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我有以下结构:! U2 k' D `: S |" C# L0 Y
CREATE 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:
( D& k0 U/ X; s- Z8 h% oSELECT 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。
7 v; S8 D/ V& [' W8 i# v但是postgres给出这个错误:) g+ o u" z: p( {# V) E6 d5 ?9 a
错误:“ book.id必须出现列GROUP BY用于子句或聚合函数:SELECT
, \/ G. t ]$ w+ m& y3 S# m9 Nbook.id,author.id,author.name,book.title为last_book FROM author JOIN book
4 t/ J! k) o( T$ J( t# _1 a3 zbook on book.author_id = author.id GROUP BY author.id ORDER BY book.id ASC
) H! r1 E, H2 [ z0 ? _这是因为:
) I& j1 p7 s. V# i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。
0 A; F" M( n G% H4 F. d( G我如何指定postgres:“仅joined_table.id在连接表中按顺序给我最后一行?# e2 j$ u4 R3 P0 A# B/ L! j0 S
编辑:使用此数据:4 `1 y& S2 \( v
INSERT 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:
2 F- M. R1 k% q0 Ibook_id author_id name last_book3 3 1 "Bob" "3rd book from bob"5 2 "David" "2nd book from David" - H, {2 r3 |) r# H2 S' w4 d/ Y
解决方案:
+ s2 q0 E5 s, y 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" N, k N6 B9 m
SELECT DISTINCT ON(expression [,…])每组行的第一行只保留给定表达式。ORDER2 A! M: Q+ W5 a
BY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
6 e# ~' l! {; HBY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。
6 j. y3 h* Z# y; L9 Q! ]3 h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( O5 \5 A/ h* D% O7 i( _& A
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! D+ G/ m( ?8 S+ r$ _
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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