Question

我有两个表订阅者和家属

 Subscribers
 1 John
 2 Joe
 3 Jean
 4 Alan
 5 Elena

 Dependents
 1 John (Same Person)
 1 Betty
 1 Hannah
 3 Grey
 4 Sumen
 4 Kevin

SELECT * FROM Subscribers LEFT JOIN Dependents ON Subscribers.Id = Dependents.Id

这给了我

John, John
John, Betty
John, Hannah
Joe,
Jean, Grey
Alan, Sumen
Alan, Kevin
Elena

我需要：

John, John
John, Betty
John, Hannah
Joe
Jean (Added)
Jean, Grey
Alan (Added)
Alan, Sumen
Alan, Kevin
Elena

基本上我希望订阅者记录独立于自己，如果依赖者本身不像约翰。 Joe，Jean，Alan，Elena都会有与家属无关的记录。

我认为我需要这样的东西

 SELECT * FROM SUBSCRIBERS LEFT JOIN Dependents ON Subscribers.Id = Depedents.Id
 Where Subscribers.First_Name IS NOT Dependents.First_Name

谢谢！

Answer 1

如果没有工会，我想不出任何办法。

SELECT S.name as subscriber, D.name as dependent FROM Subscribers S left JOIN Dependents D ON S.Id = D.Id
    UNION 
    SELECT S.name as subscriber, D.name as dependent FROM Subscribers S left JOIN Dependents D ON S.name = D.name
    ORDER BY subscriber, dependent

http://sqlfiddle.com/#!2/dc7ba/11

Answer 2

您还可以使用UNION ALL like this：

SELECT s.Name AS subscriber, d.Name AS dependent
FROM Subscribers AS s
INNER JOIN Dependents AS d ON s.Id = d.Id

UNION ALL

SELECT s.Name, NULL
FROM Subscribers AS s
LEFT JOIN Dependents AS d ON s.Name = d.Name
WHERE d.Name IS NULL

ORDER BY subscriber, dependent
;

基本上，第一个SELECT通过Id获取所有匹配，而第二个SELECT添加了dependents中找不到名称的订阅者。

订阅者依赖表mysql

2 个答案: