It works for me.

$bar=$ARGV[0];
$lak=`$bar`;
print "$lak";
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<?php
$foo="ls";
$cmd="perl split.pl $foo";
$out=`$cmd`;
print "$out";
?>
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Now the $out has the ls output. I think I understood your question. If not clarify me.

actually, in the php file, $foo="ID_106", its a string which has to be used in the system command. Example:

$foo="ID_106";
system("perl split.pl $foo");
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now, the $foo variable is passed to the perl program. the variable $foo is used for system command in perl. example:

$bar=$ARGV[0]; (the argument input, so now $bar="ID_106")
$var=system("grep -A 1 $bar *.txt");
print $var;
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Over here, $var doesnt work :(

well, that reply is from me.

I assume Over here, $var doesnt work means $var has nothing in it.

Did you tried printing the $var to a file?

When you said print $var, it prints the contents to the STDOUT, which is captured by the system function in PHP. So print $var seems not printing anything, but actually it has the output.

$bar=$ARGV[0];
$lak=system($bar);
print $lak; #This will be captured by the system function in the PHP f
+ile.
open FH,">> llll";
print FH $lak;
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Now the file will also have the output