Question

任何人都可以告诉我使用Sequelize获得与此SQL查询相同结果的最佳方法是什么：

SELECT DISTINCT (c.id),
    c.*
FROM table1 a
INNER JOIN table2 b
    ON b.id = a."table2Id" 
INNER JOIN table3 c
    ON c.id = b."table3Id"
ORDER BY c.date DESC
LIMIT 10;

现在，＆＃34;包括＆＃34;几乎给出了内连接所需的结果，但是＆＃34; distinct＆＃34;在一张连接的桌子上看起来更难获得......

目标是加入3个表，按第三个连接的日期排序，避免多次从第三个连接中输入相同的条目，最后只获得第三个连接中的不同条目（总是有一个＆＃34; table1＆＃34;关系，因为我们从这个表开始得到它们。）

Answer 1

SELECT DISTINCT (b.table3Id), c.*
FROM table1 a
INNER JOIN table2 b ON b.id = a."table2Id" 
INNER JOIN table3 c ON c.id = b."table3Id"
ORDER BY c.date DESC
LIMIT 10;

Answer 2

Table1.findAll({ 
    attributes: [ [sequelize.fn('DISTINCT', sequelize.col('table1.table3Id')), 'c.id'] ],
    include: [ [Table2, "b"], [Table3, "c" ] ],
    order: [ [Table1.associations.Table3, 'date', 'DESC'] ],
    limit: 10
})

与续集

2 个答案: