Question

我有以下查询：

$items = UserItems::join('items','items.id','=','user_items.item_id')
            ->where('user_id','=',$this->id)
            ->where('quantity','>',0)
            ->where('items.type',"stones")
            ->orWhere('items.type',"fish")
            ->get();

我的问题是此查询返回所有用户的项目而不是给定的user_Id.。这是因为orWhere适用于所有where's，但我需要它仅适用于此：

->where('items.type',"stones")
                ->orWhere('items.type',"fish")

如何修复查询以仅返回给定的$ this-＆gt; id用户项？

Answer 1

Laravel supports Advanced Where Clauses like such:

$items = UserItems::join('items','items.id','=','user_items.item_id')
    ->where('user_id','=',$this->id)
    ->where('quantity','>',0)
    ->where(function($query) {
        $query->where('items.type',"stones")
            ->orWhere('items.type',"fish");
    })
    ->get();

I guess another option (cleaner, easier to read IMO), as opposed to using orWhere, could be to use the whereIn operator like:

$items = UserItems::join('items','items.id','=','user_items.item_id')
    ->where('user_id','=',$this->id)
    ->where('quantity','>',0)
    ->whereIn('items.type', ["stones", "fish"])
    ->get();

Answer 2

使用闭包，如下所示：

$userId = $this->id;
$items = UserItems::join('items','items.id','=','user_items.item_id')
->where('user_id','=', $userId)
->where('quantity','>',0)
->where(function($query) {
    $query
    ->where('items.type',"stones")
    ->orWhere('items.type',"fish");
})
->get();

Laravel Where和OrWhere有多种条件

2 个答案: