blocks|key|5435|text|使用上面ron's+function的提示，我创建了这个公式，它运行得很好：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|5436|=LEN(A1)+-+LEN(SUBSTITUTE(A1,+"-",+""))|code-block|syntax|javascript|5437|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/21372437/4074148^0|4|E|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Using hint from <a href="https://stackoverflow.com/a/21372437/4074148">ron's function</a> above I've created this formula and it worked fine :

<pre><code>=LEN(A1) - LEN(SUBSTITUTE(A1, "-", ""))
</code></pre>

blocks|key|2470562|text|这将计算活动单元中连字符的数量。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2470563|Sub+test()
++++a+=+Len(ActiveCell)
++++my_txt+=+Replace(ActiveCell,+"-",+"",+1,+-1,+vbTextCompare)
++++b+=+Len(my_txt)
++++numb_occur+=+a+-+b
End+Sub|code-block|syntax|javascript|2470564|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This will count the number of hyphens in the activecell

<pre><code>Sub test()
 a = Len(ActiveCell)
 my_txt = Replace(ActiveCell, "-", "", 1, -1, vbTextCompare)
 b = Len(my_txt)
 numb_occur = a - b
End Sub
</code></pre>

blocks|key|5448|text|我找到了这个答案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5449|Sub+xcountCHARtestb()+
+++++'If+countCHAR(RANGE("aq528"),+".")+>+0+Then++++'YES
++++If+countCHAR(Selection,+".")+>+0+Then+'YES
++++++++MsgBox+"YES"+&+Space(10),+vbQuestion+',+"title"
++++Else+
++++++++MsgBox+"NO"+&+Space(10),+vbQuestion+',+"title"
++++End+If+
End+Sub+


Sub+xcountCHARtesta()+'YES
++++MsgBox+"There+are+"+&+countCHAR(Selection,+"test")+&+"+repetitions+of+the+character+string",+vbQuestion+'YES
End+Sub+

Function+countCHAR(myString+As+String,+myCHAR+As+String)+As+Integer+'as:+If+countCHAR(Selection,+".")+>+1+Then++selection+OR+RANGE("aq528")+'"any+char+string"
++++countCHAR+=+UBound(split(myString,+myCHAR))+'YES
End+Function+|code-block|syntax|javascript|5450|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I found this answer:

<pre><code>Sub xcountCHARtestb() 
 'If countCHAR(RANGE("aq528"), ".") &gt; 0 Then 'YES
 If countCHAR(Selection, ".") &gt; 0 Then 'YES
 MsgBox "YES" &amp; Space(10), vbQuestion ', "title"
 Else 
 MsgBox "NO" &amp; Space(10), vbQuestion ', "title"
 End If 
End Sub 


Sub xcountCHARtesta() 'YES
 MsgBox "There are " &amp; countCHAR(Selection, "test") &amp; " repetitions of the character string", vbQuestion 'YES
End Sub 

Function countCHAR(myString As String, myCHAR As String) As Integer 'as: If countCHAR(Selection, ".") &gt; 1 Then selection OR RANGE("aq528") '"any char string"
 countCHAR = UBound(split(myString, myCHAR)) 'YES
End Function 
</code></pre>

blocks|key|5410|text|这段代码可能对你有帮助。你也可以用它作为一个UDF..。:)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5411|Function+CountHypens(rng_Src+As+Range)+As+Long

'A+VARIANT+FOR+SPLITTING+CELL+CONTENTS
Dim+var+As+Variant

On+Error+Resume+Next
var+=+Split(rng_Src.Value,+"-",+,+vbTextCompare)

If+Err.Number+<>+0+Then
++++Debug.Print+"This+cell+does+not+have+any+hyphens."
Else
++++CountHypens+=+UBound(var)
++++End+If
Err.Clear:+On+Error+GoTo+0

End+Function|code-block|syntax|javascript|5412|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This code might be of your help .. you can also use it as a UDF... :)

<pre><code>Function CountHypens(rng_Src As Range) As Long

'A VARIANT FOR SPLITTING CELL CONTENTS
Dim var As Variant

On Error Resume Next
var = Split(rng_Src.Value, "-", , vbTextCompare)

If Err.Number &lt;&gt; 0 Then
 Debug.Print "This cell does not have any hyphens."
Else
 CountHypens = UBound(var)
 End If
Err.Clear: On Error GoTo 0

