Question

嗨伙计们，阅读约20个问题，但我仍有问题因为它看起来各不相同，或者我看不出什么解决方案正在解决我的问题。

我想写信给我的所有客户，他们每年都会搬到另一个公寓我做了一个表用户和一个表地址和一个mton加入他们。但是当用户有2个地址时，我总是得到第一个地址和group by 或者两者都在没有分组的情况下加入。

有人能告诉我答案，我敢打赌，我已经找不到解决方案了

我的表的快捷方式

CREATE DATABASE DL;

CREATE TABLE DL.users (
  uidx int PRIMARY KEY auto_increment,
  email varchar(255) 
);
INSERT INTO DL.users VALUES (NULL,'myemail'),(NULL,'heremail');
CREATE TABLE DL.address (
  aidx int PRIMARY KEY auto_increment,
  address varchar(255) 
);
INSERT INTO DL.address VALUES (NULL,'myadress'),(NULL,'heraddress');
CREATE TABLE DL.ntom (
  nidx int PRIMARY KEY auto_increment,
  fk_uidx int ,
  fk_aidx int ,
  movein date
);
INSERT INTO DL.ntom VALUES (NULL,1,1,'2016-01-01'),(NULL,2,2,'2016-01-01'),(NULL,1,2,'2017-01-01');

现在它给了我

SELECT email,address,movein FROM DL.users 
JOIN DL.ntom ON uidx = fk_uidx
JOIN DL.address ON aidx = fk_aidx

myemail    myadress     2016-01-01
heremail   heraddress   2016-01-01
myemail    heraddress   2017-01-01

当我分组并按日期排序时 myemail仍然住在myadress而不是heraddress

SELECT email,address,movein FROM DL.users 
JOIN DL.ntom ON uidx = fk_uidx
JOIN DL.address ON aidx = fk_aidx
GROUP BY DL.users.email ORDER BY DL.ntom.movein desc

myemail    myadress     2016-01-01
heremail   heraddress   2016-01-01

请帮助我，他想搬到她的位置，他爱她如此mutch

预期结果：

myemail    heraddress   2017-01-01
heremail   heraddress   2016-01-01

:)＆lt; 3

谢谢。

Answer 1

尝试这样：

SELECT email,address,movein FROM DL.users u 
JOIN (SELECT fk_uidx,max(movein) as maxmovein from DL.ntom group by fk_uidx) as t ON u.uidx = t.fk_uidx
JOIN DL.ntom t2 on t2.fk_uidx=t.fk_uidx and t2.movein=t.maxmovein
JOIN DL.address a ON a.aidx = fk_aidx
ORDER BY t2.movein desc

GROUPY中的MYSQL最后一个条目，从m到n JOIN

1 个答案: