Hello,
If you are working on a Linux system, you can try appending:
2>&1
To the end of your command, so that you end up with:
Mysql -u $user -p{$pass} 2>&1
What this does is tell the shell to redirect anything from stderr to go
through stdout.
If you want to get -really- fancy you can use proc_open, which will give you
handles for the process' stdout, stderr, and stdin.
HTH,
K. Bear
> -----Original Message-----
> From: Brad Bonkoski [mailto:[EMAIL PROTECTED]
> Sent: Tuesday, August 15, 2006 10:38 AM
> To: Stut
> Cc: PHP List
> Subject: Re: [PHP] Capturing System output
>
>
>
> Stut wrote:
> > Brad Bonkoski wrote:
> >> Had this problem in the past, and always programmed around it, but
> >> wondering if there is an easier way.
> >>
> >> Good Example:
> >> Creating a setup for connecting to a mysql database. Want to do
> >> something simple to make sure they have entered a valid
> >> username/password for the database.
> >> So, the idea is something like:
> >> $rc = exec("mysql -u $user -p{$pass}", $output); The
> problem is one
> >> error, the stderr does not go to the output array, but
> rather to the
> >> screen.
> >>
> >> Previously I would redirect the stderr to a file, and then
> evaluate
> >> the contents of the file, but is there an easier way to
> get this into
> >> the PHP variable with no risk of having the output make it
> through to
> >> the screen?
> >
> > I may be missing something, but why in the name of all that is holy
> > would you want to shell out to try connecting to mysql? Why not use
> > mysql_connect and avoid the potentially massive security
> hole you're
> > building?
> >
> > -Stut
> >
> Perhaps poor illustration of the question...the question
> being how to issue system like commands in PHP which would
> allow you to trap not only stdout, but also stderr.
> -Brad
>
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