我有mongo集合'orders'包含一个具有orderid和时间的用户列表，如下所示：
user orderid time has_pending
10001 1 1510489123 0
10002 2 1510489125 0
10003 3 1510489127 0
10001 5 1510489131 1
10001 6 1510489133 1
10002 7 1510489135 0
10003 8 1510489137 0
10001 9 1510489139 1
10001 10 1510489141 0
10002 11 1510489143 1
10001 12 1510489145 0 <<<<<
10002 13 1510489147 0 <<<<<
10001 14 1510489149 1
10002 15 1510489151 1
10003 16 1510489153 1
10003 17 1510489155 1
10003 18 1510489157 1
10003 21 1510489163 1
10003 22 1510489165 0 <<<<<
我正在尝试获取每个用户的订单列表，其中订单时间＆gt; =上次出现has_pending = 0的时间
例如：如果我们查看用户10001数据：
user orderid time has_pending
10001 1 1510489123 0
10001 5 1510489131 1
10001 6 1510489133 1
10001 9 1510489139 1
10001 10 1510489141 0
10001 12 1510489145 0
10001 14 1510489149 1
因此此查询对此用户的结果为：
10001 12 1510489145 0
10001 14 1510489149 1
所需的查询应该为所有用户获取数据，结果应如下所示：
user orderid time has_pending
10001 12 1510489145 0
10002 13 1510489147 0
10001 14 1510489149 1
10002 15 1510489151 1
10003 22 1510489165 0
MYSQL QUERY：
SELECT
t1.*
FROM
test AS t1
LEFT JOIN test AS t2 ON t1.time >= t2.time AND t1.user = t2.user
WHERE
t2.orderid= (SELECT max(orderid) FROM test WHERE user= t1.user AND has_pending = 0)
任何想法如何在一个mongo查询中获得结果？
由于
答案 0 :(得分：2)
鉴于以下输入文件：
{ "user" : 10001, "orderid" : 1, "time" : 1510489123, "has_pending" : 0 }
{ "user" : 10002, "orderid" : 2, "time" : 1510489125, "has_pending" : 0 }
{ "user" : 10003, "orderid" : 3, "time" : 1510489127, "has_pending" : 0 }
{ "user" : 10001, "orderid" : 5, "time" : 1510489131, "has_pending" : 1 }
{ "user" : 10001, "orderid" : 6, "time" : 1510489133, "has_pending" : 1 }
{ "user" : 10002, "orderid" : 7, "time" : 1510489135, "has_pending" : 0 }
{ "user" : 10003, "orderid" : 8, "time" : 1510489137, "has_pending" : 0 }
{ "user" : 10001, "orderid" : 9, "time" : 1510489139, "has_pending" : 1 }
{ "user" : 10001, "orderid" : 10, "time" : 1510489141, "has_pending" : 0 }
{ "user" : 10002, "orderid" : 11, "time" : 1510489143, "has_pending" : 1 }
{ "user" : 10001, "orderid" : 12, "time" : 1510489145, "has_pending" : 0 }
{ "user" : 10002, "orderid" : 13, "time" : 1510489147, "has_pending" : 0 }
{ "user" : 10001, "orderid" : 14, "time" : 1510489149, "has_pending" : 1 }
{ "user" : 10002, "orderid" : 15, "time" : 1510489151, "has_pending" : 1 }
{ "user" : 10003, "orderid" : 16, "time" : 1510489153, "has_pending" : 1 }
{ "user" : 10003, "orderid" : 17, "time" : 1510489155, "has_pending" : 1 }
{ "user" : 10003, "orderid" : 18, "time" : 1510489157, "has_pending" : 1 }
{ "user" : 10003, "orderid" : 21, "time" : 1510489163, "has_pending" : 1 }
{ "user" : 10003, "orderid" : 22, "time" : 1510489165, "has_pending" : 0 }
您的查询需要如下所示：
db.collection.aggregate([
{
$sort: {
"time": -1 // sort by "time" descending
}
}, {
$group: { // we want to slice our data per "user" so let's group by that field
_id: "$user",
"orders": {
$push: "$$ROOT" // remember each document in an array per each "user" group (entries still sorted by "time" descending)
}
}
}, {
$project: {
"orders": { // our orders array shall only contain...
