Question

update `access_tokens` as at3 set at3.`expires` = 1393995576 
where 
`access_token_expires` > 1393995576
and  at3.`id` in (select * from (
select `at2`.`id` from `refresh_tokens` as `rt` 
left join `access_tokens` as `at` on `at`.`id` = `rt`.`access_token_id` 
inner join `access_tokens` as `at2` on `at2`.`session_id` = `at`.`session_id` 
where `rt`.`refresh_token` = 's2kF5Ev6NXncnTPwVz99ksgsCGXfwPIDzXJMZJqz') as t)

我正在尝试使用查询构建器在laravel 4中进行上述查询。但是无法找到如何写这个。我写到这个：

DB::table("access_tokens as at3")
->whereIn('id', function($query) use ($authParams) 
     { 

                $query->select('at2.id')
                ->from('refresh_tokens as rt')
                ->leftJoin("access_tokens as at", "at.id", "=", "rt.session_access_token_id")
                ->join("access_tokens as at2", "at2.session_id", "=", "at.session_id")
                ->where('rt.refresh_token', '=', $authParams['refresh_token']);

            })
        //->delete();
        ->where('at3.access_token_expires', '>', time())
        ->update(array('at3.access_token_expires' =>  time()));

Answer 1

如果您只需要查询，则可以使用以下语法：

$affected_rows = DB::update( DB::raw("UPDATE users set username = 'wadus'  WHERE username = :somevariable"), array('somevariable' => $someVariable));

它有效并且安全。

如何使用查询生成器在laravel 4中编写此查询

1 个答案: