在 laravel 中加入 table - 分页不起作用

Question

//get all the data of both table

foreach($tables as $table){

 $table_history = $table.'_history';
 $result[] = DB::table($table)->join($table_history, $table.user_id, '=',$table_history.user_id )->paginate(5);

}

当我尝试时

<?php echo $result->links(); ?>

它没有 return 任何东西。

并且还试图print_r结果

<?php print_r($result); ?>

它return编辑：

Illuminate\Pagination\Paginator Object ( [factory:protected] => Illuminate\Pagination\Factory Object ( [request:protected] => Illuminate\Http\Request Object ( [json:protected] => [sessionStore:protected] => [attributes] => Symfony\Component\HttpFoundation\ParameterBag Object ( [parameters:protected] => Array ( ) ) [request] => Symfony\Component\HttpFoundation\ParameterBag Object ( [parameters:protected] => Array ( ) ) [query] => Symfony\Component\HttpFoundation\ParameterBag Object ( [parameters:protected] => Array ( ) ) [server] => Symfony\Component\HttpFoundation\ServerBag Object ( [parameters:protected] => Array ( [DOCUMENT_ROOT] => C:\xampp\htdocs\system101\public [REMOTE_ADDR] => ::1 [REMOTE_PORT] => 53241 [SERVER_SOFTWARE] => PHP 5.4.7 Development Server [SERVER_PROTOCOL]

我试着把变量变成数组：

$result[] = DB::table('user')->join('position', user.user_id, '=',position.user_id )->get();

它显示了所有需要的数据。但是，我还是把它改成了 ->paginate(5) 它 return 和上面的一样。

如何显示分页链接？或者有没有其他方法可以实现分页？

Answer 1

Join 仅接受字符串参数

Laravel 加入语法

join(string $table, string $one, string $operator = null, string $two = null, string $type = 'inner', bool $where = false)

修复查询中的数据库属性字符串

$result = DB::table('user')
             ->join('position', 'user.user_id', '=', 'position.user_id')->paginate(5);

此外...

foreach($tables as $table):

    $table_history = $table.'_history';

    $result = DB::table($table)
                    ->join("$table_history", "$table.user_id", '=', "$table_history.user_id")
                    ->paginate(5);

endforeach;

在 laravel 中加入 table - 分页不起作用

Join table in laravel - Pagination not working

php

foreach

pagination

join

laravel