从PHP传递变量到Bash

Question

我似乎无法从php将变量传递到我的bash脚本中。无论我尝试什么，$uaddress和$upassword都为空。

********************* 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 = '[e-mail]';
$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************

[e-mail]1234

+ 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: > " 

- brad

Answer 1

您需要将变量作为参数传递给shell脚本，shell脚本必须读取其参数。

因此，在PHP中：

$useraddress = escapeshellarg('[e-mail]');
$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

- Barmar

---

您肯定也想使用 escapeshellarg 转义参数，特别是密码字段可能包含特殊字符。 - tangrs

@tangrs 谢谢。我还忘记将字符串的单引号改为双引号，这样插值才能正常工作。 - Barmar

@Barmar 不先生，$useraddress = escapeshellarg($useraddress); 会导致未定义的变量错误，而escapeshellarg('$useraddress') 实际上将以字符串形式传递给BASH。http://paste.ee/p/arOIF http://bpaste.net/show/141824/ - brad

@Barmar 但这就是问题所在，我的朋友。我需要一个变量，而不是静态数据。那个项目每次都会更改。传递静态数据是可以的，但动态数据会出错。 - brad

来吧，自己写点代码。你不必完全复制我的代码。理解它在做什么，并将其应用到你的需求中。 - Barmar

Answer 2

明白了。

<?php
$upassword = 'test1234'; $uaddress = '[e-mail]';
$uaddress = escapeshellarg($uaddress);
$upassword = escapeshellarg($upassword);
$addr = shell_exec("sudo /home/tpcmedia/cgi-bin/member_add_postfixadmin $uaddress $upassword 2>&1");
?>


#!/bin/bash -x
uaddress=$1
upassword=$2
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

- brad