MongoDB中的多组是什么意思？

Question

我的收藏看起来像这样。

{
"_id" : ObjectId("572c4ed33c1b5f51215219a8"),

"name" : "This is an angular course, and integeration with php",
"description" : "After we connected we can query or update the database just how we would using the mongo API with the exception that we use a callback. The format for callbacks is always callback(error, value) where error is null if no exception has occured. The update methods save, remove, update and findAndModify also pass the lastErrorObject as the last argument to the callback function.",
"difficulty_level" : "Beginner",
"type" : "Fast Track",
"tagged_skills" : [ 
    {
        "_id" : "5714e894e09a0f7d804b2254",
        "name" : "PHP"
    }, 
    {
        "_id" : "5717355806313b1f1715fa50",
        "name" : "c++"
    }, 
    {
        "_id" : "5715025bc2c5dbb4675180da",
        "name" : "java"
    }, 
    {
        "_id" : "5714f188ec325f5359979e33",
        "name" : "symphony"
    }
]}

我希望能够按照类型、难度级别和标记技能对集合进行分组，并在一个查询中获取计数。

我无法添加技能计数。

我的查询如下：

db.course.aggregate([
{$unwind:"$tagged_skills"},
{$group:{
    _id:null,
skills: { $addToSet: "$tagged_skills.name" },
Normal_df:{$sum:{ 
                "$cond": [
                    { "$eq":[ "$difficulty_level","Normal"] },
                    1,
                    0
                ]
            }},
Beginner_df:{$sum:{ 
                "$cond": [
                    { "$eq":[ "$difficulty_level","Beginner"] },
                    1,
                    0
                ]
            }},
Intermediate_df:{$sum:{ 
                "$cond": [
                    { "$eq":[ "$difficulty_level","Intermediate"] },
                    1,
                    0
                ]
            }},
Advanced_df:{$sum:{ 
                "$cond": [
                    { "$eq":[ "$difficulty_level","Advanced"] },
                    1,
                    0
                ]
            }},
Fast_Track_type:{$sum:{ 
                "$cond": [
                    { "$eq":[ "$type","Fast Track"] },
                    1,
                    0
                ]
            }},
Normal_type:{$sum:{ 
                "$cond": [
                    { "$eq":[ "$type","Normal"] },
                    1,
                    0
                ]
            }},
Beginner_type:{$sum:{ 
                "$cond": [
                    { "$eq":[ "$type","Beginner"] },
                    1,
                    0
                ]
            }},
Normal_Track_type:{$sum:{ 
                "$cond": [
                    { "$eq":[ "$type","Normal Track"] },
                    1,
                    0
                ]
            }},
    }}
])

结果如下所示：-

{
    "_id" : null,

    "skills" : [ 
        "SQL", 
        "PHP", 
        "java", 
        "Angular Js", 
        "Laravel 23", 
        "c++", 
        "Node Js", 
        "symphony", 
        "Mysql", 
        "Express Js", 
        "JAVA"
    ],
    "Normal_df" : 1,
    "Beginner_df" : 14,
    "Intermediate_df" : 7,
    "Advanced_df" : 2,
    "Fast_Track_type" : 8,
    "Normal_type" : 6,
    "Beginner_type" : 1,
    "Normal_Track_type" : 9
}

我希望你能翻译出所有技能及其数量。

- jarry jafery

Answer 1

要获取所有技能及其数量，您需要首先获取所有技能的列表。您可以通过在适当的字段上运行distinct命令来获得该列表。有了这个列表，您就可以构建适当的$group管道文档，使用$sum和$cond运算符。

考虑以下用例：

var difficultyLevels = db.course.distinct("difficulty_level"),
    types = db.course.distinct("type"),
    skills = db.course.distinct("tagged_skills.name"),
    unwindOperator = { "$unwind": "$tagged_skills" },
    groupOperator = { 
        "$group": { 
            "_id": null,
            "skills": { "$addToSet": "$tagged_skills.name" }
        }       
    };

difficultyLevels.forEach(function (df){ 
    groupOperator["$group"][df+"_df"] = { 
       "$sum": { 
           "$cond": [ { "$eq": ["$difficulty_level",  df] }, 1, 0] 
       }
   }
});

types.forEach(function (type){ 
    groupOperator["$group"][type.replace(" ", "_")+"_type"] = { 
       "$sum": { 
           "$cond": [ { "$eq": ["$type",  type] }, 1, 0] 
       }
   }
});

skills.forEach(function (skill){ 
    groupOperator["$group"][skill] = { 
       "$sum": { 
           "$cond": [ { "$eq": ["$tagged_skills.name",  skill] }, 1, 0] 
       }
   }
});

//printjson(groupOperator);
db.course.aggregate([unwindOperator, groupOperator]);

---

在第一行中，我们通过在difficulty_level字段上运行distinct命令来获取一个难度级别的数组。

db.course.distinct("difficulty_level")

这将产生数组。

var difficultyLevels = ["Normal", "Beginner", "Intermediate", "Advanced"]

同样地，之前的distinct操作将返回该键可能的唯一值列表。

获取这些列表后，您可以使用forEach()方法创建管道对象，以填充列表中每个给定项目的文档键。然后，您可以使用生成的文档，它看起来像这样：

printjson(groupOperator);
{
    "$group" : {
        "_id" : null,
        "skills" : {
            "$addToSet" : "$tagged_skills.name"
        },
        "Beginner_df" : {
            "$sum" : {
                "$cond" : [
                    {
                        "$eq" : [
                            "$difficulty_level",
                            "Beginner"
                        ]
                    },
                    1,
                    0
                ]
            }
        },
        "Intermediate_df" : {
            "$sum" : {
                "$cond" : [
                    {
                        "$eq" : [
                            "$difficulty_level",
                            "Intermediate"
                        ]
                    },
                    1,
                    0
                ]
            }
        },
        "Fast_Track_type" : {
            "$sum" : {
                "$cond" : [
                    {
                        "$eq" : [
                            "$type",
                            "Fast Track"
                        ]
                    },
                    1,
                    0
                ]
            }
        },
        "PHP" : {
            "$sum" : {
                "$cond" : [
                    {
                        "$eq" : [
                            "$tagged_skills.name",
                            "PHP"
                        ]
                    },
                    1,
                    0
                ]
            }
        },
        "c++" : {
            "$sum" : {
                "$cond" : [
                    {
                        "$eq" : [
                            "$tagged_skills.name",
                            "c++"
                        ]
                    },
                    1,
                    0
                ]
            }
        },
        "java" : {
            "$sum" : {
                "$cond" : [
                    {
                        "$eq" : [
                            "$tagged_skills.name",
                            "java"
                        ]
                    },
                    1,
                    0
                ]
            }
        },
        "symphony" : {
            "$sum" : {
                "$cond" : [
                    {
                        "$eq" : [
                            "$tagged_skills.name",
                            "symphony"
                        ]
                    },
                    1,
                    0
                ]
            }
        },
        "C#" : {
            "$sum" : {
                "$cond" : [
                    {
                        "$eq" : [
                            "$tagged_skills.name",
                            "C#"
                        ]
                    },
                    1,
                    0
                ]
            }
        },
        "Scala" : {
            "$sum" : {
                "$cond" : [
                    {
                        "$eq" : [
                            "$tagged_skills.name",
                            "Scala"
                        ]
                    },
                    1,
                    0
                ]
            }
        },
        "javascript" : {
            "$sum" : {
                "$cond" : [
                    {
                        "$eq" : [
                            "$tagged_skills.name",
                            "javascript"
                        ]
                    },
                    1,
                    0
                ]
            }
        }
    }
}

- chridam