Re: php variables in a backtick command

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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" <rory.browne@xxxxxxxxx> wrote in message 
news:fa07888a04121118204f55784e@xxxxxxxxxxxxxxxxx
> 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 <jonathan@xxxxxxxxxx> 
> 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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>> 



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