InStr issue
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From: WB on 22 Jan 2010 19:34 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: Karl E. Peterson on 22 Jan 2010 19:53 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 -- ..NET: It's About Trust! http://vfred.mvps.org
From: DanS on 22 Jan 2010 20:08 =?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 ?
From: DanS on 22 Jan 2010 20:13 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".
From: Karl E. Peterson on 22 Jan 2010 20:34
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> -- ..NET: It's About Trust! http://vfred.mvps.org |