Re: [PHP] php variables in a backtick command
well $user['password'] has no double quotes around it. - Original Message - From: Jonathan Duncan [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Saturday, December 11, 2004 8:09 PM Subject: [PHP] php variables in a backtick command I am trying to run a shell command with backticks. However, I get a parse error. Is it because I have an array variable in there? $result = `adduser -l=$dist_id -p=$user['password'] --f=$user['name_first'] $user['name_last']`; Do I need to assign the value to a regular variable before I put it in there? Like this? $pword=$user['password'] $fname =$user['name_first'] $lname =$user['name_last'] $result = `adduser -l=$dist_id -p=$pword --f=$fname $lname`; Thanks, Jonathan -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] php variables in a backtick command
Ah, that is a good idea, putting the command in a variable and then
executing the variable. I have doen that before but did not think of it
now. Too many things going on. Thanks!
Jonathan
Rory Browne [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
I'm not sure about variable expansion with backticks(I don't use
backticks, if necessary I use shell_exec() instead). I'm just after
installing a fresh SuSE 9.1, and php is giving me a segfault at the
minute, with that particular code, but if you were using double quotes
you could:
$command = adduser -l=$dist_id -p={$user['password']}
--f=\{$user['name_first']} {$user['name_last']}\;
bearing in mind that the variables have been {}'ed, and your double
quote around $user['name_first'] and ['name_last']) has been escaped
to \
you can use shell_exec($command), which is identical to the backtick
operation.
Having that said, you should consider using alternative program
execution instead, instead of using the shell. What do you need the
shell for? You might(albeit unlikely) also come across a situation
where the shell is something like /bin/false, or /bin/falselogin, or
/bin/scponly, or basicly something that doesn't particularly work that
well as a shell.
Rory
On Sat, 11 Dec 2004 18:09:17 -0700, Jonathan Duncan [EMAIL PROTECTED]
wrote:
I am trying to run a shell command with backticks. However, I get a
parse
error. Is it because I have an array variable in there?
$result =
`adduser -l=$dist_id -p=$user['password'] --f=$user['name_first']
$user['name_last']`;
Do I need to assign the value to a regular variable before I put it in
there? Like this?
$pword=$user['password']
$fname =$user['name_first']
$lname =$user['name_last']
$result = `adduser -l=$dist_id -p=$pword --f=$fname $lname`;
Thanks,
Jonathan
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Re: [PHP] php variables in a backtick command
The quotes are only for the shell command which does not use quotes around password. Thanks for the feedback. Jonathan Sebastian [EMAIL PROTECTED] wrote in message news:[EMAIL PROTECTED] well $user['password'] has no double quotes around it. - Original Message - From: Jonathan Duncan [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Saturday, December 11, 2004 8:09 PM Subject: [PHP] php variables in a backtick command I am trying to run a shell command with backticks. However, I get a parse error. Is it because I have an array variable in there? $result = `adduser -l=$dist_id -p=$user['password'] --f=$user['name_first'] $user['name_last']`; Do I need to assign the value to a regular variable before I put it in there? Like this? $pword=$user['password'] $fname =$user['name_first'] $lname =$user['name_last'] $result = `adduser -l=$dist_id -p=$pword --f=$fname $lname`; Thanks, Jonathan -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] php variables in a backtick command
I'm not sure about variable expansion with backticks(I don't use
backticks, if necessary I use shell_exec() instead). I'm just after
installing a fresh SuSE 9.1, and php is giving me a segfault at the
minute, with that particular code, but if you were using double quotes
you could:
$command = adduser -l=$dist_id -p={$user['password']}
--f=\{$user['name_first']} {$user['name_last']}\;
bearing in mind that the variables have been {}'ed, and your double
quote around $user['name_first'] and ['name_last']) has been escaped
to \
you can use shell_exec($command), which is identical to the backtick operation.
Having that said, you should consider using alternative program
execution instead, instead of using the shell. What do you need the
shell for? You might(albeit unlikely) also come across a situation
where the shell is something like /bin/false, or /bin/falselogin, or
/bin/scponly, or basicly something that doesn't particularly work that
well as a shell.
Rory
On Sat, 11 Dec 2004 18:09:17 -0700, Jonathan Duncan [EMAIL PROTECTED] wrote:
I am trying to run a shell command with backticks. However, I get a parse
error. Is it because I have an array variable in there?
$result = `adduser -l=$dist_id -p=$user['password'] --f=$user['name_first']
$user['name_last']`;
Do I need to assign the value to a regular variable before I put it in
there? Like this?
$pword=$user['password']
$fname =$user['name_first']
$lname =$user['name_last']
$result = `adduser -l=$dist_id -p=$pword --f=$fname $lname`;
Thanks,
Jonathan
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