從2個連接表中使用datediff

我想在兩個表格中找出兩個日期之間的差異。相比completed_guides.created_at從2個連接表中使用datediff

的uploads.published_at我不斷收到這說的錯誤：

SQLSTATE[42000]: Syntax error or access violation: 1582 Incorrect parameter count in the call to native function 'datediff' (SQL: select users.*,completed_guides.created_at as completed_at,uploads.published_at,datediff(HOUR, published_at, completed_at) from `users` inner join `uploads` on `users`.`id` = `uploads`.`user_id` inner join `completed_guides` on `uploads`.`id` = `completed_guides`.`upload_id` where `completed_guides`.`upload_id` = 2839 limit 10)

這裏是我的代碼。任何幫助，將不勝感激。

$select = [ 
    'users.*', 
    'completed_guides.created_at as completed_at', 
    'uploads.published_at', 
    'datediff(HOUR, published_at, completed_at) as date_diff' 
]; 

return User::select(DB::raw(join(',', $select))) 
    ->join('uploads', 'users.id', '=', 'uploads.user_id') 
    ->join('completed_guides', 'uploads.id', '=', 'completed_guides.upload_id') 
    ->where('completed_guides.upload_id', $this->id) 
    ->take(10) 
    ->get();

來源

2016-12-09 Ben

您可以使用Laravel查詢生成器的whereRaw()這樣的：

return User::select(DB::raw(join(',', $select))) 
      ->join('uploads', 'users.id', '=', 'uploads.user_id') 
      ->join('completed_guides', 'uploads.id', '=', 'completed_guides.upload_id') 
      ->where('completed_guides.upload_id', $this->id) 
      ->whereRaw('datediff(uploads.published_at, completed_guides.completed_at) as date_diff') 
      ->take(10) 
      ->get();

設置日期，您可以使用SQL方法的格式 - date_format(date, format)這樣的：

->select(DB::raw("DATE_FORMAT(date_diff, '%b %d %Y %h:%i %p') as formatted_date_diff"));

見更多about SQL Date Format

希望這有助於！

來源

2016-12-09 15:41:15

管理得到它的工作，需要刪除'HOUR'參數。現在它在0天的差異中返回0，是否有可能以'Y-m-d H：i'格式返回？ – Ben

您可以使用'DATE_FORMAT（date，format）'sql方法，試試這個。我從我的答案中刪除了'HOUR'參數！ –

@BenShepherd - 我已經更新了我的答案，請看看！ –

帶單位的那個，是timestampdiff，不是日期。

timestampdiff(HOUR, published_at, completed_at) as date_diff

來源

2016-12-09 16:12:45

從2個連接表中使用datediff

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