How to display Apache's default 404 page in PH

问题:

I have a webapp that needs to process the URI to find if a page exists in a database. I have no problem directing the URI to the app with .htaccess:

Options +FollowSymlinks
RewriteEngine on
RewriteCond %{SCRIPT_FILENAME} !-f
RewriteRule ^(.*)$ index.php?p=$1 [NC]

My problem is that if the page does not exist, I do not want to use a custom 404 handler written in PHP, I would like do show the default Apache 404 page. Is there any way to get PHP to hand execution back to Apache when it has determined that the page does not exist?

回答1:

The only possible way I am aware of for the above scenario is to have this type of php code in your index.php:

<?php
if (pageNotInDatabase) {
   header('Location: ' . $_SERVER["REQUEST_URI"] . '?notFound=1');
   exit;
}

And then slightly modify your .htaccess like this:

Options +FollowSymlinks -MultiViews
RewriteEngine on
RewriteCond %{SCRIPT_FILENAME} !-f
RewriteCond %{QUERY_STRING} !notFound=1 [NC]
RewriteRule ^(.*)$ index.php?p=$1 [NC,L,QSA]

That way Apache will show default 404 page for this special case because of extra query parameter ?notFound=1 added from php code and with the negative check for the same in .htaccess page it will not be forwarded to index.php next time.

PS: A URI like /foo, if not found in database will become /foo?notFound=1 in the browser.

回答2:

I don't think you can "hand it back" to Apache, but you can send the appropriate HTTP header and then explicitly include your 404 file like this:

if (! $exists) {
    header("HTTP/1.0 404 Not Found");
    include_once("404.php");
    exit;
}

Update

PHP 5.4 introduced the http_response_code function which makes this a little easier to remember.

if (! $exists) {
    http_response_code(404);
    include_once("404.php");
    exit;
}

回答3:

Call this function:

http_send_status(404);