如何使用同一个雄辩查询两次，但使用一个不同的where子句？

Question

我有这个疑问：

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

我想两次使用此查询，但使用不同的where子句。是否可以在不复制整个代码的情况下重用此查询？如果是这样，怎么样？

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 
    ]); 
}); 

增加额外哪里现有的查询

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


$query->get(); 

检查此链接为另一个例子

Example

Answer 2

我使用的PHP克隆keyword。我不确定这是否是最好的解决方案，但它有帮助。欢迎任何其他建议。

$friends_query = User::leftJoin('friends', function ($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);