首页 文章

Symfony2 Mongodb聚合 - 获取字段的最大值

提问于
浏览 次
0

想要获得该字段的最大值 speed 及其相应的其他字段 .

有这个MongoDB查询：

db.Neo.aggregate({
    $group: {
        _id: '$name',
        date: { $first: '$date' },
        neo_reference_id: { $first: '$neo_reference_id' },
        name: { $first: '$name' },
        speed: {
            $max: "$speed"
        },
        is_potentially_hazardous_asteroid: { $first: '$is_potentially_hazardous_asteroid' }        
    }        
});

现在因为我使用的是Symfony2和mongo-odm-aggregation-bundle，所以想要执行这个查询：

public function neofastestAction() {

    $expr = new \Solution\MongoAggregation\Pipeline\Operators\Expr;
    $aq = $this->get('doctrine_mongodb.odm.default_aggregation_query')->getCollection('NeoNasaBundle:Neo')->createAggregateQuery();

    $result = $aq->group(['_id'=>'$name' ])
        ->getQuery()->aggregate()->toArray();        
    $serializer = $this->get('jms_serializer');
    $response = $serializer->serialize($result,'json');
    return new Response($response);
}

这会执行它，但只打印 _id . 如何从上面构建我的MondoDB查询到Symfony2 mongo-odm-aggregation-bundle（https://packagist.org/packages/solution/mongo-odm-aggregation-bundle）并获取最后一行？它应该是 speed 的最高值，所以最快的小行星 .

谢谢 .

mongodb symfony doctrine-odm

1 回答

  • 0

    这样做：

    public function neofastestAction() {

      $expr = new \Solution\MongoAggregation\Pipeline\Operators\Expr;
      $aq = $this->get('doctrine_mongodb.odm.default_aggregation_query')->getCollection('NeoNasaBundle:Neo')->createAggregateQuery();

      $result = $aq->group(['_id'=>'$name', 'date'=> $expr->first('$date'),  'neo_reference_id'=> $expr->first('$neo_reference_id'), 'name'=> $expr->first('$name'), 'speed'=> $expr->max('$speed'),'is_potentially_hazardous_asteroid'=> $expr->first('$is_potentially_hazardous_asteroid') ])
             ->sort(['speed' => -1])->limit(1)->getQuery()->aggregate()->toArray();


      $serializer = $this->get('jms_serializer');
      $response = $serializer->serialize($result,'json');

      return new Response($response);
    }
    回复于

相关问题