你如何从PHP返回CGI标题？
How does php return CGI headers?
I've been writing CGI 'scripts' in C++, and this is straightforward: I just manually output a Content-type and a Status line. However, I need to use phpmailer, so I also have some php code. The problem is that my simple php code just works, even though I'm not explicitly returning any headers. I've got to the point where I now have to return a failure status code to the client, and I can't see how I'm meant to do this. Any ideas? Is Apache doing this automatically for me and, if so, how do I persuade it to return a 400 status?
Thanks.
php如何返回CGI标题？ p>
我一直在编写CGI C ++中的'scripts'，这很简单：我只需手动输出 谢谢。 p>
div> Content-type code>和 Status code>行。 但是，我需要使用phpmailer，所以我也有一些PHP代码。 问题是我的简单PHP代码正常工作，即使我没有显式返回任何标题。 我已经到了我现在必须将失败状态代码返回给客户端的地步，我无法看到我是如何做到这一点的。 有任何想法吗？ Apache是否为我自动执行此操作，如果是，我如何说服它返回400状态？ p>
Php has this function for things like you need:
Example:
header("HTTP/1.0 404 Not Found");
You may use the header function to add custom headers. Please refer to the header function documentation.