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Accepted
ilyaplot
Asked:2020-07-29 19:58:01 +0000 UTC2020-07-29 19:58:01 +0000 UTC

如何为ON添加条件？

  • 772

没有办法在on上为依赖添加条件，有这么一个模型：

class Parent extends ActiveRecord
{
    public function getChild()
    {
        return $this->hasMany(Child::class, ['parent_id' => 'parent_id'])->alias('c');
    }
}

我得到这样的数据：

$query = Parent::find()
    ->alias('p')
    ->distinct()
    ->joinWith([
        'child c'
    ], true, 'inner join');

结果，请求看起来像这样：

select 
  p.* 
from parent p
inner join child c 
  on c.parent_id = p.parent_id

问题是我不知道如何获得这样的请求：

select 
  p.* 
from parent p
inner join child c 
  on c.parent_id = p.parent_id and c.is_disabled = false
php

1 个回答

  1. Best Answer

    ActiveQuery 文档解决了这个问题\ http://www.yiiframework.com/doc-2.0/yii-db-activequery.html#onCondition()-detail 

    class Parent extends ActiveRecord
{
    public function getChild()
    {
        return $this->hasMany(Child::class, ['parent_id' => 'parent_id'])
            ->alias('c')
            ->onCondition('c.is_disabled = false');
    }
};
    • 1

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