help with comparing strings....

[Paul M]

hi there,

i have some code which in VB6, worked fine, but in VB.NET does not match the
following:

If Mid(sBuffer, nI, 2) = vbCrLf Then
the MID string returns the 2 squares that resemble VbCrlF, and when i put
both variables into the watch window, they return the same string, but the
match returns FALSE.

Any help appreciated,
Paul.

[Bryan Martin]

Why not test to see what is being returned?

Dim yourArray()as byte = System.Text.Encoding.Ascii.GetBytes(Mid(sBuffer,
nI, 2))

Now check what yourarray is from the locals window.

"Paul M" <mi******@hotmail.com> wrote in message
news:eV****************@TK2MSFTNGP12.phx.gbl...

hi there,

i have some code which in VB6, worked fine, but in VB.NET does not match the following:

If Mid(sBuffer, nI, 2) = vbCrLf Then
the MID string returns the 2 squares that resemble VbCrlF, and when i put
both variables into the watch window, they return the same string, but the
match returns FALSE.

Any help appreciated,
Paul.

[Cor]

Hi Paul,

I would do in this case while debugging
dim a as integer = asc(mid(sBuffer, nI, 1))
dim b as integer = asc(mid(sBuffer, nI+1, 1))

To be sure that it is really 13 and 10.

Cor

[Herfried K. Wagner [MVP]]

* "Paul M" <mi******@hotmail.com> scripsit:

i have some code which in VB6, worked fine, but in VB.NET does not match the
following:
Should still work fine.
If Mid(sBuffer, nI, 2) = vbCrLf Then
the MID string returns the 2 squares that resemble VbCrlF, and when i put
both variables into the watch window, they return the same string, but the
match returns FALSE.

Are you really sure that you cut out the right part?

--
Herfried K. Wagner [MVP]
<http://www.mvps.org/dotnet/>

..NET FAQs: <http://www.mvps.org/dotnet/dotnet/faqs/> (German)