Question

假设以下结构：

'kitchenThings': [
    {'foods': [
        {'name': 'cakes',
         'ingredients': [
            {name: 'Sugar', healthy: false},
            {name: 'Butter', healthy: false},
            {name: 'Flour', healthy: true}
          ]
        }
    }
]

我将运行什么查询来仅检索包含healthy: false的kitchenThings？我的第一个猜测是

{ { 'foods.ingredients.healthy': false } }

但返回0条记录。我不知道从哪里去。有人可以帮忙吗？我的预期产量完全如上，但不匹配的成分（面粉）不存在。只需返回匹配的成分（可能是通过投影？）也可以。

由于

Answer 1

要获取包含至少一种 healthy : false成分的食物清单，您可以使用aggregation pipeline。

一种可能的解决方案可能如下：

db.test.aggregate([
    {"$unwind": "$foods"},
    {"$match": {
        "foods.ingredients.healthy": false
    }},
    {"$project":{
        "_id": 0,
        "food": "$foods"
    }}
])

测试用例：

db.test.insert({
    'foods': [
        {'name': 'cakes',
         'ingredients': [
            {name: 'Sugar', healthy: false},
            {name: 'Butter', healthy: false},
            {name: 'Flour', healthy: true}
          ]
        },
        {'name': 'fuits',
         'ingredients': [
            {name: 'Fiber', healthy: true},
            {name: 'Vitamins', healthy: true}
          ]
        },
        {'name': 'cookies',
         'ingredients': [
            {name: 'Sugar', healthy: false},
            {name: 'Butter', healthy: false},
            {name: 'Flour', healthy: true}
          ]
        },
    ]
})

结果：

{ "food" : { "name" : "cakes", "ingredients" : [ { "name" : "Sugar", "healthy" : false }, { "name" : "Butter", "healthy" : false }, { "name" : "Flour", "healthy" : true } ] } }
{ "food" : { "name" : "cookies", "ingredients" : [ { "name" : "Sugar", "healthy" : false }, { "name" : "Butter", "healthy" : false }, { "name" : "Flour", "healthy" : true } ] } }

查询Mongo中的列表

1 个答案: