如何使用查詢生成器在laravel 4中編寫此查詢

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) 

我正在使用查詢生成器在laravel 4中查詢以上查詢。但無法找到如何寫這個。我寫這高達：

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

如果您只需要查詢工作，你可以使用這個語法：

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

它的工作原理，它是安全的。