更新:
我使用MySQL语句创建视图:
如果出版了50多本书,我需要出示编辑名,姓和城。我有三张桌子:
create table editors (
ed_id char(11),
ed_lname varchar(20),
ed_fname varchar(20),
ed_pos varchar(12),
phone varchar(10),
address varchar(30),
city varchar(20),
state char(2),
zip char(5),
ed_boss char(11));
create table titleditors (
ed_id char(11),
title_id char(6),
ed_ord integer);
create table salesdetails (
sonum integer,
qty_ordered integer,
qty_shipped integer,
title_id char(6),
date_shipped date);
有谁能告诉我创建此结果的代码是什么? 我没有制作表格,我只需要处理我给出的内容。
答案 0 :(得分:18)
过时的语法(注意连接条件和过滤条件的混合):
CREATE VIEW qtyorderedview AS
SELECT
salesdetails.title_id, salesdetails.qty_shipped,
editors.ed_id, editors.ed_lname, editors.ed_fname, editors.city
FROM
titleditors, salesdetails, editors
WHERE
titleditors.title_id = salesdetails.title_id
AND editors.ed_id = titleditors.ed_id
AND salesdetails.qty_ordered > 50
现代语法(连接条件和过滤条件是分开的):
CREATE VIEW qtyorderedview AS
SELECT
salesdetails.title_id, salesdetails.qty_shipped,
editors.ed_id, editors.ed_lname, editors.ed_fname, editors.city
FROM
titleditors
INNER JOIN salesdetails ON titleditors.title_id = salesdetails.title_id
INNER JOIN editors ON editors.ed_id = titleditors.ed_id
WHERE
salesdetails.qty_ordered > 50
加入反对视图的工作方式与对表的连接完全相同。只需使用视图名称代替常规表名。
答案 1 :(得分:5)
SELECT e.*
FROM (
SELECT DISTINCT te.ed_id
FROM (
SELECT title_id
FROM sales_details
GROUP BY
title_id
HAVING SUM(qty_shipped) > 50
) t
JOIN titleditors te
ON te.title_id = t.title_id
) te
JOIN editors e
ON e.ed_id = te.ed_id