如何将多个PHP参数传递给bash

setip.php :

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

示例输入：ip = 10，ip2 = 20，ip3 = 30，ip4 = 40，dns1 = 10

ipset.sh :

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 :

导致ifcfg-eth0：

IPADDR=1020304010
NETMASK=
NETWORK=
BROADCAST=
DNS=

any sugestion how to pass argument correctly ?

任何消化如何正确传递参数？

1 个解决方案

#1

Solved, add space between argument.

解决了，在参数之间添加空格。

  <?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");