在数组和子数组中使用 $size 和 $sort
Use $size with $sort in array and sub array
这是我的 collection 的结构部分:
_id: ObjectId("W"),
names: [
{
number: 1,
subnames: [ { id: "X", day: 1 }, { id: "Y", day: 10 }, { id: "Z", day: 2 } ],
list: ["A","B","C"],
day: 1
},
{
number: 2,
day: 5
},
{
number: 3,
subnames: [ { id: "X", day: 8 }, { id: "Z", day: 5 } ],
list: ["A","C"],
day: 2
},
...
],
...
我使用这个请求:
db.publication.aggregate( [ { $match: { _id: ObjectId("W") } }, { $group: { _id: "$_id", SizeName: { $first: { $size: { $ifNull: [ "$names", [] ] } } }, names: { $first: "$names" } } }, { $unwind: "$names" }, { $sort: { "names.day": 1 } }, { $group: { _id: "$_id", SzNames: { $sum: 1 }, names: { $push: { number: "$names.number", subnames: "$names.subnames", list: "$names.list", SizeList: { $size: { $ifNull: [ "$names.list", [] ] } } } } } } ] );
但我现在会为我的 names 数组和 subnames 数组使用 $sort获得此结果(subnames 可能不存在):
_id: ObjectId("W"),
names: [
{
number: 2,
SizeList: 0,
day: 5
},
{
number: 3,
subnames: [ { id: "Z", day: 5 }, { id: "X", day: 8 } ],
list: ["A","C"],
SizeList: 2,
day: 2
},
{
number: 1,
subnames: [ { id: "X", day: 1 }, { id: "Z", day: 2 }, { id: "Y", day: 10 } ],
list: ["A","B","C"],
SizeList: 3,
day: 1
}
...
],
...
你能帮帮我吗?
你可以做到,但难度很大。我个人很乐意投票支持 $sort
along the lines of the $map
运算符的内联版本。那会让事情变得容易得多。
目前您需要在排序后解构并重建数组。你必须对此非常小心。因此在处理 $unwind
:
之前用单个条目制作假数组
db.publication.aggregate([
{ "$project": {
"SizeNames": {
"$size": {
"$ifNull": [ "$names", [] ]
}
},
"names": { "$ifNull": [{ "$map": {
"input": "$names",
"as": "el",
"in": {
"SizeList": {
"$size": {
"$ifNull": [ "$$el.list", [] ]
}
},
"SizeSubnames": {
"$size": {
"$ifNull": [ "$$el.subnames", [] ]
}
},
"number": "$$el.number",
"day": "$$el.day",
"subnames": { "$ifNull": [ "$$el.subnames", [0] ] },
"list": "$$el.list"
}
}}, [0] ] }
}},
{ "$unwind": "$names" },
{ "$unwind": "$names.subnames" },
{ "$sort": { "_id": 1, "names.subnames.day": 1 } },
{ "$group": {
"_id": {
"_id": "$_id",
"SizeNames": "$SizeNames",
"names": {
"SizeList": "$names.SizeList",
"SizeSubnames": "$names.SizeSubnames",
"number": "$names.number",
"list": "$names.list",
"day": "$names.day"
}
},
"subnames": { "$push": "$names.subnames" }
}},
{ "$sort": { "_id._id": 1, "_id.names.day": 1 } },
{ "$group": {
"_id": "$_id._id",
"SizeNames": { "$first": "$_id.SizeNames" },
"names": {
"$push": { "$cond": [
{ "$ne": [ "$_id.names.SizeSubnames", 0 ] },
{
"number": "$_id.names.number",
"subnames": "$subnames",
"list": "$_id.names.list",
"SizeList": "$_id.names.SizeList",
"day": "$_id.names.day"
},
{
"number": "$_id.names.number",
"list": "$_id.names.list",
"SizeList": "$_id.names.SizeList",
"day": "$_id.names.day"
}
]}
}
}},
{ "$project": {
"SizeNames": 1,
"names": {
"$cond": [
{ "$ne": [ "$SizeNames", 0 ] },
"$names",
[]
]
}
}}
])
您可以 "hide away" 内部文档中的原始空数组,如图所示,但是如果不拉出类似的条件数组 "push" 技术,这确实不是一种实用的方法。
如果所有这些只是关于对单个文档中的数组元素进行排序,那么聚合框架不应该是执行此操作的工具。它可以如图所示完成,但根据文档,这在客户端代码中更容易完成。
输出:
{
"_id" : ObjectId("54b5cff8102f292553ce9bb5"),
"SizeNames" : 3,
"names" : [
{
"number" : 1,
"subnames" : [
{
"id" : "X",
"day" : 1
},
{
"id" : "Z",
"day" : 2
},
{
"id" : "Y",
"day" : 10
}
],
"list" : [
"A",
"B",
"C"
],
"SizeList" : 3,
"day" : 1
},
{
"number" : 3,
"subnames" : [
{
"id" : "Z",
"day" : 5
},
{
"id" : "X",
"day" : 8
}
],
"list" : [
"A",
"C"
],
"SizeList" : 2,
"day" : 2
},
{
"number" : 2,
"SizeList" : 0,
"day" : 5
}
]
}
这是我的 collection 的结构部分:
_id: ObjectId("W"),
names: [
{
number: 1,
subnames: [ { id: "X", day: 1 }, { id: "Y", day: 10 }, { id: "Z", day: 2 } ],
list: ["A","B","C"],
day: 1
},
{
number: 2,
day: 5
},
{
number: 3,
subnames: [ { id: "X", day: 8 }, { id: "Z", day: 5 } ],
list: ["A","C"],
day: 2
},
...
