Passing variables from PHP to PERL

Question

i have written this program to pass a variable in the embedded perl script but it gives blank output :-( why it is not working. Please fix this problem.

the php script

<?php
$var1='high'; 
exec('C:/xampp/htdocs/WORK/hello.pl'.' '.EscapeShellArg("$var1"),$output);
echo ($output);
?>

the perl script

#!/usr/bin/perl -w

    $var1=<>; 

    print $var1;

Are you sure that your *.pl files are executable? Try to execute hello.pl from the command line, without php — Vitaly Dyatlov, May 29 '12 at 11:49
Ok, recreated it locally. `shift` does work, and `<>` not. Im not sure why it this way, but.. You have to use either `shift` or `$ARGV[0]` to get the value — Vitaly Dyatlov, May 29 '12 at 12:19
http://stackoverflow.com/questions/3745623/why-should-i-use-argv-or-instead-of-stdin-in-perl - so.. is only for piping, for arguments you have to use ARGV — Vitaly Dyatlov, May 29 '12 at 12:23

score 1 · Answer 1 · answered May 29 '12 at 11:47

1

Because youre trying to send an array as a string... Youll have to serialize the array into a string that perl can deserialize, like JSON, XML, CSV, etc..

answered May 29 '12 at 11:47

prodigitalson

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No, because "$var1" becomes "Array" (raw command is like this: C:/xampp/htdocs/WORK/hello.pl "Array" ) and there should be the output anyway. Issues somewhere else – Vitaly Dyatlov May 29 '12 at 11:53
Thats true... kinda jumped the gun... regardless hes not going to get the result hes expecting, though he needs to get the var passing correct first. – prodigitalson May 29 '12 at 11:58
sorry, i m new to this field.... could you please tell me where is the problem actually.. – user1083096 May 29 '12 at 12:05

score 1 · Accepted Answer · answered May 29 '12 at 12:00

1

What about

 $var1 = shift;

rather than reading the value from standard input or argument files?

answered May 29 '12 at 12:00

choroba

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i want to take the input as a string, instead of an array – user1083096 May 29 '12 at 12:08
@user1083096: OK. But let's process step by step. By now, you are not taking it at all. – choroba May 29 '12 at 12:09

score 0 · Answer 3 · answered May 29 '12 at 11:48

0

One reason is that EscapeShellArg expects the $arg parameter to be of type string, while, in your case, it's an array...

answered May 29 '12 at 11:48

Mihai Todor

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Actually at that point he should be sending the string representation of the array i think because hes got it in double quotes when he passes it... – prodigitalson May 29 '12 at 11:52
i changed "$var1" to '$var1' but its not working :( – user1083096 May 29 '12 at 12:10
Are you able to call the PERL script via command line? Does it produce the expected output? Could we get a sample of the parameters that you are using when calling the script and the output that it produces? – Mihai Todor May 29 '12 at 12:27
yes i m able to call the script and it does produces the output. i tried using the same script posted above and its printing "array" as the output. – user1083096 May 29 '12 at 12:34

Passing variables from PHP to PERL

3 Answers3

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