Laravel Pagination与漂亮的URL比查询字符串 [英] Laravel Pagination with pretty urls than query string

问题描述

在freenode中被称为 Kindari @ irc room #laravel的用户，还可归功于用户 iampseudo 和 Debolaz 。

with credit to the user called Kindari @ irc room #laravel in freenode , also credit goes to user iampseudo and Debolaz.

以下laravel路径代码，

with following laravel route code,

Route::bind('key_pairs', function($s) {
// some logic to transform string to associative array
$arr = explode("/",$s);
$arr2 = array();
if(count($arr)%2 == 0)
{
    for($i=0;$i<count($arr);$i+=2)
    {
        $arr2[$arr[$i]] = $arr[$i+1];
    }
}
return $arr2;   
});

Route::get('foo/{key_pairs}', function($key_pairs) {
var_dump($key_pairs);
})->where('key_pairs', '.*'); 

现在我们可以将/ foo / page / 1用于Laravel读取为/ foo？page = 1 ，但前者比后者更漂亮。

now we can get /foo/page/1 for Laravel to read as /foo?page=1 , but former is more prettier than latter.

现在需要的是Laravel的分页实例读取/ page / 1而不是？page = 1，所以漂亮的分页网址将顺利运行。

now what is needed here is for Laravel's pagination instance to read /page/1 rather than ?page=1 , so pretty pagination urls will work smoothly.

现在有人知道这样做，而不改变基本代码吗？

Does anyone know now to do this , without altering the base code ?

如果我们能有这样的话Users :: paginate（5） - > page（$ page）或任何其他功能（如果已经存在）（我无法找到），这很棒。

if we can have something like Users::paginate(5)->page($page) or any other functionality if already exists (which i am unable to find) , that is great.

欢呼

推荐答案

好的问题解决了，现在Laravel的分页网址正在运行，这就是解决方案。

Okay problem solved , now pretty pagination urls for Laravel is working and here is the solution.

我将getByPage方法添加到以下链接中发布的相关模型类中 http://culttt.com/2014/02/24/working-pagination-laravel-4/ （信用证给他）

i added the getByPage method into the relevant model class posted in the following link http://culttt.com/2014/02/24/working-pagination-laravel-4/ (Credit goes to him)

并在路线中调用 $ this-> user-> getByPage（$ page，$ limit）; ，

我们有漂亮的分页网址！

There we have Pretty pagination urls !

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