我想在同一个文档中找到基于groupby、count和createdate最大的详细信息的所有记录，下面是一个示例集合：
/* 0 */
{
"_id" : 1,
"name" : "pradeep",
"age" : 26,
"createDate" : ISODate("2015-06-20T22:14:41.423-18:30")
}
/* 1 */
{
"_id" : 2,
"name" : "pradeep",
"age" : 26,
"createDate" : ISODate("2015-05-21T22:14:50.598-18:30")
}
/* 2 */
{
"_id" : 3,
"name" : "pradeep",
"age" : 26,
"createDate" : ISODate("2015-05-21T22:14:54.250-18:30")
}
/* 3 */
{
"_id" : 4,
"name" : "pradeep",
"age" : 24,
"createDate" : ISODate("2015-04-21T22:15:21.720-18:30")
}
/* 4 */
{
"_id" : 5,
"name" : "pradeep",
"age" : 21,
"createDate" : ISODate("2015-12-21T22:16:31.676-18:30")
}
/* 5 */
{
"_id" : 6,
"name" : "pradeep",
"age" : 24,
"createDate" : ISODate("2015-05-21T22:16:31.678-18:30")
}
/* 6 */
{
"_id" : 7,
"name" : "pradeep",
"age" : 24,
"createDate" : ISODate("2015-05-21T22:16:31.678-18:30")
}
/* 7 */
{
"_id" : 8,
"name" : "pradeep",
"age" : 24,
"createDate" : ISODate("2015-02-21T22:16:31.678-18:30")
}
/* 8 */
{
"_id" : 9,
"name" : "pradeep",
"age" : 24,
"createDate" : ISODate("2015-11-21T22:16:31.679-18:30")
}
/* 9 */
{
"_id" : 10,
"name" : "pradeep",
"age" : 23,
"createDate" : ISODate("2015-05-21T22:16:31.679-18:30")
}
/* 10 */
{
"_id" : 11,
"name" : "pradeep gupta",
"age" : 27,
"createDate" : ISODate("2015-05-21T22:16:31.679-18:30")
}
/* 11 */
{
"_id" : 12,
"name" : "ram",
"age" : 21,
"createDate" : ISODate("2015-08-21T22:17:21.014-18:30")
}
/* 12 */
{
"_id" : 13,
"name" : "ram",
"age" : 27,
"createDate" : ISODate("2015-05-21T22:17:21.015-18:30")
}
/* 13 */
{
"_id" : 14,
"name" : "ram",
"age" : 28,
"createDate" : ISODate("2015-05-21T22:17:21.016-18:30")
}
/* 14 */
{
"_id" : 16,
"name" : "lucky",
"age" : 27,
"createDate" : ISODate("2015-09-21T22:18:09.985-18:30")
}
/* 15 */
{
"_id" : 17,
"name" : "zawahir",
"age" : 28,
"createDate" : ISODate("2015-05-21T22:18:09.986-18:30")
}
/* 16 */
{
"_id" : 15,
"name" : "lucky",
"age" : 21,
"createDate" : ISODate("2015-05-21T22:18:23.134-18:30")
}
如何使用groupby名称查找createdate最大的所有记录，以及使用mongodb查询查找分组的总数？
想要这样的输出：
[{
"_id" : "lucky",
"countTotal" : 2,
"MaxCreateDate" : ISODate("2015-09-22T22:18:09.985+05:30"),
"MaxCreateDateUserDetails" :{_id: "lucky", name:"",age:""}
},{
......
}]
最佳答案
使用aggregation framework来达到所需的结果。以下聚合管道执行此操作。它使用$group运算符按name标识符表达式对输入文档进行分组，并将累加器表达式$max、$sum和$first应用于每个组：
--编辑--
获取MaxCreateDateUserDetails需要对系统变量Group Accumulator Operator应用$first$$ROOT，该变量返回对根文档（即当前正在聚合管道阶段处理的顶级文档）的引用。
db.collection.aggregate([
{
"$group": {
"_id": "$name",
"MaxCreateDate": {
"$max": "$createDate"
},
"count": {
"$sum": 1
},
"MaxCreateDateUserDetails": {
"$first": "$$ROOT"
}
}
},
{
"$project": {
"MaxCreateDate": 1,
"count": 1,
"MaxCreateDateUserDetails": {
"_id": "$_id",
"name": "$MaxCreateDateUserDetails.name",
"age" : "$MaxCreateDateUserDetails.age",
}
}
}
])
结果：
/* 1 */
{
"result" : [
{
"_id" : "zawahir",
"MaxCreateDate" : ISODate("2015-05-22T16:48:09.986Z"),
"count" : 1,
"MaxCreateDateUserDetails" : {
"_id" : "zawahir",
"name" : "zawahir",
"age" : 28
}
},
{
"_id" : "lucky",
"MaxCreateDate" : ISODate("2015-09-22T16:48:09.985Z"),
"count" : 2,
"MaxCreateDateUserDetails" : {
"_id" : "lucky",
"name" : "lucky",
"age" : 27
}
},
{
"_id" : "ram",
"MaxCreateDate" : ISODate("2015-08-22T16:47:21.014Z"),
"count" : 3,
"MaxCreateDateUserDetails" : {
"_id" : "ram",
"name" : "ram",
"age" : 21
}
},
{
"_id" : "pradeep gupta",
"MaxCreateDate" : ISODate("2015-05-22T16:46:31.679Z"),
"count" : 1,
"MaxCreateDateUserDetails" : {
"_id" : "pradeep gupta",
"name" : "pradeep gupta",
"age" : 27
}
},
{
"_id" : "pradeep",
"MaxCreateDate" : ISODate("2015-12-22T16:46:31.676Z"),
"count" : 10,
"MaxCreateDateUserDetails" : {
"_id" : "pradeep",
"name" : "pradeep",
"age" : 26
}
}
],
"ok" : 1
}
关于mongodb - 查找基于groupby，count和max createDate的所有记录，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/30403930/