我有两个这样的表
表 1:
CREATE TABLE `table_1` (
`id` int(11) NOT NULL,
`name` varchar(100) NOT NULL,
`status` int(11) NOT NULL,
`date_added` datetime NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
INSERT INTO `table_1` (`id`, `name`, `status`, `date_added`) VALUES
(1, 'CNN', 1, '2018-07-01 00:00:01'),
(2, 'BBC', 1, '2018-07-03 00:00:01');
表 2:
CREATE TABLE `table_2` (
`id` int(11) NOT NULL,
`title` varchar(100) NOT NULL,
`status` int(11) NOT NULL,
`date_added` datetime NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
INSERT INTO `table_2` (`id`, `title`, `status`, `date_added`) VALUES
(1, 'Windows', 1, '2018-07-02 00:00:01'),
(2, 'Linux', 1, '2018-07-04 00:00:01');
我的 SQL 查询是这样的
SELECT * FROM
(
SELECT id, name, date_added FROM `table_1`
UNION
SELECT id, title as name, date_added FROM `table_2`
)
as new_table order by date_added ASC
我想要的结果如下:
id name date_added Ascending 1
1 CNN 2018-07-01 00:00:01
1 Windows 2018-07-02 00:00:01
2 BBC 2018-07-03 00:00:01
2 Linux 2018-07-04 00:00:01
我无法在 mongodb 中得到这样的结果。
最佳答案
经过更多研究,我能够找到在 mongodb 中使用 union 的解决方案,如下所示:
db.table_1.aggregate([
{
$lookup: {
from: "table_2",
pipeline: [],
as: "table_2"
}
},
{
$addFields: {
table_2: {
$map: {
input: "$table_2",
as: "tbl2",
in: {
"_id":"$$tbl2._id",
"type": "table_2",
"title": "$$tbl2.title",
"status": "$$tbl2.status",
"date_added": "$$tbl2.date_added",
}
}
}
}
},
{
$group: {
_id: null,
table_1: {
$push: {
_id:"$_id",
type: "table_1",
name: "$name",
status: "$status",
date_added: "$date_added"
}
},
table_2: {
$first: "$table_2"
}
}
},
{
$project: {
items: {
$setUnion: ["$table_1", "$table_2"]
}
}
},
{
$unwind: "$items"
},
{
$replaceRoot: {
newRoot: "$items"
}
},
{
'$match': { status: true}
},
{ '$sort': { date_added: -1 } },
{ '$limit': 10 }
])
关于mongodb - 如何从 Mongodb 中的两个不同集合合并?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/51222861/