$slice: [ "$orders", 0, { // ...all items from the last one up until...
$add: [ { $indexOfArray: [ "$orders.has_pending", 0 ] }, 1 ] // ...the first appearance of an "has_pending" == 0 entry
// the $add makes sure that we include the found element with "has_pending" == 0, too
}]
}
}
}, {
$unwind: "$orders" // restore original documents again by flattening the "orders" array
}, {
$replaceRoot: { // move the (single) entry of the orders array to the root level of each document
"newRoot": "$orders"
}
}, {
$sort: {
"time": 1 // your example output was sorted by date so that's why we do that here, too...
}
}])
这将为您提供您要求的确切顺序和内容（加上我为了简洁而省略的_id字段）：
{ "user" : 10001, "orderid" : 12, "time" : 1510489145, "has_pending" : 0 }
{ "user" : 10002, "orderid" : 13, "time" : 1510489147, "has_pending" : 0 }
{ "user" : 10001, "orderid" : 14, "time" : 1510489149, "has_pending" : 1 }
{ "user" : 10002, "orderid" : 15, "time" : 1510489151, "has_pending" : 1 }
{ "user" : 10003, "orderid" : 22, "time" : 1510489165, "has_pending" : 0 }
答案 1 :(得分：0)
db.getCollection('order').aggregate([
{ $sort: {"time": -1}},
{
$group:{
_id: {
user: "$user",
has_pending: "$has_pending"
},
time: { $first: "$time"},
orderid: { $first: "$orderid"}
}
},
{
$project: {
_id: 0,
user: "$_id.user",
orderid: "$orderid",
time: "$time",
has_pending: "$_id.has_pending"
}
}
])
如果您想了解自己在每个聚合管道中正在做什么，可以继续阅读。
为了解释每个管道中发生的事情，我将采用您发布的内容的子集。所以我们假设我们有这些文件：
user orderid time has_pending
10001 1 1510489123 0
10002 2 1510489125 0
10001 5 1510489131 1
10002 7 1510489135 0
10002 11 1510489143 1
10001 12 1510489145 0
10002 13 1510489147 0
10001 14 1510489149 1
10002 15 1510489151 1
按时间{ $sort: {"time": -1}}排序，按时间降序排列结果。这会使你的结果看起来像这样
user orderid time has_pending
10002 15 1510489151 1
10001 14 1510489149 1
10002 13 1510489147 0
10001 12 1510489145 0
10002 11 1510489143 1
10002 7 1510489135 0
10001 5 1510489131 1
10002 2 1510489125 0
10001 1 1510489123 0
现在，我们可以按user和has_pending对结果进行分组。因为我们每个user和每个has_pending只需要一个结果。所以我们只需要一个
user: 1001 with has_pending: 0,
user: 1001 with has_pending: 1,
user: 1002 with has_pending: 0,
user: 1002 with has_pending: 1
这发生在您的群组聚合中：
_id: {
user: "$user",
has_pending: "$has_pending"
}
您论坛中的字段_id是强制性的，您可以根据想要分组的内容进行描述。
注意我添加了：
time: { $first: "$time"},
orderid: { $first: "$orderid"}
我使用$first，因为我知道我的文档已经排序。所以我绝对肯定第一个
user: 1001 with has_pending: 0 will take "time" : 1510489145 and "orderid" : 12
user: 1001 with has_pending: 1 will take "time" : 1510489149 and "orderid" : 14
user: 1002 with has_pending: 0 will take "time" : 1510489147 and "orderid" : 13
user: 1002 with has_pending: 1 will take "time" : 1510489151 and "orderid" : 15
在这种情况下，$project仅用于＆＃34;标准化＆＃34;你的结果。所以我们可以得到你要求的最终结果。
答案 2 :(得分：-1)
您可以使用sort属性。 MongoDB shell中的示例，其输出与SQL查询相同：
db.collection.find({}).sort({ user: 1, orderid: 1, time: 1, has_pending: 1 }).pretty()