php yii2活动记录查找列不相等

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时间:2020-08-25 16:24:00  来源:igfitidea点击:

yii2 active record find column not equal

phpyii2

提问by Sarvar Nishonboev

I have this code to find a user from db which status is active and role is user

我有此代码可以从数据库中查找状态为活动且角色为用户的用户

public static function findByUsername($username)
{
 return static::find(['username' => $username, 'status' => static::STATUS_ACTIVE, 'role'=>'user']);
}

I need to find a user that role is not equal to "user". How can I do this?

我需要找到一个角色不等于“用户”的用户。我怎样才能做到这一点?

P.S: I'm using YII2

PS:我正在使用 YII2

回答by Alex

I want to offer another solution, it's more elegant for me. I usually use this approach since Yii 1 when i need check not equal operation.

我想提供另一种解决方案,它对我来说更优雅。当我需要检查不相等操作时,我通常从 Yii 1 开始使用这种方法。

$models = Book::find()->where('id != :id and type != :type', ['id'=>1, 'type'=>1])->all();

回答by Stanimir Stoyanov

You can pass custom where clause like this:

您可以像这样传递自定义 where 子句:

andWhere(['!=', 'field', 'value'])

Your final function will look like:

您的最终函数将如下所示:

public static function findByUsername($username)
{
    return static::find()
        ->where([
            'username' => $username,
            'status' => static::STATUS_ACTIVE
        ])
        ->andWhere(['!=', 'role', 'user'])
        ->all();
}

回答by Sarvar Nishonboev

Ok, i've done by this way:

好的,我已经这样做了:

public static function findByUsername($username)
{
    $sql="select * from tbl_user where username=:uname and status=:st and role != 'user'";
    return static::findBySql($sql,[":uname"=>$username,":st"=>static::STATUS_ACTIVE])->one();
}

回答by Manoj Kumar

you can try this one . just an example

你可以试试这个。只是一个例子

 $existEmail = Users::model()->findByAttributes(
        array('email'=>$this->email),
        array(
            'condition'=>'user_id!=:id',
        'params'=>array('id'=>$this->user_id),
        ));

回答by Wit Wikky

You can try:

你可以试试:

public static function findByUsername($username)
{
 $criteria = new CDbCriteria;  
 $criteria->addCondition("username != 'username'");
 $criteria->addCondition("role != 'user'");
 $result = User::model()->find($criteria);
 return $result;
}

Or:

或者:

public static function findByUsername($username)
{
  $result=Users::model()->findByAttributes(array('condition'=>"role != 'user',username = '$username'"));
}