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MongoDB 查询嵌套文档列表

转载 作者:可可西里 更新时间:2023-11-01 09:52:22 26 4
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我不是 MongoDB 的新手,而是聚合概念的新手......我有看起来像这样的集合数据,目前它包含 2 个文档

 {
"_id" : ObjectId("52cc0b079f0ae55e9fb770f8"),
"uid" : 100,
"data" : {
"mi" : [
{
"miId" : NumberLong(1),
"name" : "ABC",
"severity" : "HIGH",
"failures" : NumberLong(2),
"description" : "Some description",
"remediation" : "Some remedy"
},
{
"miId" : NumberLong(10),
"name" : "PQR",
"severity" : "HIGH",
"failures" : NumberLong(3),
"description" : "Some description",
"remediation" : "Some remedy"
}
}

{
"_id" : ObjectId("52cc0b079f0ae55easdas8"),
"uid" : 200,
"data" : {
"mi" : [
{
"miId" : NumberLong(10),
"name" : "ABC",
"severity" : "HIGH",
"failures" : NumberLong(20),
"description" : "Some description",
"remediation" : "Some remedy"
},
{
"miId" : NumberLong(18),
"name" : "PQR",
"severity" : "HIGH",
"failures" : NumberLong(30),
"description" : "Some description",
"remediation" : "Some remedy"
}
}
}

我如何在 MongoDB shell 或 Java 中提出一个查询,该查询基于“名称”执行 groupby() 并总结所有“失败”,结果还应包含具有最高的“名称”的“uid” “失败”。结果应该是这样的:

{
{
"_id" : ObjectId("508894efd4197aa2b9490741"),
"name" : "ABC",
"sum_total_of_failures" : 22,
"uid" : 200
}

{
"_id" : ObjectId("508894efd4197aa2b9490741"),
"name" : "PQR",
"sum_total_of_failures" : 33,
"uid" : 200
}
}

任何帮助将不胜感激,我用 $unwind 编写了一个查询,因为“mi”文档存储在列表中,但它返回空结果。查询如下:

    db.temp.aggregate(
{$unwind: "$mi"},
{$project: {mi : "$mi"}},
{$group: { _id: "$name",total: { $sum: "$failures" }}})

最佳答案

尝试以下查询:

db.collection.aggregate(
{$unwind : "$data.mi"},
{$sort : {"data.mi.failures" : -1}},
{$group : {_id : "$data.mi.name",
sum_total_of_failures : {$sum : "$data.mi.failures"},
uid : {$first : "$uid"}}}
)

结果会是这样的:

"result" : [
{
"_id" : "PQR",
"sum_total_of_failures" : NumberLong(33),
"uid" : 200
},
{
"_id" : "ABC",
"sum_total_of_failures" : NumberLong(22),
"uid" : 200
}
]

使用 Java 驱动程序,您可以按如下方式执行此操作:

    DBCollection coll = ...

DBObject unwind = new BasicDBObject("$unwind", "$data.mi");
DBObject sort = new BasicDBObject("$sort", new BasicDBObject("data.mi.failures", -1));

DBObject groupObj = new BasicDBObject();
groupObj.put("_id", "$data.mi.name");
groupObj.put("sum_total_of_failures", new BasicDBObject("$sum", "$data.mi.failures"));
groupObj.put("uid", new BasicDBObject("$first", "$uid"));

DBObject group = new BasicDBObject("$group", groupObj);

AggregationOutput output = coll.aggregate(unwind, sort, group);
if (output != null) {
for (DBObject result : output.results()) {
String name = (String) result.get("_id");
Long sumTotalOfFailures = (Long) result.get("sum_total_of_failures");
Integer uid = (Integer) result.get("uid");
}
}

关于MongoDB 查询嵌套文档列表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20998822/

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