Yii2 model::find() 使用默认 where 条件 [英] Yii2 model::find() with default where conditions

问题描述

在视图中使用 Yii2...

Using Yii2 in the View...

Products::find()->asArray()->all()

将所有产品作为数组返回.我正在寻找一种方法让它返回所有产品 WHERE id != 1我只想让一个地方修改->all()"为每个模型返回的内容.我知道 Product::find()->where('id != 1')->... 是可能的，但我不想编写和维护它不止一处.

returns all products as array. I'm looking for a way to make it return all products WHERE id != 1 I want to have only one place do modify what "->all()" returns for every model. I know that Product::find()->where('id != 1')->... is possible, but I don't want to write and maintain it in more than one place.

推荐答案

1) 您可以简单地覆盖模型中的 find() 方法:

1) You can simply override find() method in your model:

/**
 * @return \yii\db\ActiveQuery
 */
public static function find()
{
    return parent::find()->where(['<>', 'id', 1]);
}

用法:

$products = Products::find()->all();

2) 使用范围.

创建自定义查询类:

namespace app\models;

use yii\db\ActiveQuery;

class ProductQuery extends ActiveQuery
{
    public function withoutFirst()
    {
        $this->andWhere(['<>', 'id', 1]);

        return $this;
    }
}

覆盖模型中的 find() 方法:

Override find() method in your model:

namespace app\models;

use yii\db\ActiveRecord;

class Product extends ActiveRecord
{
    /**
     * @inheritdoc
     * @return ProductQuery
     */
    public static function find()
    {
        return new ProductQuery(get_called_class());
    }
}

然后你可以这样使用它:

Then you can use it like this:

$products = Products::find()->withoutFirst()->all();

我认为使用第二种方法更灵活，因为它使代码更清晰.

I think using second method is more flexible, because it makes code more clear.

附加说明:

• 硬编码 id 不是好习惯.最好用等效条件替换它.

• Hardcoded id is not good practice. Better replace it with equivalent condition.

对于这个例子，我使用了不同的方式来指定条件.查看在 官方文档.

For this examples I used different way of specifying condition. See different ways of specifying condition in where statement in official documentation.

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