本文介绍了嵌套选择或连接查询?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我是 sql 新手,所以这可能有一个非常基本的答案,但要回答的问题如下.....

i am new to sql so this may have a very basic answer but the question to answer is as follows .....

哪部电影的表演费用最少?在结果中包含电影名称和电影名称.?

电影名称电影名称在两个不同的表电影电影中.收入表现表中.我不知道如何完成这个查询.这是我到目前为止所得到的,但它在第 3 行第 7 列中出现了错误.

film name and cinema name are in two different table film and cinema. takings are in the performance table. i cant figure out how to complete this query. this is what i have got so far but it comes with an error in line 3 column 7.

select cinema_no,film_no
from CINEMA, film
where takings ( select min(takings)
from performance);

推荐答案

因为 oracle 有多个标签,我忽略了标签 mysql(其中一个你应该去掉.例如,请决定你使用的是哪个 DBMS,顺便说一下,我已经删除了不相关的 oracle-sqldeveloper ).

Because of multiple tags with oracle, I ignored the tag mysql ( of one which you should get rid of. e.g. please decide which DBMS are you using, by the way I already removed the irrelevant one oracle-sqldeveloper ).

似乎您需要这样一个 select 语句(更喜欢使用现代 ANSI-92 JOIN 语法,易于维护和理解)按降序 sumrow_number 函数的贡献为:

It seems you need such a select statement ( prefer using modern ANSI-92 JOIN syntax, easily maintained and understandable ) with ordering by descending sum and contribution of row_number function as :

SELECT Name, Sum_Takings
  FROM
  (
   SELECT f.Name, sum(p.Takings) Sum_Takings,
          row_number() over (ORDER BY sum(p.Takings)) as rn
     FROM Film f
     LEFT JOIN Cinema c ON f.Cinema_ID = c.ID
     LEFT JOIN Performance p ON f.ID = p.id_film
    GROUP BY f.Name
   )
  WHERE rn = 1;

添加了 DDL 语句如下:

SQL>  CREATE TABLE Cinema (
  2     ID     integer PRIMARY KEY NOT NULL,
  3     Title  varchar2(100) NOT NULL
  4   );

Table created

SQL>  CREATE TABLE Film (
  2     ID   integer PRIMARY KEY NOT NULL,
  3     Name varchar2(100) NOT NULL,
  4     Cinema_ID integer
  5       CONSTRAINT fk_Cinema_ID REFERENCES Cinema(ID)
  6   );

Table created

SQL>   CREATE TABLE Performance (
  2     ID      integer PRIMARY KEY NOT NULL,
  3     ID_Film integer
  4          CONSTRAINT fk_Film_ID REFERENCES Film(ID),
  5     Takings integer
  6   );

Table created

SQL>  INSERT ALL
  2         INTO Cinema(ID,Title) VALUES(1,'NiteHawk')
  3         INTO Cinema(ID,Title) VALUES(2,'Symphony Space')
  4         INTO Cinema(ID,Title) VALUES(3,'The Ziegfeld')
  5         INTO Cinema(ID,Title) VALUES(4,'Cinema Village')
  6       SELECT * FROM dual;

4 rows inserted

SQL>  INSERT ALL
  2         INTO Film(ID,Name,Cinema_ID) VALUES(1,'Citizen Kane',1)
  3         INTO Film(ID,Name,Cinema_ID) VALUES(2,'Titanic',2)
  4         INTO Film(ID,Name,Cinema_ID) VALUES(3,'Brave Heart',4)
  5         INTO Film(ID,Name,Cinema_ID) VALUES(4,'Dumb and Dummer',3)
  6         INTO Film(ID,Name,Cinema_ID) VALUES(5,'How To Train Your Dragon',2)
  7         INTO Film(ID,Name,Cinema_ID) VALUES(6,'Beetle Juice',3)
  8       SELECT * FROM dual;

6 rows inserted

SQL>  INSERT ALL
  2         INTO Performance VALUES(1,1,15)
  3         INTO Performance VALUES(2,1,4)
  4         INTO Performance VALUES(3,2,10)
  5         INTO Performance VALUES(4,3,1)
  6         INTO Performance VALUES(5,4,5)
  7         INTO Performance VALUES(6,3,3)
  8         INTO Performance VALUES(7,2,7)
  9         INTO Performance VALUES(8,5,7)
 10         INTO Performance VALUES(9,6,6)
 11       SELECT * FROM dual;

9 rows inserted

SQL> commit;

Commit complete

SQL> SELECT Name, Sum_Takings
  2    FROM
  3    (
  4     SELECT f.Name, sum(p.Takings) Sum_Takings,
  5            row_number() over (ORDER BY sum(p.Takings)) as rn
  6       FROM Film f
  7       LEFT JOIN Cinema c ON f.Cinema_ID = c.ID
  8       LEFT JOIN Performance p ON f.ID = p.id_film
  9      GROUP BY f.Name
 10     )
 11    WHERE rn = 1
 12  ;

NAME                                                                  SUM_TAKINGS
--------------------------------------------------------------------- -----------
Brave Heart                                                                     4

dbfiddle.uk 演示

这篇关于嵌套选择或连接查询?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

08-01 14:59