Re: 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" <sebastian@broadbandgaming.net> wrote in message news:004101c4dfeb$6a71d490$77a13f44@GAMING... > well $user['password'] has no double quotes around it. > > ----- Original Message ----- > From: "Jonathan Duncan" <jonathan@nacnud.com> > To: <php-general@lists.php.net> > 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 >> >> >>