Cyber-EE
asked on
Find a string in another string
What function can help me find a string in another string.
Example :
string str1 = "i am in my house"
string str2 = "in my"
string str3 = "inmy"
compare_function(str1,str2 ) will return true because str2 was found in str1.
compare_function(str1,str3 ) will return false.
Example :
string str1 = "i am in my house"
string str2 = "in my"
string str3 = "inmy"
compare_function(str1,str2
compare_function(str1,str3
ASKER CERTIFIED SOLUTION
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kaliyugkaarjun, the optional start argument is the first argument, not the third:
InStr([start, ]string1, string2[, compare])
Jon
InStr([start, ]string1, string2[, compare])
Jon
jmundsack is right. Arthur, is too, he should probably look up something so simple, but he did offer up points that he may or may not have payed for so he might as well get a full answer. To make Kaliyugkaarjun's answer more legible and correct:
Function compare_function(byVal str1 As String, str2 As String) As Boolean
compare_function = InStr$(str1, str2) <> 0
End Function
(To Kaliyugkaarjun's: != does not make sense in visual basic and dim i = 0 was not necessary and 'ur' is not a word.)
Cheers :D
Alain
Function compare_function(byVal str1 As String, str2 As String) As Boolean
compare_function = InStr$(str1, str2) <> 0
End Function
(To Kaliyugkaarjun's: != does not make sense in visual basic and dim i = 0 was not necessary and 'ur' is not a word.)
Cheers :D
Alain
alainbryden's apprach is the easiest
-Dean
-Dean
Hey bro just chk out i have written comment and not statement in vb :D
//substring found so do ur required operation
cheers
//substring found so do ur required operation
cheers
Hey bro, the proper way to comment in vb is with ' not //, this isn't java :D
cheers
cheers
ya kalyugkaarjun ... ur comment was good but a bit "Java"ish ...
be happy
be happy
Forget it ...it was just to mention that the line has nothing to do with code...
InStrRev (VB 6 Only)
>Returns the first occurance of one string in another, starting from the right to the left.
>InstrRev(string1, string2[, start[, compare]])
InStr
>Returns the first occurance of one string in another, starting from the left to the right.
>InStr([start, ]string1, string2[, compare])
InStrRev (VB 6 Only)
>Returns the first occurance of one string in another, starting from the right to the left.
>InstrRev(string1, string2[, start[, compare]])
InStr
>Returns the first occurance of one string in another, starting from the left to the right.
>InStr([start, ]string1, string2[, compare])
I'd object, while Arthur_wood gave a partial answer, leading in the right direction, what the user wanted was a working solution, and it was not he, but kaliyugkaarjun (almost) and I that gave a working solution.
InStr(sCheck, sMatch[, Start[, Compare]])
Arguments:
sCheck => Required. String expression being searched.
sMatch =>Required. String expression being searched for.
Start Optional. => Numeric expression that sets the starting position for each search. If omitted, search begins at the first character position (Start = 1).
Compare Optional. => Numeric value indicating the kind of comparison to use when evaluating substrings. If omitted, a binary comparison is performed.
Example:
InStr("abc", "a") => 1
For ur code :
dim i=0;
if Instr(str1,str2) != 0 then
//substring found so do ur required operation
end if
Cheers