我需要按发件人分组(fleche)并从每个消息中获取最后一条消息。我通过 $ push 和 $ group 成功实现了这个想法,但它返回了每个'fleche'(发件人)的所有消息。与消息日期相同的pb,应该是最后一个元素。是否可以在mongo 2.6.6 上使用聚合框架?
预期结果,实际查询,实际查询和消息收集样本的实际结果如下:
预期结果:
"result" : [
{
"_id" : 1,
"date" : [
1421948851800
],
"mess" : [
"dernier mess ffsdfsdfsd 2"
]
},
{
"_id" : 3,
"date" : [
1421948859593
],
"mess" : [
"dernier mess ffsdfsdfsd 2"
]
}
实际查询的实际结果:
"result" : [
{
"_id" : 1,
"date" : [
1421948851800,
1421947200606,
1421947163929,
1421947153646,
1421946853567
],
"mess" : [
"dernier mess ffsdfsdfsd 2",
"fssdfencore didon ! fds 2",
"fsd fds fds 2",
"safdsfdsfs sdf fie2",
"salut la companie"
]
},
{
"_id" : 3,
"date" : [
1421948859593,
1421947176679,
1421947098168
],
"mess" : [
"dernier mess ffsdfsdfsd 2",
"fssdfsdffsd ss fds 2",
"salut la companie"
]
}
实际查询:
db.messages.aggregate([
{$match :{cible :2 , lu:0} } ,
{ $sort : { date : -1 } },
{ $group : {_id : "$fleche", date : { $push : "$date" } , mess :{$push: "$mess" } } } ,
{$limit:20}
])
邮件收集示例:
{
"_id" : ObjectId("54c12fe5a48da55ee08c7804"),
"fleche" : 1,
"cible" : 2,
"date" : 1421946853567,
"mess" : "salut la companie",
"lu" : 0
}
{
"_id" : ObjectId("54c130daa48da55ee08c7805"),
"fleche" : 3,
"cible" : 2,
"date" : 1421947098168,
"mess" : "salut la companie",
"lu" : 0
}
{
"_id" : ObjectId("54c13111a48da55ee08c7807"),
"fleche" : 1,
"cible" : 2,
"date" : 1421947153646,
"mess" : "safdsfdsfs sdf fie2",
"lu" : 0
}
{
"_id" : ObjectId("54c1311ba48da55ee08c7808"),
"fleche" : 1,
"cible" : 2,
"date" : 1421947163929,
"mess" : "fsd fds fds 2",
"lu" : 0
}
{
"_id" : ObjectId("54c13128a48da55ee08c7809"),
"fleche" : 3,
"cible" : 2,
"date" : 1421947176679,
"mess" : "fssdfsdffsd ss fds 2",
"lu" : 0
}
答案 0 :(得分:4)
db.messages.aggregate([
{$match :{cible :2 , lu:0} } ,
{ $sort : { date : -1 } },
{ $group : {_id : "$fleche", date : { $first : "$date" } , mess :{$first: "$mess" } } } ,
{$limit:20}
])
输出将是:
{
"result" : [
{
"_id" : 1,
"date" : 1421947163929,
"mess" : "fsd fds fds 2"
},
{
"_id" : 3,
"date" : 1421947176679,
"mess" : "fssdfsdffsd ss fds 2"
}
],
"ok" : 1
}