匹配功能？VBA 无法正常工作
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Match function? VBA not working properly
提问by user2703472
I have a following code and i have "name" in cells B11 , B19 and B25 and i was expecting a result of 19 or 25 with the code but it returns Y=1. Could anyone tell me what is the problem?.
我有以下代码，并且在单元格 B11、B19 和 B25 中有“名称”，我期望代码的结果为 19 或 25，但它返回 Y=1。谁能告诉我是什么问题？
Sub mat()
Y = Application.Match("name", ActiveSheet.Range("B19:B30"), 0)
MsgBox Y
End Sub
回答by jpw
According to the documentation for MATCH:
MATCH returns the position of the matched value within lookup_array, not the value itself.
MATCH 返回匹配值在 lookup_array 中的位置，而不是值本身。
and with 0as the optional third argument (match_type):
并0作为可选的第三个参数 ( match_type)：
If match_type is 0, MATCH finds the first value that is exactly equal to lookup_value. Lookup_array can be in any order.
如果 match_type 为 0，MATCH 会找到与 lookup_value 完全相等的第一个值。Lookup_array 可以按任何顺序排列。
So the returned 1refers to the position on B19in the array Range("B19:B30")and the code sample is indeed behaving as expected.
所以返回的1指的B19是数组中的位置，Range("B19:B30")代码示例确实按预期运行。
回答by Mark Fitzgerald
Application.Match("name", ActiveSheet.Range("B19:B30"), 0)
The MATCH function searches for a specified item in a range of cells, and then returns the relative positionof that item in the range
MATCH 函数在单元格范围内搜索指定的项目，然后返回该项目在该范围内的相对位置
So parsing the parameter ActiveSheet.Range("B19:B30")means that B19equals relative position=1.
所以解析参数ActiveSheet.Range("B19:B30")意味着B19等于相对位置=1。
回答by ayadprof
add slgn
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Sub mat()
Y = Application.Match(slng(range("a4").value), ActiveSheet.Range("B19:B30"), 0)
MsgBox Y
End Sub
ex2:
例2：
Application.WorksheetFunction.Match(CLng(TextBox1.Text), sheet110.Range("B6:B" & ls), 0)
