Question

我的结果设置似乎很好，除了我得到了我想避免的其他重复结果，就是这种情况：

我正在使用两张桌子：

顺序

id|companyname|routeid|orderdate
1 |company1   | 23dsa | 
2 |company1   | 23dsa | 2011-04-28
3 |company1   | 23dsa | 2011-04-05
4 |company2   | 24gcc | 2011-22-05

车辆

id| vehiclename | routeid
1 | van1        | 23dsa 
2 | van2        | 24gcc

这是我的查询1：

select t1.companyname, max(t1.orderdate) as orderdate
from order t1, vehicle t2 where t1.routeid=t2.routeid
group by t1.companyname, t1.orderdate
order by t1.id

这是查询2：

 select t1.companyname, max(t1.orderdate) as orderdate
    from order t1, vehicle t2 where t1.routeid=t2.routeid
    group by t1.companyname, t1.orderdate having count(1) = 1
    order by t1.id

这是我的结果：

companyname | orderdate
company1    |  
company1    | 2011-04-28
company1    | 2011-04-05
company2    | 2011-22-05

我希望它是这样的：

companyname | orderdate
company1    | 2011-04-28
company2    | 2011-22-05

我失踪不是空的：

select t1.companyname, max(t1.orderdate) as orderdate
    from order t1, vehicle t2 where t1.orderdate is not null and t1.routeid=t2.routeid
    group by t1.companyname, t1.orderdate
    order by t1.id

Answer 1

select t1.companyname, max(t1.orderdate) as orderdate
from order t1, vehicle t2 where t1.routeid=t2.routeid
group by t1.companyname
order by t1.id

无需按顺序分组，因为您只选择一个。

Answer 2

怎么样：

SELECT
    companyname, MAX(orderdate) AS orderdate
FROM
    `order`
GROUP BY
    companyname
ORDER BY
    id

如果你需要那些车辆：

SELECT
    companyname, MAX(orderdate) AS orderdate
FROM
    `order` o, vehicle v
WHERE
    o.routeid=v.routeid
GROUP BY
    companyname
ORDER BY
    o.id

需要有关复杂查询的帮助

2 个答案: