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我有以下结构:
5 T- W& U: `% v- M5 O, W9 vCREATE TABLE author id integer ,name varchar(255));CREATE TABLE book id integer ,author_id integer ,title varchar(255) ,rating integer);我希望每个作者都能读到最后一本书:
: K' l( U# ]# O: u! ]. x) M' tSELECT book.id,author.id,author.name,book.title as last_bookFROM authorJOIN book book ON book.author_id = author.idGROUP BY author.idORDER BY book.id ASC显然,你可以在那里mysql执行此操作:在MySQL连接两个表,只从第二个表返回一行。
0 k& F$ F N$ w) p但是postgres给出这个错误:9 `1 B) a0 g) ~ q
错误:“ book.id必须出现列GROUP BY用于子句或聚合函数:SELECT
, n6 R A5 [# i4 H9 abook.id,author.id,author.name,book.title为last_book FROM author JOIN book
% ?6 q& f4 x9 xbook on book.author_id = author.id GROUP BY author.id ORDER BY book.id ASC
/ r! ^: F: I3 K这是因为:% `! W* g. U. m% \; L/ Q% Y W
如果存在GROUP BY,则SELECT未分组列(聚合函数除外)不能引用列表表达式,因为未分组列将返回多个可能值。
- K6 d5 x- X" E+ C* v; Z3 O/ H我如何指定postgres:“仅joined_table.id在连接表中按顺序给我最后一行?
5 z- A8 _8 J5 ^) s8 ^/ j$ _编辑:使用此数据:
9 C' ~9 w; ~ u* tINSERT INTO author (id,name) VALUES (1,'Bob(二),David(3),John');INSERT INTO book (id,author_id,title,rating) VALUES 1、1、1、1st book from bob五、nd book from bob六、rd book from bob七、st book from David(、nd book from David',6);我应该看到:8 ^7 t3 v% C4 N3 X4 t
book_id author_id name last_book3 3 1 "Bob" "3rd book from bob"5 2 "David" "2nd book from David"
. e$ `4 I. q1 Z- v# p3 e 解决方案: # K1 I: a; d/ W0 M) Z5 A- U* ^
select distinct on (author.id) book.id,author.id,author.name,book.title as last_bookfrom author inner join book on book.author_id = author.idorder by author.id,book.id desc查看 distinct on- f$ r4 E2 y2 j9 d: u
SELECT DISTINCT ON(expression [,…])每组行的第一行只保留给定表达式。ORDER
) F3 X6 V: o! R7 R. ^( Q1 aBY解释相同的规则DISTINCT ON表达式(请参见上面)。请注意,除非使用。ORDER
" B/ T: F3 @) l$ U1 q- MBY确保所需的行先出现,否则每一集的第一行都是不可预测的。4 L% V. ]' Q( E
内容独特,有必要在其中添加独特列order by。如果不是你想要的顺序,你需要包装查询并重新排序
& f9 C8 S3 `2 \select *from select distinct on (author.id) book.id,author.id,author.name,book.title as last_book from author inner join book on book.author_id = author.id order by author.id,book.id desc) authors_with_first_bookorder by authors_with_first_book.name另一种解决方案是使用Lennart答案中的窗口函数。另一个很常见的是
0 L% t( y0 g( `select book.id,author.id,author.name,book.title as last_bookfrom book inner join ( select author.id as author_id,max(book.id) as book_id from author inner join book on author.id = book.author_id group by author.id ) s on s.book_id = book.id inner join author on book.author_id = author.id |
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