您好,我有一列包含字符串(产品名称)。现在这些被格式化为Name LenghtxWidth,例如绿色框 20x30。现在我需要在这个例子中将 20 更改为 30,这样我就得到了 30x20 的绿框,有什么想法可以实现这一点吗?谢谢
3 回答
4
这是一个公式解决方案,以及一个使用正则表达式的 VBA 解决方案:
公式
=LEFT(A1,FIND(TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)),A1)-1)&
MID(TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)),SEARCH("x",TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)))+1,99)&
"x"&
LEFT(TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)),SEARCH("x",TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)))-1)
UDF
Option Explicit
Function RevWL(S As String)
Dim RE As Object
Const sPat As String = "(\d+.?\d*)x(\d+.?\d*)"
'If L or W might start with a decimal point, and not a digit,
'Then change sPat to: (\d*.?\d+)x(\d*.?\d+)
Set RE = CreateObject("vbscript.regexp")
With RE
.Global = True
.ignorecase = True
.Pattern = sPat
RevWL = .Replace(S, "$2x$1")
End With
End Function
这是我测试的数据类型的示例:
该公式的工作原理是查看最后一个以空格分隔的子字符串,即LxW
,然后反转 之前和之前的部分x
,然后将所有内容连接在一起。
正则表达式模式捕获两个数字(可以是整数或小数,只要以整数开头 - 尽管如果需要可以更改),并将它们反转。
以下是正则表达式(和替换字符串)的更详细说明,并附有教程链接:
(\d+.?\d*)x(\d+.?\d*)
(\d+.?\d*)x(\d+.?\d*)
选项:不区分大小写;^$ 在换行符处不匹配
- 匹配下面的正则表达式并将其匹配捕获到反向引用编号 1
(\d+.?\d*)
- 从字面上匹配字符“x”
x
- 匹配下面的正则表达式并将其匹配捕获到反向引用编号 2
(\d+.?\d*)
$2x$1
于 2016-08-09T12:16:08.910 回答
2
好的,所以使用 VBA 确实更容易,但如果你只想要一些公式,你可以使用一些列来分割你的文本,然后连接你的单元格。
当然 B1-4 是可选的。它在这里只是为了有更易读的东西,但你只能使用一个公式
=CONCATENATE(LEFT(A1, SEARCH(" ",A1,1)-1)," ",RIGHT(RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)),LEN(RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)))-SEARCH("x",RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)),1)),"x",LEFT(RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)), SEARCH("x",RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)),1)-1))
如果你的名字中有多个空格,你可以使用这个公式来搜索文本中的最后一个空格
=CONCATENATE(LEFT(A1, SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))-1)," ",RIGHT(RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))),LEN(RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))))-SEARCH("x",RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))),1)),"x",LEFT(RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))), SEARCH("x",RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))),1)-1))
于 2016-08-09T12:02:13.167 回答
2
这是一个适合您的 VBA 解决方案:
Option Explicit
Function Switch(r As Range) As String
Dim measurement As String
Dim firstPart As String
Dim secondPart As String
measurement = Right(r, Len(r) - InStrRev(r, " "))
secondPart = Right(measurement, Len(measurement) - InStr(1, measurement, "x"))
firstPart = Left(measurement, InStr(1, measurement, "x") - 1)
Switch = Left(r, InStrRev(r, " ") - 1) & " " & secondPart & "x" & firstPart
End Function
您可以将其粘贴到常规模块中VBE (Visual Basic Editor)
并将其用作常规函数/公式。如果您的值在单元格 A1 中,则=Switch(A1)
在单元格 B1 中键入。希望能帮助到你!
于 2016-08-09T12:03:19.710 回答