[英]Restler 404 Error
i have just started to create an api with Restler and have the follwoing code. 我刚刚开始使用Restler创建一个api,并具有以下代码。
index.php index.php
require_once '../vendor/restler.php';
use Luracast\Restler\Defaults;
use Luracast\Restler\Restler;
Defaults::$useUrlBasedVersioning = true;
$r = new Restler();
$r->setAPIVersion(1);
$r->setSupportedFormats('XmlFormat', 'JsonFormat');
$r->addAPIClass('Info', '');
$r->addAPIClass('Info');
$r->addAPIClass('getList');
$r->handle();
info.php info.php
<?php
class Info {
function index() {
return "Version 1.0.0";
}
}
?>
v1/getList.php v1 / getList.php
<?php
namespace v1;
use stdClass;
class getList
{
function index()
{
return "hello";
}
function newest() {
return "hello2";
}
}
I'm testing it with xampp on a windows machine and localhost/api/ is mapped to the public folder. 我正在Windows机器上使用xampp测试它,并且localhost / api /映射到公用文件夹。 When i open localhost/api/ in the browser it will show Version 1.0 as expected.
当我在浏览器中打开localhost / api /时,它将按预期显示版本1.0 。 But when i open info or getList i get a 404 error from apache.
但是当我打开info或getList时,我从apache收到404错误。
How can i fix it? 我该如何解决? Many thanks for your help
非常感谢您的帮助
You should add this to your /api/.htaccess
file: 您应该将此添加到您的
/api/.htaccess
文件中:
DirectoryIndex index.php
<IfModule mod_rewrite.c>
RewriteEngine On
RewriteRule ^$ index.php [QSA,L]
RewriteCond %{REQUEST_FILENAME} !-f
RewriteCond %{REQUEST_FILENAME} !-d
RewriteRule ^(.*)$ index.php [QSA,L]
</IfModule>
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