RE: Capturing System output

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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:bbonkoski@xxxxxxxxxxxxxx] 
> Sent: Tuesday, August 15, 2006 10:38 AM
> To: Stut
> Cc: PHP List
> Subject: Re:  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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> 

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