how to pass multiple PHP arguments to bash

Question

setip.php :

<?php
$IP_Addr = $_GET['ip'];
$NetMask = $_GET['ip2'];
$NetWork = $_GET['ip3'];
$Broad = $_GET['ip4'];
$DNS = $_GET['dns1'];

$cmd="sh /var/www/cgi-bin/ipset.sh ".escapeshellarg($IP_Addr)."".escapeshellarg($NetMask)."".escapeshellarg($NetWork)."".escapeshellarg($Broad)."".escapeshellarg($DNS);
 exec("$cmd");

example input : ip=10, ip2=20, ip3=30, ip4=40, dns1=10

ipset.sh :

#!/bin/bash -x
echo "IPADDR=$1"       >>/etc/sysconfig/network-scripts/ifcfg-eth0
echo "NETMASK=$2"     >>/etc/sysconfig/network-scripts/ifcfg-eth0
echo "NETWORK=$3"     >>/etc/sysconfig/network-scripts/ifcfg-eth0
echo "Broadcast=$4"   >>/etc/sysconfig/network-scripts/ifcfg-eth0
echo "DNS=$5"        >>/etc/sysconfig/network-scripts/ifcfg-eth0

result in ifcfg-eth0 :

IPADDR=1020304010
NETMASK=
NETWORK=
BROADCAST=
DNS=

any sugestion how to pass argument correctly ?

Try echoing out your command - does it look like it should? Does it work when you run it directly? It looks like you need to add spaces between the arguments.... — andrewsi
– andrewsi, Commented Dec 19, 2013 at 16:30

Haruru · Accepted Answer · 2013-12-29 17:08:20Z

1

Solved, add space between argument.

  <?php
    $IP_Addr = $_GET['ip'];
    $NetMask = $_GET['ip2'];
    $NetWork = $_GET['ip3'];
    $Broad = $_GET['ip4'];
    $DNS = $_GET['dns1'];

    $cmd="sh /var/www/cgi-bin/ipset.sh ".escapeshellarg($IP_Addr)." ".escapeshellarg($NetMask)." ".escapeshellarg($NetWork)." ".escapeshellarg($Broad)." ".escapeshellarg($DNS);
     exec("$cmd");

answered Dec 29, 2013 at 17:08

Haruru

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