- Foxtable(狐表) (http://foxtable.com/bbs/index.asp)
-- 专家坐堂 (http://foxtable.com/bbs/list.asp?boardid=2)
---- (请教)帮助修改1下代码 (http://foxtable.com/bbs/dispbbs.asp?boardid=2&id=1297)
-- 发布时间：2008/12/1 22:04:00
-- (请教)帮助修改1下代码
正确的结果应是图1：
此主题相关图片如下：未命名.jpg
文件：
| 下载信息 [文件大小： 下载次数： ] | |
 点击浏览该文件:管理项目4.table |
错误的情况：
此主题相关图片如下：未命名3.jpg
-- 发布时间：2008/12/1 22:04:00
--
单独生成的代码：
\'\'\'
dim dackj as string
With tables("表A")
Dim ba As Row = .current
for each biaoc as datarow in datatables("表C").select("[编号] = \'"& ba("编号") &"\'")
if biaoc("结果") = "检出" then
dackj = dackj & "," & "编号" & biaoc("序号")
.current("结论") = dackj & "不合格"
end if
next
End With
-- 发布时间：2008/12/1 22:08:00
--
全部生成的代码：（本来的思路是希望移动1行，就生成1行的结论，但没有想到：是移动，但会加上上一行的结论）
\'\'\'
dim dackj as string
With tables("表A")
for i as integer = 0 to .count
.Position = i
Dim ba As Row = .current
for each biaoc as datarow in datatables("表C").select("[编号] = \'"& ba("编号") &"\'")
if biaoc("结果") = "检出" then
dackj = dackj & "," & "编号" & biaoc("序号")
.current("结论") = dackj & "不合格"
end if
next
next
End With
-- 发布时间：2008/12/1 22:17:00
--
请帮助修写，或重新写也行(能实现需要的结果就行了，又是试了好久不成功）
-- 发布时间：2008/12/1 23:36:00
--
终于找到1种可以实现的方法（用计算代码）
但想不通，（以下的代码也不行，晕中....）
\'\'\'
With tables("表A")
for i as integer = 0 to .count
.Position = i
dim dackj as string
with tables("表A.表C")
For r As integer = 0 To .Rows.Count - 1
if .rows(r)("结果") = "检出" then
dackj = dackj & "," & "序号" & .rows(r)("序号")
tables("表A").current("结论") = dackj & "不合格"
end if
next
end with
next
End With
-- 发布时间：2008/12/2 0:35:00
--
\'\'\'
dim dr1,dr2 as datarow
dim bj,dackj as string
dim dt1,dt2 as datatable
dt1 = datatables("表A")
dt2 = datatables("表C")
for Each dr1 in dt1.select("[编号] >\'\'")
bj = "[编号] = \'" & dr1("编号")
bj = bj & "\'And [结果] = \'检出\'"
if dt2.Find(bj) Is Nothing then
dr1("结论") = "全部合格"
Else
for each dr2 in Dt2.select(bj)
dackj = dackj & "," & "编号" & dr2("序号")
next
dr1("结论") = dackj.substring(1) & "不合格"
End If
next
-- 发布时间：2008/12/2 0:57:00
--
可能我说得不清楚，我要的是1楼图1的结果：
如20080001 我要的是1，38不合格
200800002 我要的是47不合格 （而不是要1，38，47不合格。这是2008001和20080002二个结果合并了）
200800003 我要的是53不合格 （而不是要1，38，47，53不合格，这是20080001、20080002、20080003三个结果的合并了）
-- 发布时间：2008/12/2 8:57:00
--
\'\'\'
dim dr1,dr2 as datarow
dim bj,dackj as string
dim dt1,dt2 as datatable
dt1 = datatables("表A")
dt2 = datatables("表C")
for Each dr1 in dt1.select("[编号] >\'\'")
bj = "[编号] = \'" & dr1("编号")
bj = bj & "\'And [结果] = \'检出\'"
dackj = ""
for each dr2 in Dt2.select(bj)
dackj = dackj & "," & "编号" & dr2("序号")
next
dr1("结论") = dackj.substring(1) & "不合格"
Next
-- 发布时间：2008/12/2 12:44:00
--
先顶1下，8楼符合要求.....
没有想到6楼知我想要的，写法应是这个：
\'\'\'
dim dr1,dr2 as datarow
dim bj,dackj as string
dim dt1,dt2 as datatable
dt1 = datatables("表A")
dt2 = datatables("表C")
for Each dr1 in dt1.select("[编号] >\'\'")
bj = "[编号] = \'" & dr1("编号")
bj = bj & "\'And [结果] = \'检出\'"
dackj = ""
for each dr2 in Dt2.select(bj)
dackj = dackj & "," & "编号" & dr2("序号")
next
if dackj <> "" then
dr1("结论") = dackj.substring(1) & "不合格"
else
dr1("结论") = "全部合格"
end if
Next
-- 发布时间：2008/12/2 12:45:00
--
流程加多了1个 dackj = "" 才能实现.....请问它的作用是？