
      CONCATENATE c1 c2 INTO c2.
    ELSE.
      SHIFT c2.
      CONCATENATE '-' c2 INTO c2.
    ENDIF.
    CONCATENATE '-' c2 INTO c2.
    MOVE c2 TO p_temp.
    CLEAR i_count.
  ENDIF.
ENDFORM.                    " move_dig

0 人点赞