Question

我有这张桌子

Antecedent           Consequent
I1                   I2                   
I1                   I1,I2,I3
I1                   I4,I1,I3,I4
I1,I2                I1
I1,I2                I1,I4
I1,I2                I1,I3
I1,I4                I3,I2
I1,I2,I3             I1,I4
I1,I3,I4             I4

你可以看到它很乱。无论如何我可以删除行，如果结果中的项目存在于先行（1行）

例如： INPUT：

Antecedent           Consequent
I1                   I2                   
I1                   I1,I2,I3    <---- DELETE since I1 exist in antecedent
I1                   I4,I1,I3,I4 <---- DELETE since I1 exist in antecedent
I1,I2                I1          <---- DELETE since I1 exist in antecedent
I1,I2                I1,I4       <---- DELETE since I1 exist in antecedent
I1,I2                I1,I3       <---- DELETE since I1 exist in antecedent
I1,I4                I3,I2       
I1,I2,I3             I1,I4       <---- DELETE since I1 exist in antecedent
I1,I3,I4             I4          <---- DELETE since I4 exist in antecedent

输出：

Antecedent           Consequent
I1                   I2  
I1,I4                I3,I2

无论如何我可以通过查询来做到这一点吗？

Answer 1

我看不出如何通过没有用户定义函数的Access查询来完成这项工作，这表明可以更简单地使用VBA删除问题行。

Answer 2

Rico，每个人都对UDF是对的，用查询做这件事可能会慢一点。将此功能添加到模块，看看它有多快。我在几秒钟内在500K记录上运行它。

Public Function IsWithin(ByVal vAntecedent As Variant, ByVal vConsequent As Variant) As Boolean
    On Error GoTo ErrQuit
    IsWithin = False
    If Not IsNull(vAntecedent) Then
        If Not IsNull(vConsequent) Then
            Dim aryAnt As Variant, aryCons As Variant
            Dim vAnt  As Variant, vCons  As Variant
            aryAnt = Split(vAntecedent, ",")
            aryCons = Split(vConsequent, ",")
            For Each vAnt In aryAnt
                For Each vCons In aryCons
                    If vAnt = vCons Then
                        IsWithin = True
                        Exit For
                    End If
                Next
                If IsWithin Then Exit For
            Next

        End If
    End If
    Exit Function
ErrQuit:
    IsWithin = False
End Function

您的查询可能如下所示：

SELECT Antecedent, Consequent
FROM Table1
WHERE IsWithin([Antecedent],[Consequent])=False;

查询以查找2表中的重复项

2 个答案: