SQLAlchemy ORM __init__方法与(SQLAlchemy ORM __init__ method vs)

 这里的想法是首先按version键对文档进行排序，按title键对文档进行分组，并在订购时返回组中的第一个文档。 因此，邀请到此处聚合管道的运营商将是$sort ， $group和$project （它将聚合文档重新整形为所需的结果模式）。 
 
 现在，在$group管道中，您需要$first运算符（或者$last具体取决于您在先前的$sort管道中订购文档的方向），以便在订购时映射顶层文档的字段。 
 
 考虑mongo shell中的以下游戏： 
 
db.collection.aggregate([
    { "$sort": { "_id.version": -1 } },
    {
        "$group": {
            "_id": "$_id.title",
            "id": { "$first": "$_id" },
            "status": { "$first": "$status" }
            ...
        }
    },
    {
        "$project": {
            "_id": "$id",
            "status": 1,
            ...
        } 
    }
])
 
 
 对于mongoTemplate等价物： 
 
import static org.springframework.data.mongodb.core.aggregation.Aggregation.*;

Aggregation agg = newAggregation(
    sort(DESC, "_id.version"),
    group("_id.title"), 
        .first("status").as("status")
        ...
    project("id").previousOperation().and("status").and(...)
);
 
 
 您还可以选择将$$ROOT系统变量邀请到$group管道作为返回完整文档的方法。 考虑以下方法： 
 
db.collection.aggregate([
    { "$sort": { "_id.version": -1 } },
    {
        "$group": {
            "_id": "$_id.title",            
            "doc": { "$first": "$$ROOT" }
        }
    },
    {
        "$project": {
            "_id": "$doc._id",
            "status": "$doc.status",
            ...
        } 
    }
])
 
 这会转化为 
 
import static org.springframework.data.mongodb.core.aggregation.Aggregation.*;

Aggregation agg = newAggregation(
    sort(DESC, "_id.version"),
    group("_id.title"), 
        .first(ROOT).as("doc")
    project("doc.status").as("status")...
);

The idea here is to order the documents by the version key first, group the documents by the title key and return the first document in the group when ordered. So, the operators invited to the aggregation pipeline show here will be the the $sort, $group and $project (which reshapes the aggregated documents to be in the desired result schema). 
 
Now within the $group pipeline you'd need the $first operator (or $last depending on the direction which you ordered the documents in the previous $sort pipeline) to map the top document's field when ordered.
 
Consider the following play in mongo shell:
 
db.collection.aggregate([
    { "$sort": { "_id.version": -1 } },
    {
        "$group": {
            "_id": "$_id.title",
            "id": { "$first": "$_id" },
            "status": { "$first": "$status" }
            ...
        }
    },
    {
        "$project": {
            "_id": "$id",
            "status": 1,
            ...
        } 
    }
])
 
 
For the mongoTemplate equivalent:
 
import static org.springframework.data.mongodb.core.aggregation.Aggregation.*;

Aggregation agg = newAggregation(
    sort(DESC, "_id.version"),
    group("_id.title"), 
        .first("status").as("status")
        ...
    project("id").previousOperation().and("status").and(...)
);
 
 
You've got another alternative of inviting the $$ROOT system variable to your $group pipeline as means to return the full document. Consider the following approach:
 
db.collection.aggregate([
    { "$sort": { "_id.version": -1 } },
    {
        "$group": {
            "_id": "$_id.title",            
            "doc": { "$first": "$$ROOT" }
        }
    },
    {
        "$project": {
            "_id": "$doc._id",
            "status": "$doc.status",
            ...
        } 
    }
])
 
which would translate to
 
import static org.springframework.data.mongodb.core.aggregation.Aggregation.*;

Aggregation agg = newAggregation(
    sort(DESC, "_id.version"),
    group("_id.title"), 
        .first(ROOT).as("doc")
    project("doc.status").as("status")...
);