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: > "
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