Question

我有一个新闻报道的网站。在这个，当我点击任何文章，然后它的地址看起来像news.php？id = 12但我希望它看起来像http://paritynews.com/news/YYYY/MM/DD/＆lt; 4位数＆gt; / headline /。在这第一个节目的新闻类别然后文章的日期，然后文章编号，并在最后它显示帖子的标题。这里是id链接http://www.monkks.com/part/parity-newsfinalpage/index.php 日Thnx。

Answer 1

在这里，我发现了一个使用自定义URL的好文章。 http://net.tutsplus.com/tutorials/other/using-htaccess-files-for-pretty-urls/

您可以使用RewriteRule生成所需类型的网址。

Answer 2

使用此.htaccess文件和$ _REQUEST变量。

RewriteEngine on
RewriteCond %{REQUEST_FILENAME} !-d
RewriteCond %{REQUEST_FILENAME} !-f
RewriteRule . index.php [L]

这是index.php文件。我已添加注释，您必须实际运行查询并处理结果。

//Routing...
$request = substr($_SERVER['REQUEST_URI'],1); //Out of the http://www.domain.com/blog/date -> blog/date is returned
$params  = explode("/", $request); //Creates an array $params = ['blog','date'];

if($params[0] == ""){
//Default Route
    $include_path = "default.html";
}else{
//If there is something in the URL that is useful
//Parse the rest of the url here
switch ($params[0]) {//The first varible Ex: 'blog' or 'news'
    case 'news':
        $article_date = join(array($params[1],$params[2],$params[3]),"-");//Formats the date for MySQL
        $id_number = $params[4];
        //Make sure to escpe the values here
        $sql = "SELECT * FROM table_name WHERE (date_column = {$article_date} AND id = {$id_number}";
        //Run the query with your perfered sql extension, mysqli or pdo
        //Handle errors
        $include_path = "news.php";
    break;
    case 'other':
        $include_path = "something_else.php";
        break;
    }
}
if(isset($include_path)){
    require($include_path);
}
?>

使用php mysql创建.htaccess文件。

2 个答案: