Question

如何将两个条件传递给query where的方法？我尝试使用$query->where($cond[0])，它有效，但在$query->where($cond[1])上，它给出了未定义偏移量的错误。

public function search($params, $cond)
{
    $query = ServiceStatuses::find();

    $dataProvider = new ActiveDataProvider([
        'query' => $query,
        'sort' => ['defaultOrder' => ['last_hard_state' => SORT_DESC]],
    ]);

    $this->load($params);

    .
    .
    .

    $query->where($cond[0]);      //<<<<----- This works
    $query->where($cond[1]);      //<<<<----- Undefined offset: 1

    return $dataProvider;
}

public function searchIncidents($params)
{
    $cond = array (['last_hard_state' => 0, 'last_hard_state' => 2]);
    $dataProvider = $this->search($params, $cond);
    return $dataProvider;
}

Answer 1

您的$cond变量正在数组中创建一个数组;您正在使用array()和速记语法[]（在PHP 5.4中添加）。如果您只想将$cond作为单级数组，请删除其中一个数组调用。任

$cond = ['last_hard_state' => 0, 'last_hard_state' => 2];

或

$cond = array('last_hard_state' => 0, 'last_hard_state' => 2);

传递一个条件数组来查询Yii2中的方法

1 个答案: