将变量从php传递到bash

一尘不染

将变量从php传递到bash

linux

我似乎无法从php中将变量传递给我的bash脚本。无论我如何尝试，$ uaddress和$ upassword都变为空。

*_ __ _ __ __ * __ __ bash * __ __ * * **

#!/bin/bash -x
useraddress=$uaddress
upassword=$upassword
ssh -p 222 -6 2400:8900::f03c:91f:fe69:8af "/var/www/localhost/htdocs/postfixadmin/scripts/postfixadmin-cli mailbox add" $useraddress --password $upassword --password2 $upassword  .ssh

*_ __ _ __ * PHP __ * __ __ * __ **

<?php
$upassword = 'test1234'; $uaddress = 'mytestuser@tpccmedia.com';
$addr = shell_exec('sudo /home/tpccmedia/cgi-bin/member_add_postfixadmin 2>&1'); echo $uaddress; echo $upassword;
//$addr = shell_exec('ssh -p 222 -6 2400:8900::f03c:91f:fe69:8af /var/www/localhost/htdocs/postfixadmin/scripts/postfixadmin-cli mailbox add; echo $useraddress; --password; echo $upassword; --password2; echo $upassword; .ssh');
echo "<pre>$addr</pre>";
var_dump($addr);
?>

*_ __ _ _ _输出和调试 _ _ _ * *_

mytestuser@tpccmedia.comtest1234

+ useraddress=
+ upassword=
+ ssh -p 2222 -6 2400:8900::f03c:91ff:fe69:8aaf '/var/www/localhost/htdocs/postfixadmin/scripts/postfixadmin-cli mailbox add' --password --password2 .ssh

Welcome to Postfixadmin-CLI v0.2
---------------------------------------------------------------
Path: /var/www/localhost/htdocs/postfixadmin
---------------------------------------------------------------

Username:  
>

string(404) "+ useraddress= + upassword= + ssh -p 2222 -6 2400:8900::f03c:91ff:fe69:8aaf '/var/www/localhost/htdocs/postfixadmin/scripts/postfixadmin-cli mailbox add' --password --password2 .ssh Welcome to Postfixadmin-CLI v0.2 --------------------------------------------------------------- Path: /var/www/localhost/htdocs/postfixadmin --------------------------------------------------------------- Username: > "

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2020-06-07

共1个答案

一尘不染

您需要将变量作为参数传递给Shell脚本，并且Shell脚本必须读取其参数。

因此在PHP中：

$useraddress = escapeshellarg('mytestuser@tpccmedia.com');
$upassword = escapeshellarg('test1234');
$addr = shell_exec("sudo /home/tpccmedia/cgi-bin/member_add_postfixadmin $useraddress $upassword 2>&1");

在shell脚本中：

useraddress=$1
upassword=$2

2020-06-07