我有CRUD生成的bu Gii。默认情况下,应用程序中的搜索对每个表列使用一个字段。我应该如何更改我的搜索模型,以便通过单独的字段在所有列中进行搜索?
那是我的模特:
class UserSearch extends User
{
/**
* @inheritdoc
*/
public function rules()
{
return [
[['id', 'status', 'created_at', 'updated_at'], 'integer'],
[['username', 'auth_key', 'password_hash', 'password_reset_token', 'email'], 'safe'],
];
}
/**
* @inheritdoc
*/
public function scenarios()
{
// bypass scenarios() implementation in the parent class
return Model::scenarios();
}
/**
* Creates data provider instance with search query applied
*
* @param array $params
*
* @return ActiveDataProvider
*/
public function search($params)
{
$query = User::find();
$dataProvider = new ActiveDataProvider([
'query' => $query,
]);
$this->load($params);
if (!$this->validate()) {
// uncomment the following line if you do not want to any records when validation fails
// $query->where('0=1');
return $dataProvider;
}
$query->andFilterWhere([
'id' => $this->id,
'status' => $this->status,
'created_at' => $this->created_at,
'updated_at' => $this->updated_at,
]);
$query->andFilterWhere(['like', 'username', $this->username])
->andFilterWhere(['like', 'auth_key', $this->auth_key])
->andFilterWhere(['like', 'password_hash', $this->password_hash])
->andFilterWhere(['like', 'password_reset_token', $this->password_reset_token])
->andFilterWhere(['like', 'email', $this->email]);
return $dataProvider;
}
}
答案 0 :(得分:1)
我就是这样做的:
public function search($params)
{
..............................
if($this->keyword) {
if(preg_match("/[A-Za-z]+/", $this->keyword) == true) {
$query->andFilterWhere(['like', 'LOWER(CONCAT(name, age, WHATEVERFIELDS)), ', strtolower($this->keyword)]);
} else {
$query->andFilterWhere(['id' => $this->keyword]);
}
}
关键字实际上不是数据库中的列,它只是我附加到搜索模型的变量。如果关键字是数字,我认为它是一个ID,如果需要,可以忽略此部分。否则我会连接一些字段并搜索结果。您也不能连接字段,只需为每个字段添加一个条件,由您决定。
您可以在以下位置查看完整文件:https://github.com/Mihai-P/yii2-core/blob/master/models/ContactSearch.php