Solved: How to count occurences of a string in a doc?

[RonMcK]

Hi, All,

I need some help getting the following shippet of code to count the number of instances of a string in a document. My test document has 4 instances of the string scattered through it, but my code finds none of them because .FOUND is returning False.

Sub CountAWord()
    Dim intRowCount As Long     ' was integer
    Dim aRange As Range
    Dim response As Long
    intRowCount = 0
    Set aRange = ActiveDocument.Range
    With aRange.Find
        Do
            .Text = "ID3" ' the word I am looking for
Debug.Print aRange.Find.Text
Debug.Print ActiveDocument.Name
Debug.Print .Found
            .Execute
            If .Found Then
                intRowCount = intRowCount + 1
            End If
        Loop While .Found
    End With
    response = MsgBox("I counted " & Str(intRowCount) & vbCrLf & " in doc: " & ActiveDocument.Name, vbOKOnly, "Number of Instances")
    Set aRange = Nothing
End Sub

Is there an easier way, than the above code, to count the number of occurrences of a keyword (or phrase or string) in a document?

Thanks!

[RonMcK]

Here is a solution to my counting question that works better, based on work by Malcolm (mdmackillop) in the [thread=http://vbaexpress.com/forum/showthread.php?t=21214]Excel forum[/thread] and an assist he got from here.

Sub CountSpecificStringInWordDoc2()
    Dim txt As String, Lgth As Long, Strt As Long
    Dim i As Long
    Dim oRng As Range
    Dim Tgt As String
    Dim arr()
    ReDim arr(10)
    Dim response As Long
'Set parameters
    txt = InputBox("String to find")
    Lgth = InputBox("Length of string to return")
    Strt = Len(txt)
'Return data to array
    With Selection
   .HomeKey unit:=wdStory
      With .Find
         .ClearFormatting
         .Forward = True
         .Text = txt
         .Execute
         While .Found
         arr(1) = arr(1) + 1
         Set oRng = ActiveDocument.Range _
         (Start:=Selection.Range.Start + Strt, _
         End:=Selection.Range.End + Lgth)
         oRng.Start = oRng.End
         .Execute
         Wend
      End With
    End With
response = MsgBox("I found: " & Str(arr(1)) & " instances of the string: " & txt & _
                vbCrLf & " in the document: " & ActiveDocument.Name, vbOKOnly, "Number of occurences")
End Sub

I'm still curious to know if there is a more elegant way to get the count of the number of occurences of a string in a Word doc.

Thanks!

[mdmackillop]

I'm still curious to know if there is a more elegant way to get the count of the number of occurences of a string in a Word doc.

How about using Find/Replace? Find "Text" Replace "^&"
See here and here

[mdmackillop]

I came across this methodology. It will count strings, but also as part of words.

 
Option Compare Text
Sub CountOccurences()
    Dim MyDoc As String, txt As String, t As String
MyDoc = ActiveDocument.Range.Text
txt = InputBox("Text to find")
    t = Replace(MyDoc, txt, "")
    MsgBox (Len(MyDoc) - Len(t)) / Len(txt) & " occurrences of " & txt
End Sub

[MWE]

here is some brute force code that is simplier and seems to work. It does not differentiate between "words" and text string.

    Sub CountWords()
        Dim I           As Long
        Dim J           As Long
        Dim Num         As Long
        Dim TargetText  As String
TargetText = InputBox("target text?")
        J = 1
        I = 1
        While I > 0
   I = InStr(J, ActiveDocument.Range.Text, TargetText)
   If I > 0 Then
      Num = Num + 1
      J = I + 1
   End If
        Wend
        MsgBox Num
    End Sub

[fumei]

Or...

Dim r As Range
Dim j As Long
Set r = ActiveDocument.Range
With r.Find
   .Text = InputBox("What word(s)?")
   Do While .Execute(Forward:=True) = True
       j = j + 1
    Loop
End With
MsgBox "Given word(s) was found " & j & " times."

No error trapping on the input string. If there is no string found (because of spelling errors for example) the message will be "0 times"

[nim81]

Here is a simple way to do it, using the Split() function, where "Needle" is the word/string you want to count:

Dim Haystack As String
Dim Needle As String

Dim a As Variant

a = Split(Haystack, Needle)

MsgBox UBound(a)

[RonMcK]

Here is a simple way to do it, using the Split() function, where "Needle" is the word/string you want to count:

Dim Haystack As String
Dim Needle As String
Dim a As Variant
a = Split(Haystack, Needle)
MsgBox UBound(a)

Thanks for the suggestion. Not stated in my query is the requirement that the code run on Excel 2004 in Office for Mac where we have no 'split' function.

[fumei]

There is no Split() in Excel for the Mac? Ai caramba! Why on earth would it be different for the Mac?