使用嵌套选择进行 Sequelize 的 SQL

Question

需要在Sequelize Node js中编写如下SQL Query

select a, b, c
from (
    select tbl.*, 
       count(*) over(partition by a) cnt,
       row_number() over (partition by a order by b) brn,
       row_number() over (partition by a order by c) crn
    from tbl
    where c in (4, 5)
) t
where cnt = 2 and brn = crn;

这是我想出的。我不知道把条件 where cnt = 2 and brn = crn; 放在哪里

t.findAll({
    attributes: {
        include: [
            [sequelize.literal('count(*) over(partition by a)'),'cnt'],
            [sequelize.literal('row_number() over (partition by a order by b)'),'brn'],
            [sequelize.literal('row_number() over (partition by a order by c)'),'crn'],
        ]   
    },
    where: {
            c: {[Op.in]:[4,5]},
        }
    })

Answer 1

Sequelize 似乎不支持从子查询( https://github.com/sequelize/sequelize/issues/5354 )中进行选择
您可以改用原始查询( https://sequelize.org/master/manual/raw-queries.html ):

const { QueryTypes } = require('sequelize');
let result = await sequelize.query(`
    select a, b, c
    from (
        select tbl.*, 
           count(*) over(partition by a) cnt,
           row_number() over (partition by a order by b) brn,
           row_number() over (partition by a order by c) crn
        from tbl
        where c in (4, 5)
    ) t
    where cnt = 2 and brn = crn;
`, {type: QueryTypes.SELECT});