So I have
Dim str1
str1 = "Cat"
Dim str2
str2 = "concatenate"
I wanted to know is there a way I can match str1 with str2 and return positive non-zero number if there is match (case-insensitive) for str1 in str2 ?
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1This question has already been asked ! stackoverflow.com/questions/15585058/…titus– titus2015-09-22 20:59:14 +00:00Commented Sep 22, 2015 at 20:59
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Possible duplicate of Check if a string contains another stringniton– niton2017-03-18 00:29:17 +00:00Commented Mar 18, 2017 at 0:29
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3 Answers
For VBA, The instr is the way to go:
InStr(1, str2, str1, vbTextCompare)
The first part is where it starts looking, in this case the first character.
The second is the string in which the search is happening.
The third is the search parameter.
The fourth determines whether it looks at the text of the binary value. So for a case-insensitive we use vbtextcompare.
5 Comments
AriKari
will this work if str1 and str2 are arrays with strings and str1(i) compares for a match with all str2(j) ?
Scott Craner
No. There are many ways to do it, loops and dictionaries, depending on what is being searched. But that is not what you asked.
AriKari
which variable will return a positive value if match is found ?
Scott Craner
The result of the function that is being called will result in a positive number if found and a zero if not. So for your example the result would be 4. You would create a variable and set it = to the function above.
WltrRpo
To work with an array you have to loop in and compare each position.
Use the indexOf method:
str2.indexOf(str1)
4 Comments
AriKari
will this work if str1 and str2 are arrays with strings and str(i) compares for a match with all str(j) ?
ergonaut
no, neither will any of the other answers. You'll need to loop.
AriKari
Yeah sure I will have to loop but it will work for arrays too right
ergonaut
indexing an array returns the object at a particular location. If that object is a string, it will act like one because it is one.
Use instr or instrrev:
instr(str2, str1, 1)
See msdn for more info. This does exactly what you need.
1 Comment
Scott Craner
I may be wrong but this is for vb not vba, which is what the OP asked.