2013-03-14 132 views
2
Private Sub createSequenceDataFiles_Sequence(sPath As String, sTableName As String, iSeqNo As Integer) 
     Dim RST As DAO.Recordset 
     Dim nIL As Long, nLastIL As Long 
     Dim sSQL As String 
     ' 
     ' Get the data from the table 
     ' 
     sSQL = _ 
    " SELECT [“ & sTableName & “].IL, “ & _ 
    “ [XL]-[FIRST_XL]+1 AS XL_IDX, “ & _ 
    “ [“ & sTableName & “] “ & _ 
    “ FROM [“ & sTableName & “] “ & _ 
    “ INNER JOIN TBL_FIRST_XL_FOR_IL “ & _ 
    “ ON [“ & sTableName & “].IL = TBL_FIRST_XL_FOR_IL.IL “ & _ 
    “ ORDER BY [“ & sTableName & ”].IL, [XL]-[FIRST_XL]+1” 
     Set RST = CurrentDb.OpenRecordset(sSQL) 
     While Not RST.EOF 
      nIL = RST("IL").Value 
      If Not (nLastIL = nIL) Then 
       ' 
       ' If we've already done one, close the file 
       ' 
       If nLastIL > 0 Then Close #1 
       ' 
       ' Open the file for the current in-line 
       ' 
       Open sPath & "\WBS1\IL_" & Format(nIL, "0000") & ".pks" For Append As 1 
      End If 
      ' 
      ' Write the data 
      ' 
      Write #1, iSeqNo, RST(“XL_IDX”).value + 1, RST("TIME").Value 
      nLastIL = nIL 
      RST.MoveNext 
     Wend 
     ' 
     ' Close 
     ' 
     Set RST = Nothing 
     Close 
    End Sub 

我收到此错误Visual Basic Access:参数太少。预计1

是不是有什么问题呢?

这是整个代码现在我很确定它是SQL位,这是问题,也许我的表名是错误的。

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您确定导致错误的代码吗?你能发布更多的代码吗?什么是'sTableName'内容? – 2013-03-14 05:22:43

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我不完全相信我已经添加了更多的代码 – 2013-03-14 05:35:44

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这里是你的公式生成的SQL(使用“aTable”作为表名):'SELECT [aTable] .IL,[XL] - [FIRST_XL ] +1 AS AS XL_IDX,** [aTable] ** FROM [aTable] INNER JOIN TBL_FIRST_XL_FOR_IL ON [aTable] .IL = TBL_FIRST_XL_FOR_IL.IL ORDER BY [aTable] .IL,** [XL] - [FIRST_XL] + 1 * *'。 “**”中的项目在语句中被引用为字段,但它们不可以,对吗? – chuff 2013-03-14 05:40:28

回答

4

SQL查询中的表和/或字段名与您的数据库结构不匹配。

这包括sTableName参数中的任何值。

只要检查并确保所有的字段完全匹配,你应该找到罪魁祸首。

编辑:只需在SELECT部分​​注意到这个

“ [“ & sTableName & “] “ & _ 

你不应该引用整个表在发言的SELECT部分,如果从SQL语句删除此行发生了什么?

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在这部分之后有一个字段名称丢失,非常感谢您的帮助 – 2013-03-14 07:26:02

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很高兴为您提供帮助。自从解决了你的问题以后,接受答案?我需要我的互联网点哈哈 – 2013-03-14 07:38:23

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