如何使用同一個雄辯查詢兩次，但使用一個不同的where子句？

Question

我有這個疑問：

User::leftJoin('friends', function ($join) { 
    $join->on('friends.user_id_1', '=', 'users.id') 
     ->orOn('friends.user_id_2', '=', 'users.id'); 
}) 
->where(function ($query) use ($myID) { 
    // Group orwhere functions so the query builder knows these belong together 
    $query->where([ 
     'friends.user_id_1' => $myID, 
     'friends.accepted' => true 
    ]) 
    ->orWhere([ 
     'friends.user_id_2' => $myID, 
     'friends.accepted' => true 
    ]); 
}) 
->where('users.id', '!=', $myID) // Exclude the user with id $myID 
->get(); 

https://stackoverflow.com/a/41832867/5437864

我想兩次使用此查詢，但使用不同的where子句。是否可以在不復制整個代碼的情況下重用此查詢？如果是這樣，怎麼樣？

Answer 1

$query = User::leftJoin('friends', function ($join) { 
    $join->on('friends.user_id_1', '=', 'users.id') 
     ->orOn('friends.user_id_2', '=', 'users.id'); 
}) 
->where(function ($query) use ($myID) { 
    // Group orwhere functions so the query builder knows these belong together 
    $query->where([ 
     'friends.user_id_1' => $myID, 
     'friends.accepted' => true 
    ]) 
    ->orWhere([ 
     'friends.user_id_2' => $myID, 
     'friends.accepted' => true 
    ]); 
}); 

增加額外哪裏現有的查詢

$query->where('users.id', '!=', $myID) 


$query->get(); 

檢查此鏈接爲另一個例子

Example

Answer 2

我使用的PHP克隆keyword。我不確定這是否是最好的解決方案，但它有幫助。歡迎任何其他建議。

$friends_query = User::leftJoin('friends', function ($join) { 
    $join->on('friends.user_id_1', '=', 'users.id') 
     ->orOn('friends.user_id_2', '=', 'users.id'); 
}) 
->where(function ($query) use ($myID) { 
    // Group orwhere functions so the query builder knows these belong together 
    $query->where([ 
     'friends.user_id_1' => $myID, 
     'friends.accepted' => true 
    ]) 
    ->orWhere([ 
     'friends.user_id_2' => $myID, 
     'friends.accepted' => true 
    ]); 
}); 

$friends_me = clone $friends_query; 
$friends_me = $friends_me->where('users.id', '!=', $myID); 

$friends_others = clone $friends_query; 
$friends_others = $friends_others->where('users.id', '=', $myID);