Assuming you can feed 'admin' into the function and it works, but feeding
'fred' doesn't, it looks like there is a problem with the data.

Starting with the obvious that you have probably already covered

1.      Is the data in the database at all - could there have been a problem
inserting it?
2.      Has it been entered correctly, not forgetting case sensitivity
3.      Could there be any invisible chars (space or tab for example) in the
data. I had a script that generated entries and I accidentally left a space
at the beginning of the field. Took me hours to realise because it isn't
immediately obvious!

Worth a browse through with some sort of management tool, such as phpMyAdmin
http://www.phpwizard.net/projects/phpMyAdmin/
if you haven't already.

Not sure otherwise.
HTH

Peter

> -----Original Message-----
> From: Martin [mailto:[EMAIL PROTECTED]]
> Sent: 04 November 2001 06:03
> To: [EMAIL PROTECTED]
> Subject: RE: [PHP-DB] Nothing returned
>
>
> Peter Lovatt wrote:
>
> > $sql = "SELECT * FROM users WHERE (name='$name')"
> >
> > if you are not getting an error, it is usually worth echoing
> the query SQL
> >
> > echo $sql ;
> >
> > and perhaps trying it manually.
> >
> > Is it possible the quotes around the $name are giving
> >
> > SELECT * FROM users WHERE (name='$name')
> > and not evaluating $name
> >
> > rather than
> >
> > SELECT * FROM users WHERE (name='fred')
> >
> > which you wanted
>
> echo $sql gives me:
> SELECT * FROM users WHERE (name='admin')
> when entering admin as user. So the name is passed to the
> function but from
> then on ...
>
> M.
>
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