Php Laravel MongoDB：我们能用它拉一个特定的子文档吗'；谁的身份证？

这是我的收藏

{
    "_id": "987654322",
    "subscriptions": [
        {
            "list_id": "13"
        },
        {
            "list_id": "12"
        }
    ],
    "offers": [
        {
            "campaign_id": 47,
            "title": "MasalaBox Awesome Sunday Offer",
            "description": "50% off on ALL online purchases!!",
            "thunbnailURL": "thumb/5m9Oav_images (4).jpg",
            "detailImageURL": "img/BpWyBT_original_290019_9BoDnY6NTBGpQq84atetEsA5Z.jpg",
            "detailedDescription": "Masalabox proudly presents the Awesome Sunday offer!! More info goes here. And some more here. Blah blah blah blah blah blah blah blah blah blah blah blah blah",
            "targetURL": "http://www.altavista.com/",
            "end_date": "2015-02-28"
        }
    ]
}

我只想做的是：（1）在“offers”字段的所有子文档中搜索集合中的“campaign_id”47。（2） 如果子文档中存在“campaign_id”47，我想删除整个47文档

 {
            "campaign_id": 47,
            "title": "MasalaBox Awesome Sunday Offer",
            "description": "50% off on ALL online purchases!!",
            "thunbnailURL": "thumb/5m9Oav_images (4).jpg",
            "detailImageURL": "img/BpWyBT_original_290019_9BoDnY6NTBGpQq84atetEsA5Z.jpg",
            "detailedDescription": "Masalabox proudly presents the Awesome Sunday offer!! More info goes here. And some more here. Blah blah blah blah blah blah blah blah blah blah blah blah blah",
            "targetURL": "http://www.altavista.com/",
            "end_date": "2015-02-28"
        }

我过去常常按下面的说明推

$offer = [
        'campaign_id' => $campaign->id,
        'title'       => $template->title,
        'description' => $template->description,
        'thunbnailURL' => $template->thunbnailURL,
        'detailImageURL' => $template->detailImageURL,
        'detailedDescription' => $template->detailedDescription,
        'targetURL'     => $template->targetURL,
        'end_date'      => $template->end_date,

    ];
    $subscribers= Subscription::where('offers.campaign_id', $campaign->id)->get();
    foreach ($subscribers as $subscriber) {
        Subscription::where('_id',$subscriber->_id)->push('offers', array($offer));
    }

我认为下面的代码可以解决我的问题，但是，如何在Laravel中实现它呢

  dbh.subscription.update({"_id": "987654322"}, {"$unset":{"offers.campaign_id":47}},False,False)

我已经试着解决这个问题好几天了，但还是没办法解决。

---

可以在Laravel MongoDB中执行吗？

可以使用Jensegers软件包执行。但是，如果要对数据进行深度嵌套，则需要更改集合中文档的结构

"offers": { 
        "47" : {
           "title": "MasalaBox Awesome Sunday Offer",
           "description": "50% off on ALL online purchases!!",
           "thunbnailURL": "thumb/5m9Oav_images (4).jpg",
           "detailImageURL": "img/BpWyBT_original_290019_9BoDnY6NTBGpQq84atetEsA5Z.jpg",
           "detailedDescription": "Masalabox proudly presents the Awesome Sunday offer!! More info goes here. And some more here. Blah blah blah blah blah blah blah blah blah blah blah blah blah",
           "targetURL": "http://www.altavista.com/",
           "end_date": "2015-02-28"
    }
}

然后是查询

Subscription::where('offers.47', 'exists', true)->unset('offers.47');

这将删除特定的子文档。尽管我建议不要深入嵌套，因为这样会给搜索带来麻烦

如果你更深入一点，可能最好创建一个新的订阅模型，并使用belongtomany-belongtomany关系将两者连接起来

致以最良好的祝愿

Bdschr