1

得到这个 SQL：

SELECT DISTINCT(LocCity), LocZipCode FROM exp_course_events order by LocCity

这个数据：

INSERT INTO `exp_course_events` (`LocCity`, `LocZipCode`) VALUES 
  ('Aguadilla', '00602'), ('Akron', '44300'),('Akron', '44333'),
  ('Albany', '12205'),    ('Albuquerque', '87102'),
  ('Albuquerque', '87109'), ('Austin', '78741'),
  ('Austin', '78753'), ('Austin', '78757'),
  ('Bend', '97701'), ('San Antonio', '78200'),
  ('San Antonio', '78201'),
  ....
  ('San Antonio', '78207');

只需要为每个 LocCity 返回一个 LocZipCode 值，最好是该 LocCity 的最低编号 LocZipCode。

这是我想要的结果：

Aguadilla, 00602
Akron, 44300
Albany, 12205
Albuquerque, 87102
Austin, 78741
Bend, 97701
San Antonio, 78200
San Diego, 92108
San Francisco, 94111
San Juan, 00926
Santa Clara (San Jose), 95054
Springdale, 72762
Springfield, 62703
St. Louis. 63105
Visalia, 993291
Waco, 76705
Warwick, 02886
Waukesha, 53186
West Chester, 45069
West Des Moines, 50300
4

4 回答 4

3 
SELECT LocCity, MIN(LocZipCode) 
FROM exp_course_events 
GROUP BY LocCity
ORDER BY LocCity

在此处阅读有关GROUP BY子句的信息。

于 2011-08-22T16:53:56.493 回答
1 
select LocCity, min(LocZipCode)
from exp_course_events
group by LocCity
于 2011-08-22T16:53:45.247 回答
1

一个简单GROUP BY的MIN()聚合就可以了。只需确保MIN()使用名称（如AS LocZipCode）为列命名，以便您可以更轻松地在应用程序端获取它。

SELECT LocCity, MIN(LocZipCode) AS LocZipCode FROM exp_course_events GROUP BY LocCity;
于 2011-08-22T16:53:58.147 回答
1 
SELECT LocCity
    , MIN(LocZipCode)
FROM exp_course_events
GROUP by LocCity
order by LocCity
于 2011-08-22T16:54:12.303 回答