End Function
</code></pre>

blocks|key|4066761|text|跟进:吊艇架，用吊艇架。:)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4066762|我一直在寻找一种方法来测试公式中的查找文本字符串。这个答案似乎对两种公式都是正确的/不适用于一条线。(我还是vba的新手，请告诉我是否有更好的方法)谢谢。|4066763|If+countChar(UCase(Selection.Formula),+UCase("offset"))+>+0+Then+++'YES+(thee?+answer,+works+for+both+formulas+/+not)
'If+countChar(Selection.Formula,+"OFFSET")+>+0+Then+++'yes
'If+countChar(Cells(ActiveCell.row,+Selection.Column).Formula,+"OFFSET")+>+0+Then+++'yes
'If+countChar(Cells(ActiveCell.row,+"BG").Formula,+"OFFSET")+>+0+Then+++'yes
'If+countChar(UCase(Selection),+UCase("OffSET"))+>+0+Then++++'yes+but+not+work+on+formula
'If+Selection.Formula+Like+"*offset*"+Then+++'no+(for+eq)
MsgBox+"YES"+&+Space(15),+vbQuestion
Else
MsgBox+"NO"+&+Space(15),+vbQuestion
End+If|code-block|syntax|javascript|4066764|注:以上变量"+BG“，我使用永久工作单元来改进对列BG的使用，工作单元格A3有/显示:+BG:BG。|4066765|=SUBSTITUTE(SUBSTITUTE(CELL("address",$BG3),"$",""),ROW(),"")&":"&SUBSTITUTE(SUBSTITUTE(CELL("address",$BG3),"$",""),ROW(),"")|4066766|您还需要在vba的顶部/之前使工作单元格变暗：|4066767|Dim+A3+As+String
A3+=+RANGE("A3")|4066768|对不起，试了三次把所有的代码都放进了一个盒子里。实际上，建议在工具栏中放置一个代码停止开始图标。|4066769|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

Follow up to: davex, by davex.. :)

I had been looking all over trying to find a way to test same for find text string in a formula.
This answer seems to work correctly for both formulas / not &amp; fits in a 1 liner..
 (am still pretty novice at vba, let me know if any better way(s) ) thanks.

<pre><code>If countChar(UCase(Selection.Formula), UCase("offset")) &gt; 0 Then 'YES (thee? answer, works for both formulas / not)
'If countChar(Selection.Formula, "OFFSET") &gt; 0 Then 'yes
'If countChar(Cells(ActiveCell.row, Selection.Column).Formula, "OFFSET") &gt; 0 Then 'yes
'If countChar(Cells(ActiveCell.row, "BG").Formula, "OFFSET") &gt; 0 Then 'yes
'If countChar(UCase(Selection), UCase("OffSET")) &gt; 0 Then 'yes but not work on formula
'If Selection.Formula Like "*offset*" Then 'no (for eq)
MsgBox "YES" &amp; Space(15), vbQuestion
Else
MsgBox "NO" &amp; Space(15), vbQuestion
End If
</code></pre>

NOTE: in place of variable "BG" above, i use permanent work cells to improve use for column BG example, work cell A3 has / shows: BG:BG

<pre><code>=SUBSTITUTE(SUBSTITUTE(CELL("address",$BG3),"$",""),ROW(),"")&amp;":"&amp;SUBSTITUTE(SUBSTITUTE(CELL("address",$BG3),"$",""),ROW(),"")
</code></pre>

you will also need to dim the work cell, at the top / before the vba:

<pre><code>Dim A3 As String
A3 = RANGE("A3")
</code></pre>

pardon, tried 3 times to get all of code into 1 box. really suggest putting a code stop start icon in the toolbar.

I have a number of items in cells that are separated by dashes. I'm trying to normalize the database by splitting rows so that each row contains only one entry. How do you find/count strings in Excel VBA. I know you can do values for whole cells with

<pre><code>myVar = Application.WorksheetFunction.COUNTIF(Range("A1:Z100"),"Value")
</code></pre>

I need to search a single cell and find out how many hyphens there are. Example

<pre><code>123-456-789 = 2
9876-12 = 1
</code></pre>

How to Count the Number of a Specific Character in a Cell with Excel VBA

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我在单元格中有许多用破折号分隔的项目。我试图通过拆分行来规范数据库，以便每行只包含一个条目。如何在Excel中查找/计数字符串。我知道你可以为整个细胞做值myVar = Application.WorksheetFunction.COUNTIF(Range("A1:Z100"),"Value")我需要搜索一个单元，找出有多少连字符。示例123-456-789 = 29876-12 = 1

问如何使用Excel计算单元格中特定字符的数目
EN

回答 5

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