Laravel連接表和連接行

所以我有兩個表，組織和聯繫人。這兩個表都有列「電子郵件」，我需要做的是保持組織的名稱，但在電子郵件列中連接所有電子郵件（組織的+所有聯繫電子郵件）。Laravel連接表和連接行

這裏有一些版本我沒有運氣

1）這一個沒有試過組：

$customers = DB::table('customers') 
    ->whereRaw('LENGTH(customers.email) > 4') 
    ->select([ 
     'customers.id', 
     'customers.name', 
     'customers.email' 
    ]); 

$contacts = DB::table('contacts') 
    ->whereRaw('LENGTH(contacts.email) > 4') 
    ->leftJoin('customers', 'contacts.customer_id', '=', 'customers.id') 
    ->select([ 
     'customers.id', 
     'customers.name', 
     'contacts.email' 
    ]); 

return $customers 
    ->union($contacts) 
    ->select([ 
     'id', 
     'name', 
     DB::raw('GROUP_CONCAT(DISTINCT email, ", ") AS emails'), 
    ]) 
    ->groupBy('id') 
    ->get();

2）這個實際上是非常接近，但它不篩選出項目既沒有接觸或客戶entires有DB::raw('LENGTH(email) > 4')

return $customers = DB::table('customers') 
       ->leftJoin('contacts', 'contacts.customer_id', '=', 'customers.id') 
       ->select([ 
        'customers.id', 
        'customers.name', 
        'registration', 
        DB::raw('GROUP_CONCAT(DISTINCT contacts.email, ", ") AS contact_emails'), 
        'customers.email' 
       ]) 
       ->groupBy('customers.id') 
       ->get();

3）我試圖與子查詢更貼近（我知道這隻會過濾掉，沒有電子郵件聯繫人）

3.1）嘗試子查詢1周的結果在誤差：JoinClause::whereRaw()不存在

return $customers = DB::table('customers') 
      ->leftJoin('contacts', function($join) { 
       $join->on('contacts.customer_id', '=', 'customers.id') 
        ->whereRaw('LENGTH(email) > 4'); 
      })...

3.2）這一個產生下面的語法錯誤：

return $customers = DB::table('customers') 
      ->leftJoin('contacts', function($join) { 
       $join->on('contacts.customer_id', '=', 'customers.id') 
        ->where(DB::raw('LENGTH(email) > 4')); 
      })

1/2 PDOException in Connection.php line 333: SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '? group by customers . id ' at line 1

2/2 QueryException in Connection.php line 713: SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '? group by customers . id ' at line 1 (SQL: select customers . id , customers . name , registration , GROUP_CONCAT(DISTINCT contacts.email, ", ") AS contact_emails, customers . email from customers left join contacts on contacts . customer_id = customers . id and LENGTH(contacts.email) 4 group by customers . id)

3.3）一些實例說我應該這樣做，但這會產生錯誤Not enough arguments for the on clause.

return $customers = DB::table('customers') 
    ->leftJoin('contacts', function($join) { 
     $join->on('contacts.customer_id', '=', 'customers.id'); 
     $join->on(DB::raw('LENGTH(contacts.email) > 4')); 
    })

來源

2016-12-02 Edmund Sulzanok

我首先想到的是子查詢：http://dev.mysql.com/doc/refman/5.7/en/subqueries.html - 但問題實際上就是爲什麼要分析所有電子郵件。他們以後會不會在單獨的形式中有任何價值？ – Kjell

我正在通過yajra datatables在單個表中提供結果。 –

@Kjell我實際上試過子查詢。用失敗的例子更新了我的問題。 –

這適用於我。沒有語法錯誤和過濾器不與接觸長度小於4個字符：

DB::table('customers') 
    ->leftJoin('contacts', function ($join) { 
     $join->on('contacts.customer_id', '=', 'customers.id') 
       ->where(DB::raw('length(contacts.email)'), '>', 4); 
    }) 
    ->select([ 
     'customers.id', 
     'customers.name', 
     DB::raw('group_concat(distinct contacts.email separator ", ") AS contact_emails'), 
    ]) 
    ->groupBy('customers.id') 
    ->get();

測試在Laravel 5.3.26時，MySQL 5.6.20（沒有嚴格的模式）。

來源

2016-12-02 19:42:11

Laravel連接表和連接行

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