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InStr |
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| message from gr on 4 Jun 2004 |
intInString = VBA.InStr(aAvailable(inty),
strRowSourceSelected)
intInString returns always 0 even when ther's a matching
string.
Thank you
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| anonymous replied to gr on 4 Jun 2004 |
Hello, I find out that InStr only works when the second
string is of length 1 - it only searches the first
character -
is there any equivalent that searches the whole word?
i Want to know if a string is already contained in another
example:
strSearchIn = "Hello;GoodBye;Hola;Adios;Gruzi;Tschus"
strSearchFor = "Adios"
I want to get a True or something to tell me that "Adios"
is on strSearchIn and a False to tell me that "Ciao" is
not int strSearchIn.
Any ideas?
statement:
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| Rick Brandt replied to anonymous on 4 Jun 2004 |
You are incorrect, The "string to be found" can be any length. Try this
in the debug window...
?InStr(1,"This is the string to search","the") <Enter>
The result is 9.
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| fredg replied to anonymous on 04 Jun 2004 |
Not True! InStr() searches the entire string from the first character
to the end, unless you specifically tell it to search from a different
start point.
See VBA Help for all the InStr() arguments.
Dim intX as Integer
Dim strSearchIn as String
strSearchIn = "Hello;GoodBye;Hola;Adios;Gruzi;Tschus"
intX = InStr(strSearchIn,"Adios")
If intX = 0 then
MsgBox "Not in string"
Else
MsgBox "In string"
End If
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| gr replied to fredg on 4 Jun 2004 |
ok, I thought that was the problem.
if I type the string in the immediate window works. But
not while code execution. Could it be because I'm using an
varPos = InStr(aAvailable(inty), strRowSource)
thx
first character
from a different
another
that "Adios"
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| Dirk Goldgar replied to gr on 4 Jun 2004 |
"gr" <anonymous@discussions.microsoft.com> wrote in message
news:182f301c44a31$e7a41680$a501280a@phx.gbl
I suggest you try dumping the parameter values to the Immediate Window
to see what's happening:
Debug.Print "available", aAvailable(inty)
Debug.Print "rowsource", strRowSourceSelected
intInString = VBA.InStr(aAvailable(inty), strRowSourceSelected)
Debug.Print intInString
You may find that the values aren't what you think they are.
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Archived message: InStr (MS Access)