],
...
我使用这个请求:
db.publication.aggregate( [ { $match: { _id: ObjectId("W") } }, { $group: { _id: "$_id", SizeName: { $first: { $size: { $ifNull: [ "$names", [] ] } } }, names: { $first: "$names" } } }, { $unwind: "$names" }, { $sort: { "names.day": 1 } }, { $group: { _id: "$_id", SzNames: { $sum: 1 }, names: { $push: { number: "$names.number", subnames: "$names.subnames", list: "$names.list", SizeList: { $size: { $ifNull: [ "$names.list", [] ] } } } } } } ] );
但我现在会为我的 names 数组和 subnames 数组使用 $sort获得此结果(subnames 可能不存在):
_id: ObjectId("W"),
names: [
{
number: 2,
SizeList: 0,
day: 5
},
{
number: 3,
subnames: [ { id: "Z", day: 5 }, { id: "X", day: 8 } ],
list: ["A","C"],
SizeList: 2,
day: 2
},
{
number: 1,
subnames: [ { id: "X", day: 1 }, { id: "Z", day: 2 }, { id: "Y", day: 10 } ],
list: ["A","B","C"],
SizeList: 3,
day: 1
}
...
],
...
你能帮帮我吗?
你可以做到,但难度很大。我个人很乐意投票支持 $sort
along the lines of the $map
运算符的内联版本。那会让事情变得容易得多。
目前您需要在排序后解构并重建数组。你必须对此非常小心。因此在处理 $unwind
:
db.publication.aggregate([
{ "$project": {
"SizeNames": {
"$size": {
"$ifNull": [ "$names", [] ]
}
},
"names": { "$ifNull": [{ "$map": {
"input": "$names",
"as": "el",
"in": {
"SizeList": {
"$size": {
"$ifNull": [ "$$el.list", [] ]
}
},
"SizeSubnames": {
"$size": {
"$ifNull": [ "$$el.subnames", [] ]
}
},
"number": "$$el.number",
"day": "$$el.day",
"subnames": { "$ifNull": [ "$$el.subnames", [0] ] },
"list": "$$el.list"
}
}}, [0] ] }
}},
{ "$unwind": "$names" },
{ "$unwind": "$names.subnames" },
{ "$sort": { "_id": 1, "names.subnames.day": 1 } },
{ "$group": {
"_id": {
"_id": "$_id",
"SizeNames": "$SizeNames",
"names": {
"SizeList": "$names.SizeList",
"SizeSubnames": "$names.SizeSubnames",
"number": "$names.number",
"list": "$names.list",
"day": "$names.day"
}
},
"subnames": { "$push": "$names.subnames" }
}},
{ "$sort": { "_id._id": 1, "_id.names.day": 1 } },
{ "$group": {
"_id": "$_id._id",
"SizeNames": { "$first": "$_id.SizeNames" },
"names": {
"$push": { "$cond": [
{ "$ne": [ "$_id.names.SizeSubnames", 0 ] },
{
"number": "$_id.names.number",
"subnames": "$subnames",
"list": "$_id.names.list",
"SizeList": "$_id.names.SizeList",
"day": "$_id.names.day"
},
{
"number": "$_id.names.number",
"list": "$_id.names.list",
"SizeList": "$_id.names.SizeList",
"day": "$_id.names.day"
}
]}
}
}},
{ "$project": {
"SizeNames": 1,
"names": {
"$cond": [
{ "$ne": [ "$SizeNames", 0 ] },
"$names",
[]
]
}
}}
])
您可以 "hide away" 内部文档中的原始空数组,如图所示,但是如果不拉出类似的条件数组 "push" 技术,这确实不是一种实用的方法。
如果所有这些只是关于对单个文档中的数组元素进行排序,那么聚合框架不应该是执行此操作的工具。它可以如图所示完成,但根据文档,这在客户端代码中更容易完成。
输出:
{
"_id" : ObjectId("54b5cff8102f292553ce9bb5"),
"SizeNames" : 3,
"names" : [
{
"number" : 1,
"subnames" : [
{
"id" : "X",
"day" : 1
},
{
"id" : "Z",
"day" : 2
},
{
"id" : "Y",
"day" : 10
}
],
"list" : [
"A",
"B",
"C"
],
"SizeList" : 3,
"day" : 1
},
{
"number" : 3,
"subnames" : [
{
"id" : "Z",
"day" : 5
},
{
"id" : "X",
"day" : 8
}
],
"list" : [
"A",
"C"
],
"SizeList" : 2,
"day" : 2
},
{
"number" : 2,
"SizeList" : 0,
"day" : 5
}
]
}