每个项目 mongodb 组按日期的最新记录 [英] Latest record by date for each item mongodb group
问题描述
一个集合中有很多项目,每个项目都有很多记录,有些有新日期的记录,有些有周或/和月的旧记录.我需要一个查询,它返回每个项目的最新记录.在 .aggregate() 的情况下,我需要提交完整的数据".我想要这个结果使用 mongodb $group,每个设备的记录应该是最新的.
There are many items in a collection, each has many records, some have records of new dates and some have week or/and month older. I need a query which returns latest last record of each item. In the case of .aggregate() I need that complete "data" filed. I want this result using mongodb $group, the record should be latest for each device.
{
"result" : [
{
"_id" : 29,
"gateway_id" : 1,
"data" : [
{
"r" : 203,
"v" : 3002
},
{
"r" : 221,
"v" : 3006
}
],
"device_id" : 29,
"date_time" : "a"
},
{
"_id" : 28,
"gateway_id" : 1,
"data" : [
{
"r" : 203,
"v" : 3002
},
{
"r" : 221,
"v" : 3006
}
],
"device_id" : 28,
"date_time" : "b"
},
{
"_id" : 27,
"gateway_id" : 1,
"data" : [
{
"r" : 203,
"v" : 3642
},
{
"r" : 221,
"v" : 3666
}
],
"device_id" : 27,
"date_time" : "a"
}
],
"ok" : 1
}
我希望使用 mongodb $group 得到这个结果,每个设备的记录应该是最新的.
I want this result using mongodb $group, the record should be latest for each device.
推荐答案
尝试使用以下代码段
db.collection.aggregate([
{$group: {
"_id": "$device_id",
"gateway_id": {"$last":"$gateway_id"},
"data": {"$last": '$data'},
"date": {"$last": '$date_time'},
}},
{$project: {
"device_id": "$_id",
"gateway_id": "$gateway_id",
"data": "$data",
"date_time": "$date"
}},
{$sort: {
"date": -1
}}
]);
在上面的查询组中按设备 ID 和日期,数据和网关 ID 将在每行中最新.
In above query group by device id and date, data, and gateway_id will be latest in each row.
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