将变量从 php 传递给 bash
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passing a variable from php to bash
提问by brad
I cannot seem to get a variable passed to my bash script from php. $uaddress and $upassword come up empty no matter what I try.
我似乎无法将变量从 php 传递给我的 bash 脚本。无论我尝试什么，$uaddress 和 $upassword 都为空。
*********************bash ****************
** *** *** *** *** *** *** *bash *** *** *** *** ****
#!/bin/bash -x
useraddress=$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 *** *** *** *** ****
<?php
$upassword = 'test1234'; $uaddress = '[email protected]';
$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);
?>
***********output and debug ************
** *** *** ***输出和调试 ** *** *** *** *
[email protected]
+ 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: > "
回答by Barmar
You need to pass the variables as arguments to the shell script, and the shell script has to read its arguments.
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。
So in PHP:
所以在 PHP 中：
$useraddress = escapeshellarg('[email protected]');
$upassword = escapeshellarg('test1234');
$addr = shell_exec("sudo /home/tpccmedia/cgi-bin/member_add_postfixadmin $useraddress $upassword 2>&1");
and in the shell script:
并在shell脚本中：
useraddress=
upassword=
回答by brad
Got it.
知道了。
<?php
$upassword = 'test1234'; $uaddress = '[email protected]';
$uaddress = escapeshellarg($uaddress);
$upassword = escapeshellarg($upassword);
$addr = shell_exec("sudo /home/tpcmedia/cgi-bin/member_add_postfixadmin $uaddress $upassword 2>&1");
?>
#!/bin/bash -x
uaddress=
upassword=
ssh -p 2222 -6 2400:8900::f03c:91ff:fe69:8aaf "/var/www/localhost/htdocs/postfixadmin/scripts/postfixadmin-cli mailbox add" $uaddress --password $upassword --password2 $upassword
