Question

我正处于最后一个“高级选择陈述”问题，似乎无法得到正确答案。

表1：书虫数据库的模式。主键带下划线。有一些外键引用将表连接在一起;你可以利用天然连接来利用它们。

Author(aid, alastname, afirstname, acountry, aborn, adied).
Book(bid, btitle, pid, bdate, bpages, bprice).
City(cid, cname, cstate, ccountry).
Publisher(pid, pname).
Author_Book(aid, bid).
Publisher_City(pid, cid).

问题是......“查找已经写过三本或更多书籍的作者的姓名。”

代码有效但我想要完整的作者姓名而不是作者ID ..

select count(aid) as authorBook, aid as authorName
from Author_book natural join Author
group by aid
having count(aid) > 3;

决赛桌出现了......

 authorbook | authorname 
------------+------------
      8 | dick
      4 | thar
(2 rows)

Answer 1

您需要HAVING COUNT(aid) > 3您不能使用别名authorBook。

如果您需要作者姓名及其在表格中的名称，您可以这样做：

SELECT AuthorName, COUNT(*) AS CountOfBooks FROM AuthorBooks GROUP BY AuthorName HAVING COUNT(*) > 3

如果作者的姓名是外键，请执行以下操作：

SELECT x.AuthorName, COUNT(x2.*) FROM
AuthorBooks x2 INNER JOIN Authors x1 ON x1.AuthorID = x2.AuthorID
GROUP BY x.AuthorName HAVING COUNT(x2.*) > 3

Answer 2

SELECT  A.alastname,
        A.afirstname,
        COUNT(*) as Books
FROM Author_Book AB
INNER JOIN Author A
    ON AB.aid = A.aid
GROUP BY aid
HAVING COUNT(*) > 3

SQL聚合计数语句

2 个答案: