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From: JPB on 22 Jan 2010 20:47
What happens if you change the <> to >? What's the return value that
you are getting from InStr?


On Jan 22, 7:34 pm, WB <W...(a)discussions.microsoft.com> wrote:
> I'm trying to handle multiple exit codes when running an external app in VB6.
> InStr doesn't work like I would expect it to. It says "-1" exists in the
> string "1 1030", and also says "(-1)" exists in "(1) (1030)":
>
> ' Hard-coded the return value for this sample code:
> strReturnValue = "-1"
> strAlternateExitCodes = "1 1030"
> If InStr(1, strReturnValue, strAlternateExitCodes) <> 0 Then
>   ' This says it found "-1" in "1 1030"
> End If
>
> ' With parens:
> strReturnValue = Chr(40) & "-1" & Chr(41)
> strAlternateExitCodes = "(1) (1030)"
> If InStr(1, strReturnValue, strAlternateExitCodes) <> 0 Then
>   ' This says it found "(-1)" in "(1) (1030)"
> End If
>
> Am I missing something here? Is there a better way?
>
> --
> Bill Baker

From: DanS on 22 Jan 2010 22:10
Karl E. Peterson <karl(a)exmvps.org> wrote in
news:eplR1w8mKHA.5696(a)TK2MSFTNGP04.phx.gbl:

> DanS presented the following explanation :
>> Karl E. Peterson <karl(a)exmvps.org> wrote in
>> news:elxwBa8mKHA.5692(a)TK2MSFTNGP04.phx.gbl:
>>
>>> WB formulated on Friday :
>>>> I'm trying to handle multiple exit codes when running an external
>>>> app in VB6. InStr doesn't work like I would expect it to. It says
>>>> "-1" exists in the string "1 1030", and also says "(-1)" exists in
>>>> "(1) (1030)":
>>>>
>>>> ' Hard-coded the return value for this sample code:
>>>> strReturnValue = "-1"
>>>> strAlternateExitCodes = "1 1030"
>>>> If InStr(1, strReturnValue, strAlternateExitCodes) <> 0 Then
>>>> ' This says it found "-1" in "1 1030"
>>>> End If
>>>>
>>>> ' With parens:
>>>> strReturnValue = Chr(40) & "-1" & Chr(41)
>>>> strAlternateExitCodes = "(1) (1030)"
>>>> If InStr(1, strReturnValue, strAlternateExitCodes) <> 0 Then
>>>> ' This says it found "(-1)" in "(1) (1030)"
>>>> End If
>>>>
>>>> Am I missing something here? Is there a better way?
>>>
>>> I think we need to see some real code, to say for sure. What you
>>> posted won't run "as is." Modified to look like this:
>>>
>>> Public Sub test()
>>> Dim strReturnValue As String
>>> Dim strAlternateExitCodes As String
>>> ' Hard-coded the return value for this sample code:
>>> strReturnValue = "-1"
>>> strAlternateExitCodes = "1 1030"
>>> Debug.Print InStr(1, strReturnValue, strAlternateExitCodes)
>>>
>>> ' With parens:
>>> strReturnValue = Chr(40) & "-1" & Chr(41)
>>> strAlternateExitCodes = "(1) (1030)"
>>> Debug.Print InStr(1, strReturnValue, strAlternateExitCodes)
>>> End Sub
>>>
>>> I get the expected results:
>>>
>>> 0
>>> 0
>>>
>>
>> Of course Karl.......
>>
>> .......you will *never* find the string "(1) (1030)" contained within
>> the string "-1".
>
> As I said, the *expected* results... <eg>

Yes, that is what you said, and that is in fact correct.

Touche'.
From: WB on 26 Jan 2010 19:40
Sorry, this was a combination of the reversed InStr parameters that DanS
mentioned, and also a logic error on my part.

Thanks DanS (and everyone else).
--
Bill Baker


"DanS" wrote:

> =?Utf-8?B?V0I=?= <WB(a)discussions.microsoft.com> wrote in
> news:A967A23F-6FFD-4964-8C3C-C952174FCA1F(a)microsoft.com:
>
> > I'm trying to handle multiple exit codes when running an external app
> > in VB6. InStr doesn't work like I would expect it to. It says "-1"
> > exists in the string "1 1030", and also says "(-1)" exists in "(1)
> > (1030)":
> >
> > ' Hard-coded the return value for this sample code:
> > strReturnValue = "-1"
> > strAlternateExitCodes = "1 1030"
> > If InStr(1, strReturnValue, strAlternateExitCodes) <> 0 Then
> > ' This says it found "-1" in "1 1030"
> > End If
> >
> > ' With parens:
> > strReturnValue = Chr(40) & "-1" & Chr(41)
> > strAlternateExitCodes = "(1) (1030)"
> > If InStr(1, strReturnValue, strAlternateExitCodes) <> 0 Then
> > ' This says it found "(-1)" in "(1) (1030)"
> > End If
> >
> > Am I missing something here? Is there a better way?
>
> Putting this in a command button procedure never says it found it.
>
> You've got the order of the strings wrong, but neither way said it found
> it. It's......Instr(1,[StringToSearch],[FindThisString]
>
> Private Sub Command1_Click()
>
> Dim strReturnValue As String
> Dim strAlternateExitCodes As String
> ' Hard-coded the return value for this sample code:
> strReturnValue = "-1"
> strAlternateExitCodes = "1 1030"
> If InStr(1, strReturnValue, strAlternateExitCodes) <> 0 Then
> Debug.Print "-1 found"
> ' This says it found "-1" in "1 1030"
> End If
>
> End Sub
>
> Possibly your hard-coded value shown here as "-1" is *not* "-1" in the
> production code ?
> .
>
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