how to quote a unix command having both single and double quotes

newbie271

Hi
I have a unix command with mixed type.

awk -vOFS="\t" '{$4=$1"_"$2} NR==1{$4="name"}1' file.txt

How to call it from php if it has both and single and double quotes in it. Trying different escapes did not work well for me.

I keep getting errors because of this quote and somehow I am not able to fix it.

$dir = '/home/tmpDir/'
$part2 = 'awk -vOFS="\t" \'{$4=$1"_"$2} NR==1{$4="name"}1\' . '  '. $dir." file.txt"';
$command = "/bin/bash ".$part2;

Thanks

Derokorian

$dir = '/home/tmpDir/';
$part2 = 'awk -vOFS="\t" \'{$4=$1"_"$2} NR==1{$4="name"}1\' '. $dir.'file.txt';

johanafm

Would it not be easier to use [man]escapeshellarg[/man]

Derokorian

I think the problem was misquoting when creating the string johanafm. When I tried to use the code he provided my syntax highlighting went haywire. I didn't know how to explain the quote problems other than just fix it lol.

newbie271

Thanks Derokorian and Johanafm,

I am still struggling with this shell_exec call from php.
I am being more specific with the following code. I am generating these variables in my code and that is why I am using these variables.

<?php
print(shell_exec('/bin/bash -c "ls -l"'));


$dir = '/home/tmpDir';
$idx1 = 1;
$idx2 = 2;
$idx3 = 3;
$idx4  = 4;
$inpfile= 'file.txt';
$outfile= 'fileout.txt';
$bash = '/bin/bash';

$part1 =  'awk -vOFS="\t" ' . '\'{$'.$idx4. '=$'.$idx1.'"_"$'.$idx2.'"_"$'.$idx3.'}NR==1{$'.$idx4.'='. '"name"}1\'';
$command1 = $part1. ' '. $dir. '/'.$inpfile . " > " . $dir. '/'.$outfile;
print($command1);  //This command works fine on command line.
print("<br><br>");
$command2 = $bash. ' -c "'.$command1. '"'; 
print($command2); 
shell_exec($command2);
?>

Following are the output of the above file.

awk -vOFS="\t" '{$4=$1""$2""$3}NR==1{$4="name"}1' /home/tmpDir/file.txt > /home/tmpDir/fileout.txt

/bin/bash -c "awk -vOFS="\t" '{$4=$1""$2""$3}NR==1{$4="name"}1' /home/tmpDir/file.txt > /home/tmpDir/fileout.txt"

The first command 'command1' works well on shell prompt. Somehow I want to embed them into double quotes as I did in the first line of the php script to print list of files. Upto now I remained unsuccessful.

Thanks.

bradgrafelman

Again [man]escapeshellarg/man will take care of all of the necessary escaping for you. Read the manual page - it even has code examples.

newbie271

Still not working !!!

<?php
/$dir = '/home/tmpData';
$idx1 = 1;
$idx2 = 2;
$idx3 = 3;
$idx4  = 4;
$inpfile= 'file.txt';
$outfile= 'fileout.txt';
$bash = '/bin/bash';

$part1 =  escapeshellarg('awk -vOFS="\t" ' . '\'{$'.$idx4. '=$'.$idx1.'"_"$'.$idx2.'"_"$'.$idx3.'}NR==1{$'.$idx4.'='. '"name"}1\'');
//print($part1);
$command1 = $part1. ' '. $dir. $inpfile . " > " . $dir.$outfile;
$command2 = (escapeshellarg($bash. ' -c ' . "$command1"));  //This command works fine on command line.
//$command2 = $bash. ' -c '.$command1. '';
shell_exec(escapeshellarg($command2));
//shell_exec($bash ' -c ' $command2); 

?>

$php index.php
still shows the error.

bradgrafelman

You're escaping the entire command rather than its (only) argument. Not only that, but you're doing so multiple times. You only need to call [man]escapeshellarg/man once to escape the value of the '-c' argument for the 'bash' command.

Furthermore, you aren't surrounding the value of the '-c' argument in quotes as you were before.