Excel在＆＃39; x＆＃39;之前和之后替换字符串中的字符

Question

您好我有一个包含字符串（产品名称）的列。 现在这些格式为名称LenghtxWidth ，例如绿色框20x30。现在我需要在这个例子中用30改变20，所以我得到绿盒30x20，任何想法我怎么能实现这个？ 感谢

Answer 1

以下是公式解决方案，以及使用正则表达式的VBA解决方案：

<强>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)

<强> 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

以下是我测试过的各类数据的示例：

通过查看最后一个以空格分隔的子字符串LxW，然后在x之后和之前反转该部分，然后将所有内容连接在一起，该公式工作。

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 - 尽管可以根据需要更改），然后将它们反转。

以下是正则表达式（和替换字符串）的更详细说明，其中包含指向教程的链接：

（\ d +。？\ d *）×（\ d +。？\ d *）

(\d+.?\d*)x(\d+.?\d*)

选项：不区分大小写; ^ $不匹配的换行

• Match the regex below and capture its match into backreference number 1 (\d+.?\d*)
  • Match a single character that is a “digit” \d+
    • Between one and unlimited times, as many times as possible, giving back as needed (greedy) +
  • Match any single character that is NOT a line break character .?
    • Between zero and one times, as many times as possible, giving back as needed (greedy) ?
  • Match a single character that is a “digit” \d*
    • Between zero and unlimited times, as many times as possible, giving back as needed (greedy) *
• Match the character “x” literally x
• Match the regex below and capture its match into backreference number 2 (\d+.?\d*)
  • Match a single character that is a “digit” \d+
    • Between one and unlimited times, as many times as possible, giving back as needed (greedy) +
  • Match any single character that is NOT a line break character .?
    • Between zero and one times, as many times as possible, giving back as needed (greedy) ?
  • Match a single character that is a “digit” \d*
    • Between zero and unlimited times, as many times as possible, giving back as needed (greedy) *

2×$ $ 1

• Insert the text that was last matched by capturing group number 2 $2
• Insert the character “x” literally x
• Insert the text that was last matched by capturing group number 1 $1

使用RegexBuddy

创建

Answer 2

好的，所以使用VBA真的很容易，但是如果你只想要一些公式，你可以使用一些列来分割文本，然后连接你的单元格。

这是一个小例子：

当然B1-4是可选的。这里只有更具可读性的东西，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))

如果您的名字中有多个空格，则可以使用此公式搜索文本中的最后一个空间

=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))

Answer 3

以下是适用于您的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VBE (Visual Basic Editor)的常规模块中，并将其用作常规函数/公式。如果您的值在单元格A1中，则在单元格B1中键入=Switch(A1)。希望它有所帮助！