将多个 PHP 变量传递给 shell_exec()？
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Passing multiple PHP variables to shell_exec()?
提问by user2314387
I am calling test.sh from PHP using shell_exec method.
我正在使用 shell_exec 方法从 PHP 调用 test.sh。
$my_url="http://www.somesite.com/";
$my_refer="http://www.somesite.com/";
$page = shell_exec('/tmp/my_script.php $my_url $my_refer');
However, the command line script says it only received 1 argument: the /tmp/my_script.php
但是，命令行脚本说它只收到 1 个参数：/tmp/my_script.php
When i change the call to:
当我将呼叫更改为：
Code:
代码：
$page = shell_exec('/tmp/my_script.php {$my_url} {$my_refer}');
It says it received 3 arguments but the argv[1] and argv[2] are empty.
它说它收到了 3 个参数，但 argv[1] 和 argv[2] 是空的。
When i change the call to:
当我将呼叫更改为：
Code:
代码：
$page = shell_exec('/tmp/my_script.php "http://www.somesite.com/" "http://www.somesite.com/"');
The script finally receives all 3 arguments as intended.
该脚本最终按预期接收了所有 3 个参数。
Do you always have to send just quoted text with the script and are not allowed to send a variable like $var? Or is there some special way you have to send a $var?
您是否总是必须使用脚本发送仅引用的文本，并且不允许发送像 $var 这样的变量？或者有什么特殊的方式必须发送 $var 吗？
采纳答案by Code L?ver
There is need to send the arguments with quota so you should use it like:
需要发送带有配额的参数，因此您应该像这样使用它：
$page = shell_exec("/tmp/my_script.php '".$my_url."' '".$my_refer."'");
回答by Dave Chen
Change
改变
$page = shell_exec('/tmp/my_script.php $my_url $my_refer');
$page = shell_exec('/tmp/my_script.php $my_url $my_refer');
to
到
$page = shell_exec("/tmp/my_script.php $my_url $my_refer");
$page = shell_exec("/tmp/my_script.php $my_url $my_refer");
OR
或者
$page = shell_exec('/tmp/my_script.php "'.$my_url.'" "'.$my_refer.'"');
$page = shell_exec('/tmp/my_script.php "'.$my_url.'" "'.$my_refer.'"');
Also make sure to use escapeshellargon both your values.
还要确保对escapeshellarg您的两个值都使用。
Example:
例子：
$my_url=escapeshellarg($my_url);
$my_refer=escapeshellarg($my_refer);
回答by Orangepill
Variables won't interpolate inside of a single quoted string. Also you should make sure the your arguments are properly escaped.
变量不会插入到单引号字符串中。此外，您应该确保正确转义您的参数。
$page = shell_exec('/tmp/myscript.php '.escapeshellarg($my_url).' '.escapeshellarg($my_refer));
回答by DaoWen
You might find sprintfhelpful here:
您可能会在sprintf此处找到帮助：
$my_url="http://www.somesite.com/";
$my_refer="http://www.somesite.com/";
$page = shell_exec(sprintf('/tmp/my_script.php "%s" "%s"', $my_url, $my_refer));
You should definitely use escapeshellargas recommended in the other answers if you're not the one supplying the input.
escapeshellarg如果您不是提供输入的人，则绝对应该按照其他答案中的建议使用。
回答by bluenapalm
I had difficulty with this so thought I'd share my code snippet.
我对此有困难，所以我想分享我的代码片段。
Before
前
$output = shell_exec("/var/www/sites/blah/html/blahscript.sh 2>&1 $host $command");
After
后
$output = shell_exec("/var/www/sites/blah/html/blahscript.sh 2>&1 $host {$command}");
Adding the {}brackets is what fixed it for me.
添加{}括号是为我修复它的原因。
Also, to confirm escapeshellargis also needed.
此外，escapeshellarg还需要确认。
$host=escapeshellarg($host);
$command=escapeshellarg($command);
Except script also needed:
除了还需要脚本：
set host [lindex $argv 0]
set command [lindex $argv 1]
回答by David Jashi
Change
改变
$page = shell_exec('/tmp/my_script.php $my_url $my_refer');
to
到
$page = shell_exec('/tmp/my_script.php "'.$my_url.'" "'.$my_refer.'"');
Then you code will tolerate spaces in filename.
然后你的代码将容忍文件名中的空格。
