该问题已经困惑很长时间了,大神帮忙解决一下
每位员工的满意,不满意数量 一次性统计出来(效果如右图)
[此贴子已经被作者于2021-3-22 11:23编辑过]
[此贴子已经被作者于2021-3-22 11:23编辑过]
[此贴子已经被作者于2021-3-22 22:16编辑过]
[此贴子已经被作者于2021-3-22 22:02编辑过]
CLOSE DATABASES USE pyk IN 0 AFIELDS(lazdm) LOCAL latj[ALEN(lazdm, 1), 4] FOR lnj = 1 TO ALEN(latj, 1) latj[lnj, 2] = 0 latj[lnj, 3] = 0 latj[lnj, 4] = 0 ENDFOR FOR lnj = 1 TO ALEN(lazdm, 1) latj[lnj, 1] = FIELD(lnj) SCAN IF ALLTRIM(EVALUATE(FIELD(lnj))) == "跳过" LOOP ENDIF DO CASE CASE ALLTRIM(EVALUATE(FIELD(lnj))) == "很满意" latj[lnj, 2] = latj[lnj, 2] + 1 CASE ALLTRIM(EVALUATE(FIELD(lnj))) == "满意" latj[lnj, 3] = latj[lnj, 3] + 1 CASE ALLTRIM(EVALUATE(FIELD(lnj))) == "不满意" latj[lnj, 4] = latj[lnj, 4] + 1 ENDCASE ENDSCAN ENDFOR SET ORDER TO CREATE CURSOR tj (部门 C(10), 很满意 N(3), 满意 N(3), 不满意 N(3)) APPEND FROM ARRAY latj BROWSE
[此贴子已经被作者于2021-3-23 16:39编辑过]