mysql 转写 mongo 语句.不知道算复杂不复杂！

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This topic created in 2374 days ago, the information mentioned may be changed or developed.

原 sql 语句

SELECT `*`,GROUP_CONCAT(`invoice_number`) as `invoice_number_all`,ROUND(SUM(`invoice_amount`)/100,2) as `invoice_amount_all`,ROUND(SUM(`premium_amount`)/100,2) as `premium_amount_all` FROM `t_invoice` WHERE `id` != 0 GROUP BY `insurer_id`,`region_id`,`invoice_company_id`,`data_type`,`invoice_date`,`invoice_remark` ORDER BY `invoice_date` DESC LIMIT 0,15

刚开始用 mongo 数据库很多操作不会写。简单的还会，这是 mysql 查询分组后的结果，转换到 mongo 没有写正确导致查询不出数据。想问问这样的 sql 语句转 mongo 合适吗？还是要怎么处理呢！

db.t_invoice.aggregate({ $group: { _id: "$insurer_id"} }) 我这样写只能一个，多个我怎么写都报错，查询不出结果。而且字段名也不是我想要的 我把 _id 改成字段就开始报错！

mongo

语句

MySQL

SQL

5 replies 2019-11-12 09:26:28 +08:00

Hieast

Nov 11, 2019 via iPhone

mongo 里面到处是 json，那个 id 也要写成 json

coderabbit

Nov 11, 2019

@Hieast

db.t_invoice.aggregate(
{
"$group": {
_id: {
insurer_id:"$insurer_id",
region_id: "$region_id",
invoice_company_id:"$invoice_company_id",
data_type: "$data_type",
invoice_date: "$invoice_date",
invoice_remark: "$invoice_remark",
},
count:{$sum:1}
}
}
)

这样子我看分组是出来了！但是还不知道如何加字段，count kv 我删除后就没有统计结果了。_id 我一改名就报错！我想它跟 mysql 查出来一样的结果！

Hieast

Nov 12, 2019 via iPhone

搜下 pipe

mmdsun

Nov 12, 2019 via Android

你需要一个转换工具。
用 studio 3t。直接写 SQL 可转换成 MongoDB shell，Java 代码，Python 代码。

catinsides

Nov 12, 2019

aggregate 里得是数组，db.t_invoice.aggregate([{...}, {...}]