php - 如何在 php 中有条件地链接方法？

Question

我有以下 php 代码，我想知道是否可以有条件地链接方法。如果角色为空或空 ([])，我无法调用 whereIn。这行得通，但是有没有更简洁的方法呢？

 if ($params['roles']){
        $query1 = QB::table('personnel_role')
            ->select(QB::raw('IF(personnel.nickname != "",CONCAT(IFNULL(personnel.first_name,"")," ", IFNULL(personnel.last_name, ""), " (",personnel.nickname,")"), CONCAT(IFNULL(personnel.first_name,"")," ", IFNULL(personnel.last_name,""))) AS "name", personnel.id'))
            ->leftJoin('personnel', 'personnel_role.personnel_id', '=', 'personnel.id')
            ->whereIn('personnel_role.role_id', $params['roles'])
            ->groupBy('personnel.id')->limit($params['skip'])
            ->offset($params['start']);
        }else{
        $query1 = QB::table('personnel_role')
            ->select(QB::raw('IF(personnel.nickname != "",CONCAT(IFNULL(personnel.first_name,"")," ", IFNULL(personnel.last_name, ""), " (",personnel.nickname,")"), CONCAT(IFNULL(personnel.first_name,"")," ", IFNULL(personnel.last_name,""))) AS "name", personnel.id'))
            ->leftJoin('personnel', 'personnel_role.personnel_id', '=', 'personnel.id')
            ->groupBy('personnel.id')
            ->limit($params['skip'])
            ->offset($params['start']);
 } 

Answer 1

因为除了 whereIn 之外，您的查询基本上是相同的，所以您可以在条件中单独添加它。

$query1 = QB::table('personnel_role')
    ->select(QB::raw('IF(personnel.nickname != "",CONCAT(IFNULL(personnel.first_name,"")," ", IFNULL(personnel.last_name, ""), " (",personnel.nickname,")"), CONCAT(IFNULL(personnel.first_name,"")," ", IFNULL(personnel.last_name,""))) AS "name", personnel.id'))
    ->leftJoin('personnel', 'personnel_role.personnel_id', '=', 'personnel.id')
    ->groupBy('personnel.id')->limit($params['skip'])
     ->offset($params['start']);

if($params['role']) {
    $query1 = $query1->whereIn('personnel_role.role_id', $params['roles']);
}