相互显示两个用户的角色？
Show roles of two users against each other?
我写了下面的 ABAP 程序，它可以从 AGR_USERS 中获取两个用户的角色，这些角色由用户在 selection 屏幕上指定。
输出屏幕应该有 User id 以及分配给用户的角色。程序对一个用户运行良好，但无法为第二个用户获取角色。
REPORT Z_REFERNCE_USER_PROG.
TYPE-POOLS: slis. " SLIS contains all the ALV data types
TYPE-POOLS: SLIS,KCDE.
DATA: username(20) TYPE C,
G_REPID TYPE SY-REPID,
IT_EVENTS TYPE SLIS_T_EVENT,
IT_FIELD TYPE SLIS_T_FIELDCAT_ALV,
WA_FIELD TYPE SLIS_FIELDCAT_ALV,
IT_SORT TYPE SLIS_T_SORTINFO_ALV.
INITIALIZATION.
G_REPID = SY-REPID.
DATA:BEGIN OF itab OCCURS 0,
UNAME type AGR_USERS-UNAME,
AGR_NAME type AGR_USERS-AGR_NAME,
UNAME1 type AGR_USERS-UNAME,
AGR_NAME1 type AGR_USERS-AGR_NAME,
END of itab.
start-of-selection.
SELECT-OPTIONS AFF_USER FOR USERNAME NO INTERVALS.
SELECT-OPTIONS REF_USER FOR USERNAME NO INTERVALS.
end-of-selection.
perform tosql.
perform listshow.
*--------------------------------
* form tosql
*--------------------------------
form tosql.
SELECT DISTINCT AGR_USERS~AGR_NAME AGR_USERS~UNAME
INTO corresponding fields of table itab
FROM AGR_USERS
where AGR_USERS~UNAME in AFF_USER.
SELECT DISTINCT AGR_USERS~AGR_NAME AGR_USERS~UNAME
INTO corresponding fields of table itab
FROM AGR_USERS
where AGR_USERS~UNAME in REF_USER.
endform.
*--------------------------------
* form listshow
*--------------------------------
form listshow.
DEFINE ADD_FIELD.
WA_FIELD-FIELDNAME = &1.
WA_FIELD-REPTEXT_DDIC = &2.
WA_FIELD-NO_ZERO = 'X'.
APPEND WA_FIELD TO IT_FIELD.
END-OF-DEFINITION.
ADD_FIELD 'UNAME' 'Affected user'.
ADD_FIELD 'AGR_NAME' 'Role'.
ADD_FIELD 'UNAME1' 'Reference user'.
ADD_FIELD 'AGR_NAME1' 'Role'.
CALL FUNCTION 'REUSE_ALV_GRID_DISPLAY'"
EXPORTING
I_CALLBACK_PROGRAM = G_REPID
I_BACKGROUND_ID = 'ALV_BACKGROUND'
IT_FIELDCAT = IT_FIELD
* IS_LAYOUT = GS_LAYOUT
* IT_SORT = IT_SORT
I_SAVE = 'A'
IT_EVENTS = IT_EVENTS[]
TABLES
T_OUTTAB = itab
EXCEPTIONS
PROGRAM_ERROR = 1
OTHERS = 2.
endform.
输入：
输出：
我想将第一个 SELECT 的所有值放入 table 1 并将第二个 select 的所有值放入 table 2 然后将它们合并到 table 3.我只是想看到相邻的两个用户的角色，顺序无所谓。
SELECT ... INTO 用结果覆盖 table 的内容。所以当你想得到两个SELECTS的结果时，你需要两个tables:
SELECT DISTINCT agr_name, uname
INTO TABLE @DATA(lt_aff)
FROM agr_users
WHERE uname IN @aff_user.
SELECT DISTINCT agr_name, uname
INTO TABLE @DATA(lt_ref)
FROM agr_users
WHERE uname IN @ref_user.
现在如何将这两个结果合二为一table？
问题是您不知道这两个 table 中哪一个会更长。所以我们需要使用一个带有计数器的 WHILE 循环，当计数器小于 either table 的行数时，它会向上计数。然后在该循环中，我们需要使用 READ TABLE 获取对应于计数器当前值的两个 table 的行。如果成功，我们需要将其值移动到输出 table itab 的新行，然后我们将其附加到它：
DATA lv_counter TYPE i.
DATA ls_result LIKE LINE OF itab.
lv_counter = 0.
WHILE lv_counter < lines( lt_aff ) OR lv_counter < lines( lt_ref ).
lv_counter = lv_counter + 1.
CLEAR ls_result.
READ TABLE lt_aff INDEX lv_counter INTO DATA(ls_aff).
IF sy-subrc = 0.
ls_result-uname = ls_aff-uname.
ls_result-agr_name = ls_aff-agr_name.
ENDIF.
READ TABLE lt_ref INDEX lv_counter INTO DATA(ls_ref).
IF sy-subrc = 0.
ls_result-uname1 = ls_ref-uname.
ls_result-agr_name1 = ls_ref-agr_name.
ENDIF.
INSERT ls_result INTO TABLE itab.
ENDWHILE.
实际上您不需要 SELECT-OPTIONS，根据您的任务定义，您只比较两个用户
PARAMETERS: aff_user TYPE xubname DEFAULT 'NEUMANAR',
ref_user TYPE xubname DEFAULT 'KAJAVESH'.
这是在单个数据库 SELECT 和单个 table:
中的方法
DATA ls_roles TYPE agr_users.
itab-uname = aff_user.
itab-uname1 = ref_user.
SELECT agr_name uname
FROM agr_users
INTO (ls_roles-agr_name,ls_roles-uname)
WHERE uname = aff_user OR uname = ref_user.
READ TABLE itab ASSIGNING FIELD-SYMBOL(<fs_itab>) WITH KEY agr_name = space.
IF sy-subrc = 0 AND ls_roles-uname = aff_user.
<fs_itab>-agr_name = ls_roles-agr_name.
ELSEIF ls_roles-uname = ref_user.
UNASSIGN <fs_itab>.
READ TABLE itab ASSIGNING <fs_itab> WITH KEY agr_name1 = space.
IF sy-subrc = 0.
<fs_itab>-agr_name1 = ls_roles-agr_name.
ENDIF.
ENDIF.
IF <fs_itab> IS NOT ASSIGNED.
IF ls_roles-uname = aff_user.
itab-agr_name = ls_roles-agr_name.
ELSE.
itab-agr_name1 = ls_roles-agr_name.
ENDIF.
APPEND itab.
ENDIF.
ENDSELECT.
它应该做你想做的事
我写了下面的 ABAP 程序，它可以从 AGR_USERS 中获取两个用户的角色，这些角色由用户在 selection 屏幕上指定。
输出屏幕应该有 User id 以及分配给用户的角色。程序对一个用户运行良好，但无法为第二个用户获取角色。
REPORT Z_REFERNCE_USER_PROG.
TYPE-POOLS: slis. " SLIS contains all the ALV data types
TYPE-POOLS: SLIS,KCDE.
DATA: username(20) TYPE C,
G_REPID TYPE SY-REPID,
IT_EVENTS TYPE SLIS_T_EVENT,
IT_FIELD TYPE SLIS_T_FIELDCAT_ALV,
WA_FIELD TYPE SLIS_FIELDCAT_ALV,
IT_SORT TYPE SLIS_T_SORTINFO_ALV.
INITIALIZATION.
G_REPID = SY-REPID.
DATA:BEGIN OF itab OCCURS 0,
UNAME type AGR_USERS-UNAME,
AGR_NAME type AGR_USERS-AGR_NAME,
UNAME1 type AGR_USERS-UNAME,
AGR_NAME1 type AGR_USERS-AGR_NAME,
END of itab.
start-of-selection.
SELECT-OPTIONS AFF_USER FOR USERNAME NO INTERVALS.
SELECT-OPTIONS REF_USER FOR USERNAME NO INTERVALS.
end-of-selection.
perform tosql.
perform listshow.
*--------------------------------
* form tosql
*--------------------------------
form tosql.
SELECT DISTINCT AGR_USERS~AGR_NAME AGR_USERS~UNAME
INTO corresponding fields of table itab
FROM AGR_USERS
where AGR_USERS~UNAME in AFF_USER.
SELECT DISTINCT AGR_USERS~AGR_NAME AGR_USERS~UNAME
INTO corresponding fields of table itab
FROM AGR_USERS
where AGR_USERS~UNAME in REF_USER.
endform.
*--------------------------------
* form listshow
*--------------------------------
form listshow.
DEFINE ADD_FIELD.
WA_FIELD-FIELDNAME = &1.
WA_FIELD-REPTEXT_DDIC = &2.
WA_FIELD-NO_ZERO = 'X'.
APPEND WA_FIELD TO IT_FIELD.
END-OF-DEFINITION.
ADD_FIELD 'UNAME' 'Affected user'.
ADD_FIELD 'AGR_NAME' 'Role'.
ADD_FIELD 'UNAME1' 'Reference user'.
ADD_FIELD 'AGR_NAME1' 'Role'.
CALL FUNCTION 'REUSE_ALV_GRID_DISPLAY'"
EXPORTING
I_CALLBACK_PROGRAM = G_REPID
I_BACKGROUND_ID = 'ALV_BACKGROUND'
IT_FIELDCAT = IT_FIELD
* IS_LAYOUT = GS_LAYOUT
* IT_SORT = IT_SORT
I_SAVE = 'A'
IT_EVENTS = IT_EVENTS[]
TABLES
T_OUTTAB = itab
EXCEPTIONS
PROGRAM_ERROR = 1
OTHERS = 2.
endform.
输入：
我想将第一个 SELECT 的所有值放入 table 1 并将第二个 select 的所有值放入 table 2 然后将它们合并到 table 3.我只是想看到相邻的两个用户的角色，顺序无所谓。
SELECT ... INTO 用结果覆盖 table 的内容。所以当你想得到两个SELECTS的结果时，你需要两个tables:
SELECT DISTINCT agr_name, uname
INTO TABLE @DATA(lt_aff)
FROM agr_users
WHERE uname IN @aff_user.
SELECT DISTINCT agr_name, uname
INTO TABLE @DATA(lt_ref)
FROM agr_users
WHERE uname IN @ref_user.
现在如何将这两个结果合二为一table？
问题是您不知道这两个 table 中哪一个会更长。所以我们需要使用一个带有计数器的 WHILE 循环，当计数器小于 either table 的行数时，它会向上计数。然后在该循环中，我们需要使用 READ TABLE 获取对应于计数器当前值的两个 table 的行。如果成功，我们需要将其值移动到输出 table itab 的新行，然后我们将其附加到它：
DATA lv_counter TYPE i.
DATA ls_result LIKE LINE OF itab.
lv_counter = 0.
WHILE lv_counter < lines( lt_aff ) OR lv_counter < lines( lt_ref ).
lv_counter = lv_counter + 1.
CLEAR ls_result.
READ TABLE lt_aff INDEX lv_counter INTO DATA(ls_aff).
IF sy-subrc = 0.
ls_result-uname = ls_aff-uname.
ls_result-agr_name = ls_aff-agr_name.
ENDIF.
READ TABLE lt_ref INDEX lv_counter INTO DATA(ls_ref).
IF sy-subrc = 0.
ls_result-uname1 = ls_ref-uname.
ls_result-agr_name1 = ls_ref-agr_name.
ENDIF.
INSERT ls_result INTO TABLE itab.
ENDWHILE.
实际上您不需要 SELECT-OPTIONS，根据您的任务定义，您只比较两个用户
PARAMETERS: aff_user TYPE xubname DEFAULT 'NEUMANAR',
ref_user TYPE xubname DEFAULT 'KAJAVESH'.
这是在单个数据库 SELECT 和单个 table:
中的方法DATA ls_roles TYPE agr_users.
itab-uname = aff_user.
itab-uname1 = ref_user.
SELECT agr_name uname
FROM agr_users
INTO (ls_roles-agr_name,ls_roles-uname)
WHERE uname = aff_user OR uname = ref_user.
READ TABLE itab ASSIGNING FIELD-SYMBOL(<fs_itab>) WITH KEY agr_name = space.
IF sy-subrc = 0 AND ls_roles-uname = aff_user.
<fs_itab>-agr_name = ls_roles-agr_name.
ELSEIF ls_roles-uname = ref_user.
UNASSIGN <fs_itab>.
READ TABLE itab ASSIGNING <fs_itab> WITH KEY agr_name1 = space.
IF sy-subrc = 0.
<fs_itab>-agr_name1 = ls_roles-agr_name.
ENDIF.
ENDIF.
IF <fs_itab> IS NOT ASSIGNED.
IF ls_roles-uname = aff_user.
itab-agr_name = ls_roles-agr_name.
ELSE.
itab-agr_name1 = ls_roles-agr_name.
ENDIF.
APPEND itab.
ENDIF.
ENDSELECT.
它应该做你想做的事