MongoDB对嵌套文档列表的查询
我不是MongoDB的新手,但对聚合概念还是新手。。。 我收集的数据看起来像这样,目前它包含2个文档MongoDB对嵌套文档列表的查询,mongodb,Mongodb,我不是MongoDB的新手,但对聚合概念还是新手。。。 我收集的数据看起来像这样,目前它包含2个文档 { "_id" : ObjectId("52cc0b079f0ae55e9fb770f8"), "uid" : 100, "data" : { "mi" : [ { "miId" : NumberLong(1), "name" : "ABC",
{
"_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"
}
}
}
如何在MongoDB shell或Java中生成一个查询,该查询基于“name”执行groupby(),并对所有“failures”进行汇总,结果还应包含“name”中“failures”最高的“uid”。
结果应该如下所示:
{
{
"_id" : ObjectId("508894efd4197aa2b9490741"),
"name" : "ABC",
"sum_total_of_failures" : 22,
"uid" : 200
}
{
"_id" : ObjectId("508894efd4197aa2b9490741"),
"name" : "PQR",
"sum_total_of_failures" : 33,
"uid" : 200
}
}
我用$unwind编写了一个查询,因为“mi”文档存储在一个列表中,但返回的结果为空。
查询如下:
db.temp.aggregate(
{$unwind: "$mi"},
{$project: {mi : "$mi"}},
{$group: { _id: "$name",total: { $sum: "$failures" }}})
请尝试以下查询:
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 : "$uid"}}}
)
结果如下:
"result" : [
{
"_id" : "PQR",
"sum_total_of_failures" : NumberLong(33),
"uid" : 200
},
{
"_id" : "ABC",
"sum_total_of_failures" : NumberLong(22),
"uid" : 200
}
]
使用Java驱动程序,您可以按如下方式执行:
DBCollection coll = ...
DBObject unwind = new BasicDBObject("$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");
}
}
谢谢小问题,虽然“失败总数”现在是正确的,但是“uid”总是200,“uid”应该是该名称“失败总数”最高的文档。我以为您想要最高的uid。我已经更新了查询。在分组之前,我按失败对数据进行排序,并获得第一个匹配的uid。希望这能解决您的问题。我们如何用Java重写上面的查询?