Mongodb （mongo）如何获取数组中包含值和大小的文档
我有一个mongo系列，内容如下：Mongodb （mongo）如何获取数组中包含值和大小的文档,mongodb,Mongodb,我有一个mongo系列，内容如下： { "_id" : ObjectId("59e013e83260c739f029ee21"), "createdAt" : ISODate("2017-10-13T01:16:24.653+0000"), "updatedAt" : ISODate("2017-11-11T17:13:52.956+0000"), "age" : NumberInt(34), "attributes" : [
{
"_id" : ObjectId("59e013e83260c739f029ee21"),
"createdAt" : ISODate("2017-10-13T01:16:24.653+0000"),
"updatedAt" : ISODate("2017-11-11T17:13:52.956+0000"),
"age" : NumberInt(34),
"attributes" : [
{
"year" : "2017",
"contest" : [
{
"name" : "Category1",
"division" : "Department1"
},
{
"name" : "Category2",
"division" : "Department1"
}
]
},
{
"year" : "2016",
"contest" : [
{
"name" : "Category2",
"division" : "Department1"
}
]
},
{
"year" : "2015",
"contest" : [
{
"name" : "Category1",
"division" : "Department1"
}
]
}
],
"name" : {
"id" : NumberInt(9850214),
"first" : "john",
"last" : "afham"
}
}
db.foo.aggregate([
// Start with breaking down attributes:
{$unwind: "$attributes"}
// Next, extract only name = Category1 from the contest array. This will yield
// an array of 0 or 1 because I am assuming that contest names WITHIN
// the contest array are unique. If found and we get an array of 1, turn that
// into a single doc instead of an array of a single doc by taking arrayElemAt 0.
// Otherwise, "x" is not set into the doc AT ALL. All other vars in the doc
// will go away after $project; if you want to keep them, change this to
// $addFields:
,{$project: {x: {$arrayElemAt: [ {$filter: {
input: "$attributes.contest",
as: "z",
cond: {$eq: [ "$$z.name", "Category1" ]}
}}, 0 ]}
}}
// We split up attributes before, creating multiple docs with the same _id. We
// must now "recombine" these _id (OP said he wants # of docs with name).
// We now have to capture all the single "x" that we created above; docs without
// Category1 will have NO "x" and we don't want to include them in the count.
// Also, we KNOW that name can only be Category 1 but division could vary, so
// let's capture that in the $push in case we might want it:
,{$group: {_id: "$_id", x: {$push: "$x.division"}}}
// One more pass to compute length of array:
,{$addFields: {len: {$size: "$x"}} }
// And lastly, the filter for one time or two times or n times:
,{$match: {len: {$gt: 2} }}
]);
db.foo.aggregate([
// Start with breaking down attributes:
{$unwind: "$attributes"}
// Next, extract only name = Category1 from the contest array. This will yield
// an array of 0 or 1 because I am assuming that contest names WITHIN
// the contest array are unique. If found and we get an array of 1, turn that
// into a single doc instead of an array of a single doc by taking arrayElemAt 0.
// Otherwise, "x" is not set into the doc AT ALL. All other vars in the doc
// will go away after $project; if you want to keep them, change this to
// $addFields:
,{$project: {x: {$arrayElemAt: [ {$filter: {
input: "$attributes.contest",
as: "z",
cond: {$eq: [ "$$z.name", "Category1" ]}
}}, 0 ]}
}}
// We split up attributes before, creating multiple docs with the same _id. We
// must now "recombine" these _id (OP said he wants # of docs with name).
// We now have to capture all the single "x" that we created above; docs without
// Category1 will have NO "x" and we don't want to include them in the count.
// Also, we KNOW that name can only be Category 1 but division could vary, so
// let's capture that in the $push in case we might want it:
,{$group: {_id: "$_id", x: {$push: "$x.division"}}}
// One more pass to compute length of array:
,{$addFields: {len: {$size: "$x"}} }
// And lastly, the filter for one time or two times or n times:
,{$match: {len: {$gt: 2} }}
]);
首先，我们需要按属性和竞争字段展平文档。然后根据文档首字母id和比赛名称分组，计算沿途的不同比赛。最后，我们过滤结果
db.person.aggregate([
{ $unwind: "$attributes" },
{ $unwind: "$attributes.contest" },
{$group: {
_id: {initial_id: "$_id", contest: "$attributes.contest.name"},
count: {$sum: 1}
}
},
{$match: {$and: [{"_id.contest": "Category1"}, {"count": {$gt: 1}}]}}]);
首先，我们需要按属性和竞争字段展平文档。然后根据文档首字母id和比赛名称分组，计算沿途的不同比赛。最后，我们过滤结果
db.person.aggregate([
{ $unwind: "$attributes" },
{ $unwind: "$attributes.contest" },
{$group: {
_id: {initial_id: "$_id", contest: "$attributes.contest.name"},
count: {$sum: 1}
}
},
{$match: {$and: [{"_id.contest": "Category1"}, {"count": {$gt: 1}}]}}]);
假设一次
竞赛名称展开属性db.collection.aggregate({
$project: {
"numberOfOccurrences": {
$size: { // count the number of matching contest elements
$filter: { // get rid of all contest entries that do not contain at least one entry with name "Category1"
input: "$attributes",
cond: { $in: [ "Category1", "$$this.contest.name" ] }
}
}
}
}
}, {
$match: { // filter the number of documents
"numberOfOccurrences": {
$gt: 1 // put your desired min. number of matching contest entries here
}
}
}, {
$count: "numberOfDocuments" // count the number of matching documents
})
假设一次
竞赛名称展开属性db.collection.aggregate({
$project: {
"numberOfOccurrences": {
$size: { // count the number of matching contest elements
$filter: { // get rid of all contest entries that do not contain at least one entry with name "Category1"
input: "$attributes",
cond: { $in: [ "Category1", "$$this.contest.name" ] }
}
}
}
}
}, {
$match: { // filter the number of documents
"numberOfOccurrences": {
$gt: 1 // put your desired min. number of matching contest entries here
}
}
}, {
$count: "numberOfDocuments" // count the number of matching documents
})
粘贴集合中的样本数据。这个数据看起来不正确，看起来上面的数据不正确correct@astro我已经更新了json数据，它100%类似于从collection@Sam更新json数据以澄清您的需求，示例文档是否必须计数？粘贴集合中的示例数据。这个数据看起来不正确，看起来上面的数据不正确correct@astro我已经更新了json数据，它100%类似于从collection@Samjson数据被更新以澄清您的需求，示例文档是否需要计数？这里很好地使用了数组点op（this.contest.name）！这里很好的使用了数组点op（this.contest.name）！在较大的阵列情况下，双$REWIND可能是致命的。drickless在上面发布了一个$unwind免费示例。@BuzzMoschetti可能可以，drickless对数据做了一些假设，而我没有。无论如何，有很多选择是很好的。在更大的阵列情况下，双$unwind可能是致命的。drickless在上面发布了一个$unwind免费示例。@BuzzMoschetti可能可以，drickless对数据做了一些假设，而我没有。不管怎样，很好，有很多选择。