IF statement is 'not working'

[Maurice Mertens]

Hi,

I'm using the code underneath in a function:

Dim cn As New ADODB.Connection
Dim rs As New ADODB.Recordset
Dim ds As New DataSet
Dim dv As New DataView
Dim strMsg As String, strComputerName As String
Dim row As DataRow
Dim lngCurrentMachineTeller As Long, lngI As Long

cn.ConnectionString = oleDBConn1.ConnectionString
cn.Open()
rs = cn.OpenSchema(ADODB.SchemaEnum.adSchemaProviderSpe cific, ,
"{947bb102-5d43-11d1-bdbf-00c04fb92675}")
daAdapter.Fill(ds, rs, "USERS")
ds.Tables("USERS").Columns(0).ColumnName = "Computernaam"
ds.Tables("USERS").Columns(1).ColumnName = "Loginnaam"
ds.Tables("USERS").Columns(2).ColumnName = "Connected"
ds.Tables("USERS").Columns(3).ColumnName = "State"
dv.Table = ds.Tables("USERS")
dv.RowStateFilter = DataViewRowState.CurrentRows

For lngI = 0 To dv.Count - 1
row = dv.Item(lngI).Row
If row.Item("Computernaam").ToString = gsComputerName.ToString Then
msgbox("Same!")
Else
msgbox("Not the same!")
Endif
Next
The problem is strange behaviour in the line "If row.Item
("Computernaam").ToString = gsComputerName.ToString Then
". When in the command window I type: '?row.Item("Computernaam").ToString
= gsComputerName.ToString' it returns true. But when I debug the code
step by step it doesn't go into the if statement. Like it is 'not true'
suddenly. The variable 'gsComputerName' is a global string. They both
contain the same value but the IF STATEMENT "doesn;t work" for some
reason.

--
Met vriendelijke groet / With regards / Saludos,
Moviat Automatisering
Maurice Mertens
mauricem@moviat_KillSpamWordForEMail.nl

tel: [phone]
fax: [phone]

[Galin Iliev]

this is really strange
" If row.Item("Computernaam").ToString = gsComputerName.ToString Then"
do you mean
" If row.Item("Computername").ToString = gsComputerName.ToString Then"??

be careful of spelling ("Computername")

Galin Iliev

"Maurice Mertens" <hm*****@nospam.nospam> wrote in message
news:Xn***************************@194.109.133.133 ...

Hi,

I'm using the code underneath in a function:

Dim cn As New ADODB.Connection
Dim rs As New ADODB.Recordset
Dim ds As New DataSet
Dim dv As New DataView
Dim strMsg As String, strComputerName As String
Dim row As DataRow
Dim lngCurrentMachineTeller As Long, lngI As Long

cn.ConnectionString = oleDBConn1.ConnectionString
cn.Open()
rs = cn.OpenSchema(ADODB.SchemaEnum.adSchemaProviderSpe cific, ,
"{947bb102-5d43-11d1-bdbf-00c04fb92675}")
daAdapter.Fill(ds, rs, "USERS")
ds.Tables("USERS").Columns(0).ColumnName = "Computernaam"
ds.Tables("USERS").Columns(1).ColumnName = "Loginnaam"
ds.Tables("USERS").Columns(2).ColumnName = "Connected"
ds.Tables("USERS").Columns(3).ColumnName = "State"
dv.Table = ds.Tables("USERS")
dv.RowStateFilter = DataViewRowState.CurrentRows

For lngI = 0 To dv.Count - 1
row = dv.Item(lngI).Row
If row.Item("Computernaam").ToString = gsComputerName.ToString Then
msgbox("Same!")
Else
msgbox("Not the same!")
Endif
Next
The problem is strange behaviour in the line "If row.Item
("Computernaam").ToString = gsComputerName.ToString Then
". When in the command window I type: '?row.Item("Computernaam").ToString
= gsComputerName.ToString' it returns true. But when I debug the code
step by step it doesn't go into the if statement. Like it is 'not true'
suddenly. The variable 'gsComputerName' is a global string. They both
contain the same value but the IF STATEMENT "doesn;t work" for some
reason.

--
Met vriendelijke groet / With regards / Saludos,
Moviat Automatisering
Maurice Mertens
mauricem@moviat_KillSpamWordForEMail.nl

tel: [phone]
fax: [phone]

[Maurice Mertens]

No,

ds.Tables("USERS").Columns(0).ColumnName = "Computernaam"
The columnname is "Computernaam".

But it is very strange that in the command window the comparison returns
true and in my code the IF statement returns false for exactly the same
comparison. I also did a reboot of my PC which didn't change this result (I
don't know if I would feel more relieved when the results were in fact
different after the reboot :)..)


"Galin Iliev" <iliev@_NOSPAM_.Galcho.com> wrote in
news:eV**************@TK2MSFTNGP14.phx.gbl:

this is really strange
" If row.Item("Computernaam").ToString = gsComputerName.ToString Then"
do you mean
" If row.Item("Computername").ToString = gsComputerName.ToString
Then"??

be careful of spelling ("Computername")

Galin Iliev

[Cor Ligthert]

Maurice,

Why do you than not set temporaly these two instructions in your code
dim a as string = row.Item("Computernaam").ToString
dim b as string = gsComputerName.ToString

For lngI = 0 To dv.Count - 1
row = dv.Item(lngI).Row
dim a as string = row.Item("Computernaam").ToString
dim b as string = gsComputerName.ToString
If row.Item("Computernaam").ToString = gsComputerName.ToString Then
msgbox("Same!")
Else

Than it is so easy to see when you debug step by step.

Just my thought,

Cor

[Maurice Mertens]

Hi Cor,

I already tried your example but this also didn't work. The same problem!
After a little research we found the reason for this behaviour. The value
row.item("Computernaam") contains a string which came from the kernel.
These strings are ended with asci character 0. So we now use a replace
and a trim: Replace(Trim(row.Item("Computernaam")), Chr(0), "") and this
is working fine now!

Thx for looking at this problem..

"Cor Ligthert" <no************@planet.nl> wrote in news:#EpK$gcRFHA.204
@TK2MSFTNGP15.phx.gbl:

Maurice,

Why do you than not set temporaly these two instructions in your code
dim a as string = row.Item("Computernaam").ToString
dim b as string = gsComputerName.ToString

For lngI = 0 To dv.Count - 1
row = dv.Item(lngI).Row

dim a as string = row.Item("Computernaam").ToString
dim b as string = gsComputerName.ToString

If row.Item("Computernaam").ToString = gsComputerName.ToString Then msgbox("Same!")
Else

Than it is so easy to see when you debug step by step.

Just my thought,

Cor

--
Met vriendelijke groet / With regards / Saludos,
Moviat Automatisering
Maurice Mertens
mauricem@moviat_KillSpamWordForEMail.nl

tel: [phone]
fax: [phone]

[Cor Ligthert]

Maurice,

Really a terrible situation to find.

From this newsgroup side I would probably never thought about that when you
had not told it now.

Thanks that you told what it was and congratulated as well with that.

Cor