Try using backticks, and you don't need spaces after -u and -p
$result = `MySQL -uuser -ppassword databasename < filename.sql`
Note that if it succeeds, nothing will be returned. Same as when you run
the command on the command line.
> -----Original Message-----
> From: Fargo Lee [mailto:[EMAIL PROTECTED]]
> Sent: Wednesday, July 10, 2002 1:27 AM
> To: [EMAIL PROTECTED]
> Subject: Re: [PHP] How do I import tables into MySQL from web page ...
> Thanks for pointing out the syntax error. I added the space after the
> it did not make any difference. It still gives the same result, that
> "Success", when it actually fails. What I am trying to figure out is
> can tell if it failed (did not create the tables)? The $status
> not appear to hold the output of the system() function. Anyone know
> get the output of system, passthru or exec into the $status variable
> check for success or failure?
> "Analysis & Solutions" <[EMAIL PROTECTED]> wrote in
> [EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> > On Tue, Jul 09, 2002 at 06:09:36PM -0700, Fargo Lee wrote:
> > >
> > > $status = system("mysql -umyuserid -pmypassword mydbname <
> > You need a space between "-u" and "myuserid"
> > --Dan
> > --
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