[PHP] Capturing System output
Hello All..
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?
Thanks
-Brad
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Re: [PHP] Capturing System output
On Tuesday 15 August 2006 09:19, Brad Bonkoski wrote:
Hello All..
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?
Thanks
-Brad
I'd take a look at shell_exec. There'sa comment about capturing stderr. I'm
not sure if shell_exec captures stderr, but it should at least point you in
the right direction. Just do a search for stderr and you should find some
good info.
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Ray Hauge
Programmer/Systems Administrator
American Student Loan Services
www.americanstudentloan.com
1.800.575.1099
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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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RE: [PHP] Capturing System output
Hello,
If you are working on a Linux system, you can try appending:
21
To the end of your command, so that you end up with:
Mysql -u $user -p{$pass} 21
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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Re: [PHP] Capturing System output
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
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Re: [PHP] Capturing System output
On Tuesday 15 August 2006 09:38, Brad Bonkoski wrote:
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
Best example I found was:
$shell_return = shell_exec($shell_command. 21);
that should redirect stderr to stdout and thus you'd get both.
--
Ray Hauge
Programmer/Systems Administrator
American Student Loan Services
www.americanstudentloan.com
1.800.575.1099
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PHP General Mailing List (http://www.php.net/)
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Re: [PHP] Capturing System output
On Tue, August 15, 2006 9:19 am, 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?
In some OSes, in some shells, you can use:
mysql -u $user -p{$pass} 21
The 21 is special code for redirect stdrrr (aka 2) to stdout (aka 2)
Unless it's 21 which I always forget which is which...
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