我有两张桌子:
我想相应地列出奥运奖牌表。我有这个查询,它可以工作,但它似乎扼杀了mysql。希望有人能帮助我进行优化查询。
SELECT DISTINCT country AS sc,
IFNULL(
(SELECT COUNT(silver_country)
FROM olympic_medalists
WHERE silver_country = sc AND silver_country != ''
GROUP BY silver_country),0) AS silver_medals,
IFNULL(
(SELECT COUNT(gold_country)
FROM olympic_medalists
WHERE gold_country = sc AND gold_country != ''
GROUP BY gold_country),0) AS gold_medals,
IFNULL(
(SELECT COUNT(bronze_country)
FROM olympic_medalists
WHERE bronze_country = sc AND bronze_country != ''
GROUP BY bronze_country),0) AS bronze_medals
FROM olympic_medalists, flags
GROUP BY country, gold_medals, silver_country, bronze_medals HAVING (
silver_medals >= 1 || gold_medals >= 1 || bronze_medals >= 1)
ORDER BY gold_medals DESC, silver_medals DESC, bronze_medals DESC,
SUM(gold_medals+silver_medals+bronze_medals)结果如下:
country | g | s | b | tot
---------------------------------
country1 | 9 | 5 | 2 | 16
country2 | 5 | 5 | 5 | 15诸若此类
谢谢!
olympic medalists:
`id` int(8) NOT NULL auto_increment,
`gold_country` varchar(64) collate utf8_unicode_ci default NULL,
`silver_country` varchar(64) collate utf8_unicode_ci default NULL,
`bronze_country` varchar(64) collate utf8_unicode_ci default NULL, PRIMARY KEY (`id`)
flags
`id` int(11) NOT NULL auto_increment,
`country` varchar(128) default NULL,
PRIMARY KEY (`id`)发布于 2012-08-04 06:27:59
这将比当前为交叉连接关系中的每一行执行三个不同的SELECT子查询的解决方案高效得多(您想知道为什么会停下来!):
SELECT a.country,
COALESCE(b.cnt,0) AS g,
COALESCE(c.cnt,0) AS s,
COALESCE(d.cnt,0) AS b,
COALESCE(b.cnt,0) +
COALESCE(c.cnt,0) +
COALESCE(d.cnt,0) AS tot
FROM flags a
LEFT JOIN (
SELECT gold_country, COUNT(*) AS cnt
FROM olympic_medalists
GROUP BY gold_country
) b ON a.country = b.gold_country
LEFT JOIN (
SELECT silver_country, COUNT(*) AS cnt
FROM olympic_medalists
GROUP BY silver_country
) c ON a.country = c.silver_country
LEFT JOIN (
SELECT bronze_country, COUNT(*) AS cnt
FROM olympic_medalists
GROUP BY bronze_country
) d ON a.country = d.bronze_country更快的不是将实际的文本国家名称存储在每个金、银和青铜列中,而是存储基于整数的country id。对整数的比较总是比字符串上的比较快。
此外,一旦您用相应的id替换了olympic_medalists表中的每个国家名称,您将希望在每一列(黄金、银和青铜)上创建一个索引。
将文本名称更新为相应的id是一项简单的任务,可以使用单个UPDATE语句和一些ALTER TABLE命令来完成。
发布于 2012-08-04 06:36:20
试试这个:
SELECT F.COUNTRY,IFNULL(B.G,0) AS G,IFNULL(B.S,0) AS S,
IFNULL(B.B,0) AS B,IFNULL(B.G+B.S+B.B,0) AS TOTAL
FROM FLAGS F LEFT OUTER JOIN
(SELECT A.COUNTRY,
SUM(CASE WHEN MEDAL ='G' THEN 1 ELSE 0 END) AS G,
SUM(CASE WHEN MEDAL ='S' THEN 1 ELSE 0 END) AS S,
SUM(CASE WHEN MEDAL ='B' THEN 1 ELSE 0 END) AS B
FROM
(SELECT GOLD_COUNTRY AS COUNTRY,'G' AS MEDAL
FROM OLYMPIC_MEDALISTS WHERE GOLD_COUNTRY IS NOT NULL
UNION ALL
SELECT SILVER_COUNTRY AS COUNTRY,'S' AS MEDAL
FROM OLYMPIC_MEDALISTS WHERE SILVER_COUNTRY IS NOT NULL
UNION ALL
SELECT BRONZE_COUNTRY AS COUNTRY,'B' AS MEDAL
FROM OLYMPIC_MEDALISTS WHERE BRONZE_COUNTRY IS NOT NULL)A
GROUP BY A.COUNTRY)B
ON F.COUNTRY=B.COUNTRY
ORDER BY IFNULL(B.G,0) DESC,IFNULL(B.S,0) DESC,
IFNULL(B.B,0) DESC,IFNULL(B.G+B.S+B.B,0) DESC,F.COUNTRYhttps://stackoverflow.com/questions/11806268
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