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Parsing Exact Matches In A Text File.

How do you find exact matches in string using VB6?

Example: "This is the best site in the internet"

I need to find the word "in" so it should only match in position 24. As you can see we have another position also in the word "internet". Using Instr we return two matches not one.

Thanks
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yababy
Asked:
yababy
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1 Solution
 
DrDelphiCommented:
Well, the easiest way if you are looking to get a full word match and not a substring would be to pad yourd search with a leading a trailing space:


For example:

ret = InStr("This is the best site in the internet", " in ")

Good luck!!


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yababyAuthor Commented:
Yes this would work by what if the word is at the beginning or end of a string. Does VB have any other string functions?

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okcmanCommented:
Private Sub Command1_Click()
Dim X As String
Dim Y As Integer

X = "This is the intro day we are in into town"
Y = InStr(X, "in ")
Text1.Text = Y
End Sub


In this case "in" is not just "in"
but has a space before it and a space after it so rather just to search for  in by itself we search for "in "
which will return the correct position
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RuchiCommented:
You can do what Dr. Delphi says.

Another method:

ret = InStr(1, "This is the best site in the internet", "in")

In this example, 1 indicates that the search for "in" is to begin at position 1 of "This is the best site in the internet".
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DragonWolfCommented:
Use,

ret = InStr(" " & "This is the best site in the internet" & " ", " in ")

This will look for it at the begining and end as well as in the middle...

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DragonWolfCommented:
Use,

ret = InStr(" " & "This is the best site in the internet" & " ", " in ")

This will look for it at the begining and end as well as in the middle...

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DrDelphiCommented:
or:

Mystr = "dog My  has Fleas"
ret = InStr(Mystr, "dog")
If ret = 1 Then
  MsgBox ("begining of string")
Else

If ret = (Len(Mystr) - 2) Then
  MsgBox ("Last word")
Else
If ((ret > 0) And ret < Len(Mystr)) Then
  MsgBox ("somewhere in the middle")
End If
End If
End If
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caraf_gCommented:
Use the following routine:

Private Sub FindExactMatch(ByVal strToFind As String, ByVal strToSearch As String, intPos As Integer)

Dim strTest As String
Dim blnValid As Boolean

intPos = InStr(intPos + 1, strToSearch, strToFind)

Do While intPos <> 0
    blnValid = True
    If intPos <> 1 Then
        strTest = Mid(strToSearch, intPos - 1, 1)
        If strTest >= "a" And strTest <= "z" Then
            blnValid = False
        End If
        If strTest >= "A" And strTest <= "Z" Then
            blnValid = False
        End If
    End If
    If intPos <> Len(strToSearch) Then
        strTest = Mid(strToSearch, intPos + Len(strToFind), 1)
        If strTest >= "a" And strTest <= "z" Then
            blnValid = False
        End If
        If strTest >= "A" And strTest <= "Z" Then
            blnValid = False
        End If
    End If
    If blnValid Then
        Exit Sub
    End If
    intPos = InStr(intPos + 1, strToSearch, strToFind)
Loop

End Sub


Let's say you're searching the following string

"The internet is in use by all the countries it is implemented in."

for the string

"in"

It will find "in" in "is in use", and "in" in "implemented in."

But it won't find internet, as the "in" in internet is followed by an alphabetic character.

A possible use:

Dim intPos As Integer

FindExactMatch Text1.Text, Text2.Text, intPos
Do While intPos <> 0
    MsgBox intPos
    FindExactMatch Text1.Text, Text2.Text, intPos
Loop
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caraf_gCommented:
Dragonwolf - I purposely used a sentence that ended in "in." to illustrate why your method won't work.
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caraf_gCommented:
PS - if your language can contain accented characters, include them in the if statements. That extension of my code is trivial, so I've left that out for you to do.
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yababyAuthor Commented:
Thanks, but I will be using a new object called RegExp (VBScript Regular Expressions) which works great for finding exact matches.

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