如何在不使用group by语句的情况下选择不同的行 [英] How to select distinct rows without using group by statement

问题描述

A B C
1 1 1 
1 1 1 
2 2 2 
2 2 2 
3 3 3 
3 3 3 
4 4 4 
4 4 4 
5 5 5 
5 5 5 
5 5 5 
6 6 6
6 6 6 

我只输出不重复的行，而不使用group by语句.我不能使用group by，因为它会使mysql挂起.所以它应该返回

I am to output only the distinct rows without using the group by statement. I cannot use group by because it makes mysql hang. So it should return

1 1 1
2 2 2 
3 3 3 
4 4 4 
5 5 5 
6 6 6 

我正在使用DISTINCT进行内部联接.这也不起作用:

I am using DISTINCT in for an inner join.This does not work either:

SELECT DISTINCT * FROM TABLEA inner join TABLEB on TABLEA.A = TABLEB.A 

推荐答案

SELECT DISTINCT A,B,C FROM TABLE;

根据mysql文档，DISTINCT指定从结果集中删除重复的行(

According to mysql documentation, DISTINCT specifies removal of duplicate rows from the result set (http://dev.mysql.com/doc/refman/5.0/en/select.html)

我在jsfiddle上创建了一个示例，它可以恕我直言

I created a sample on jsfiddle and it works IMHO

create table tablea (A int,B int,C int);
create table tableb (A int);

INSERT INTO tablea VALUES (1,1,1),(1,1,1),(2,2,2),(2,2,2),(3,3,3),(3,3,3),(4,4,4),(4,4,4),(5,5,5),(5,5,5),(5,5,5),(6,6,6),(6,6,6);
INSERT INTO tableb VALUES (1),(1),(1),(1),(1);

SELECT DISTINCT tablea.A,tablea.B,tablea.C FROM tablea INNER JOIN tableb ON tablea.A=tableb.A;

可以免费试用 SQLFiddle .

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