Question

我有三个加入的模型。我需要在workspace_members为1（即有效）时获取inventory.status表详细信息。我怎样才能在古怪的语言中写下这个条件。以下是我的查询。提前谢谢。

$data['inventories'] = Inventory::
    Join('category_info','inventory.category_id','=','category_info.category_id')
    ->leftJoin('workspace_member_inventory', 'inventory.inventory_id', '=', 'workspace_member_inventory.inventory_id')
    ->leftJoin('workspace_members','workspace_member_inventory.member_id','=','workspace_members.member_id')
    ->where('inventory.workspace_id', '=', $workspace_id)
    ->select('inventory.*', 'workspace_member_inventory.member_id','workspace_members.*','category_info.category_name', 'inventory.status as status')
    ->get()
    ->toArray();

Answer 1

我不知道我是否理解，但我认为您正在寻找Laravel Query Scopes

在您的库存模型上写下范围：

public function scopeIsActive($query,$status = null){
   if(!is_null($status)){
       if($status == 1){
          return $query->leftJoin('workspace_members','workspace_member_inventory.member_id','=','workspace_members.member_id')->where('inventory.status',1);
       }else{
        return $query->where('inventory.status',0);
       }
   }else{
      return $query;
   }
}

然后将其添加到您的查询中：

$data['inventories'] = Inventory::
    Join('category_info','inventory.category_id','=','category_info.category_id')
    ->leftJoin('workspace_member_inventory', 'inventory.inventory_id', '=', 'workspace_member_inventory.inventory_id')
    ->where('inventory.workspace_id', '=', $workspace_id)
    ->isActive(1)
    ->select('inventory.*', 'workspace_member_inventory.member_id','workspace_members.*','category_info.category_name', 'inventory.status as status')
    ->get()->toArray();

基于Laravel Condition的sql查询

1 个答案: