假设我的结构为；

{
    "class" : "TESTCLASS",
    "id" : "HyvF1sdZl",
    "depends" : [
        "S11T5ce",
        "BytQLN6ml",
        "Byzz8Ea7l",
        "r1TUNX58x"
    ],
    "list" : [
        {
            "class" : "DATA",
            "id" : "H113d5Pwx"
        },
        {
            "class" : "DATA",
            "id" : "H113d5Pwx"
        },
        {
            "class" : "TEST",
            "id" : "r1TUNX58x"
        },
        {
            "class" : "DATA",
            "id" : "H113d5Pwx"
        },
        {
            "class" : "TEST",
            "id" : "r1TUNX58x"
        },
        {
            "class" : "TEST",
            "id" : "r1TUNX58x"
        },
        {
            "class" : "FDSAFAS",
            "id" : "S11T5ce"
        },
        {
            "class" : "YENITEST",
            "id" : "r1TUNX58x"
        },
        {
            "class" : "FDSAFAS",
            "id" : "S11T5ce"
        },
        {
            "class" : "CALENDAR",
            "id" : "S11T5ce"
        },
        {
            "class" : "EN",
            "id" : "BytQLN6ml"
        },
        {
            "class" : "CALENDAR",
            "id" : "S11T5ce"
        },
        {
            "class" : "EN_",
            "id" : "Byzz8Ea7l"
        },
        {
            "class" : "TEST",
            "id" : "r1TUNX58x"
        },
        {
            "class" : "EN",
            "id" : "BytQLN6ml"
        },
        {
            "class" : "FDSAFAS",
            "id" : "S11T5ce"
        }
    ]
}

我想project list 项目，其id 包含在depends 数组中。

我的意思是，我只想得到；

"depends" : [
   "S11T5ce",
   "BytQLN6ml",
   "Byzz8Ea7l",
   "r1TUNX58x"
],
"list" : [
    {
        "class" : "FDSAFAS",
        "id" : "S11T5ce"
    },
    {
        "class" : "EN",
        "id" : "BytQLN6ml"
    },
    {
        "class" : "EN_",
        "id" : "Byzz8Ea7l"
    },
    {
        "class" : "TEST",
        "id" : "r1TUNX58x"
    }

]

为了能够做到这一点，我写了；

db.definition.aggregate([
    {
        $project: {
            _id: 0,
            depends :1,
            depends: {$in: ["$list.id"]},
            "list.defid": 1,
            "list.class" :1,
        }

    }
]).pretty() 

但这会引发错误，如 "errmsg": "Expression $in takes exactly 2 arguments. 1 were passed in."

我错过了什么？

最佳答案

对于 MongoDB 3.4，您可以使用 $addFields 管道来投影其他字段而不明确命名它们。作为单个管道，使用 $setUnion 的组合, $filter 和 $in 运算符返回所需的 大批。 <强> $setUnion 运算符有助于在应用 $filter 之前消除重复项 运算符。

为了遵循直觉，让我们按照这个仅使用 $filter 运行的示例管道进行操作 运算符:

db.collection.aggregate([
    {
        "$addFields": {
            "list": {
                "$filter": {
                    "input": "$list",
                    "as": "item",
                    "cond": { "$in": ["$$item.id", "$depends"] }                       
                }
            }
        }
    }
])

示例输出

{
    "_id" : ObjectId("58d0e0d97a3871921504bb69"),
    "class" : "TESTCLASS",
    "id" : "HyvF1sdZl",
    "depends" : [
        "S11T5ce",
        "BytQLN6ml",
        "Byzz8Ea7l",
        "r1TUNX58x"
    ],  
    "list" : [ 
        {
            "class" : "TEST",
            "id" : "r1TUNX58x"
        }, 
        {
            "class" : "TEST",
            "id" : "r1TUNX58x"
        }, 
        {
            "class" : "TEST",
            "id" : "r1TUNX58x"
        }, 
        {
            "class" : "FDSAFAS",
            "id" : "S11T5ce"
        }, 
        {
            "class" : "YENITEST",
            "id" : "r1TUNX58x"
        }, 
        {
            "class" : "FDSAFAS",
            "id" : "S11T5ce"
        }, 
        {
            "class" : "CALENDAR",
            "id" : "S11T5ce"
        }, 
        {
            "class" : "EN",
            "id" : "BytQLN6ml"
        }, 
        {
            "class" : "CALENDAR",
            "id" : "S11T5ce"
        }, 
        {
            "class" : "EN_",
            "id" : "Byzz8Ea7l"
        }, 
        {
            "class" : "TEST",
            "id" : "r1TUNX58x"
        }, 
        {
            "class" : "EN",
            "id" : "BytQLN6ml"
        }, 
        {
            "class" : "FDSAFAS",
            "id" : "S11T5ce"
        }
    ]
}

随着 $setUnion 运算符，在应用 $filter 之前先消除重复项 作为:

db.collection.aggregate([
    {
        "$addFields": { 

            "list": {
                "$filter": {
                    "input": { "$setUnion": ["$list", []] },
                    "as": "item",
                    "cond": { "$in": ["$$item.id", "$depends"] }                       
                }                
            }            
        }
    }
])

示例输出

{
    "class" : "TESTCLASS",
    "id" : "HyvF1sdZl",
    "depends" : [ 
        "S11T5ce", 
        "BytQLN6ml", 
        "Byzz8Ea7l", 
        "r1TUNX58x"
    ],
    "list" : [ 
        {
            "class" : "FDSAFAS",
            "id" : "S11T5ce"
        }, 
        {
            "class" : "TEST",
            "id" : "r1TUNX58x"
        }, 
        {
            "class" : "YENITEST",
            "id" : "r1TUNX58x"
        }, 
        {
            "class" : "EN_",
            "id" : "Byzz8Ea7l"
        }, 
        {
            "class" : "CALENDAR",
            "id" : "S11T5ce"
        }, 
        {
            "class" : "EN",
            "id" : "BytQLN6ml"
        }
    ]
}

关于mongodb - mongodb 查询中的 $project 和 $in，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/42920255/