php - Shell command not working via exec() in php

Question

I'm executing a php function via the PHP exec() function but it seems to only pass through 1 variable at time.

$url = "http://domain.co/test.php?phone=123&msg=testing"
exec('wget '.$url);

Within test.php file I have

$msg = "msg =".$_GET["msg"]." number=".$_GET["phone"];
mail('me@gmail.com', 'Working?', $msg);

I get an email which returns phone variable only.

But if I change the url as follows

$url = "http://domain.co/test.php?msg=testing&phone=123"

I get msg but not phone? Any ideas on what is causing this strange behavior?

Answer 1

&符号是 Unix shell 中的一个特殊字符。你需要逃避它：

exec("wget '$url'");

此外，如果您的 URL 以任何方式基于用户输入，请小心使用escapeshellarg. 否则，您的用户将能够在您的服务器上运行任意 Unix 命令。

Answer 2

$url = "http://domain.co/test.php?phone=123&msg=testing"
exec('wget "'.$url.'"');

你需要引用网址

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& 是将任务置于后台的标志