Mongodb通过id在数组中findOne对象

Question

我试图通过用户 ID 和选定的字段和项目找到我的数据库文档的特定条目。我希望返回 items 对象。这是文档结构：

{
"_id": ObjectId("58edfea4b27fd0547375eeb4"),
"user_id": ObjectId("58d2dd4c8207c28149dbc748"),
"calories": 2000,
"date": 20170312,
"snacks": [ ],
"dinner": [
 {
  "nutrients": {
    "protein": "11.6",
    "carbs": "29.4",
    "fat": "7.9"
  },
  "servings": "75",
  "calories": 750,
  "name": "Meat feast stone baked pizza"
}, 
{
  "nutrients": {
    "protein": "6.8",
    "carbs": "54",
    "fat": "30.6"
  },
  "servings": "25",
  "calories": 550,
  "name": "Mc Coy's Cheddar and onion"
}],
"lunch": [],
"breakfast": [],
}

到目前为止，我所做的是获取用户 ID、日期以及要搜索的选定餐点和项目。我返回的是整个膳食数组，但是我只想要返回食品对象。

user_food.findOne({user_id : req.session.user_id, date: today},{'dinner': 'Meat feast stone baked pizza'},function(err, item){
        if(err){
            console.log("something went wrong: " + err);
            return res.status(500).send(err);
        }
        else{
            console.log(item);
            return res.status(200).send(item);
        }
    });

我得到的回报是这样的：

"dinner": [
{
  "nutrients": {
    "protein": "11.6",
    "carbs": "29.4",
    "fat": "7.9"
  },
  "servings": "75",
  "calories": 750,
  "name": "Meat feast stone baked pizza"
}, 
{
  "nutrients": {
    "protein": "6.8",
    "carbs": "54",
    "fat": "30.6"
  },
  "servings": "25",
  "calories": 550,
  "name": "Mc Coy's Cheddar and onion"
}]

我想要的只是：

    {
  "nutrients": {
    "protein": "11.6",
    "carbs": "29.4",
    "fat": "7.9"
  },
  "servings": "75",
  "calories": 750,
  "name": "Meat feast stone baked pizza"
}

Answer 1

尝试这个：

user_food.
findOne({
        user_id : req.session.user_id, 
        date: today,
        'dinner.name': 'Meat feast stone baked pizza'
    },{
        'dinner.$' : 1
},function(err, item){
    ....
});

dinner.$只会返回符合条件的晚餐项目，即 where dinner.name : Meat feast stone baked pizza。

阅读$(positional) 运算符 以了解更多关于限制查询结果中的数组内容并返回与查询文档匹配的元素。