我想在这个路由上创建一个404错误页面。如何做到这一点？
index.php
<?php
include 'route.php';
include 'control/about.php';
include 'control/home.php';
include 'control/contact.php';
$route = new route();
$route->add('/', function(){
echo 'Hello, this is home pageeees';
});
$route->add('/about', 'about');
$route->add('/contact', 'contact');
echo '<pre>';
print_r($route);
$route->submit();
?>
route.php
<?php
class route
{
private $_uri = array();
private $_method = array();
/**
*Builds a collection of internal URL's to look for
*@parameter type $uri
*/
public function add($uri, $method = null)
{
$this->_uri[] = '/' . trim($uri, '/');
if($method != null){
$this->_method[] = $method;
}
}
/**
*Makes the thing run!
*/
public function submit()
{
$uriGetParam = isset($_GET['uri'])? '/' . $_GET['uri'] : '/';
foreach ($this->_uri as $key => $value)
{
if(preg_match("#^$value$#",$uriGetParam))
{
if(is_string($this->_method[$key]))
{
$useMethod = $this->_method[$key];
new $useMethod();
}
else
{
call_user_func($this->_method[$key]);
}
}
}
}
}
?>
记住的regex我只会在所有其他路由之后添加$route->add("/.+", function()...);。
证明：https://3v4l.org/nk5PZ
但您也可以编程一些逻辑，当无法找到相应的路由时，评估自定义404页面。
例如：
private $_notFound;
public function submit() {
$uriGetParam = isset($_GET['uri'])? '/' . $_GET['uri'] : '/';
$matched = false;
foreach ($this->_uri as $key => $value) {
if(preg_match("#^$value$#", $uriGetParam)) {
if(is_string($this->_method[$key])) {
$useMethod = $this->_method[$key];
new $useMethod();
} else {
call_user_func($this->_method[$key]);
}
$matched = true;
}
}
if(!$matched) {
if(isset($this->_notFound)) {
if(is_string($this->_notFound)) {
$action = $this->_notFound;
new $action();
} else {
call_user_func($this->_notFound);
}
}
}
}
public function notFound($callback) {
$this->_notFound = $callback;
}
然后您必须通过$route->notFound(function... or "class"); 添加404操作