優化MySQL查詢：獎牌榜

Question

我有2個表：

1. olympic_medalists與列gold_country，silver_country，bronze_country

2. 標誌與國家列

我要列出奧運獎牌表相應。我有這個查詢，它的工作原理，但它似乎殺死MySQL。希望有人可以幫助我優化查詢。

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) 

結果會是這樣的：

country | g | s | b | tot 
--------------------------------- 
country1 | 9 | 5 | 2 | 16 
country2 | 5 | 5 | 5 | 15 

等

謝謝！

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`) 

Answer 1

這將是比你目前的執行三種不同的SELECT子查詢在交叉連接關係，各行的解決方案更有效（和你想知道爲什麼它攤上了！）：

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 

---

甚至更​​快的是將實際的文本國家名稱存儲在每個黃金，白銀和古銅色列中，而不是存儲基於整數的國家id。整數比較總是比字符串比較快。

此外，一旦用對應的ID替換olympic_medalists表中的每個國家/地區名稱，您都需要在每列（黃金，白銀和銅牌）上創建索引。

更新文本名稱是他們相應的id的，而不是一個簡單的任務，可以用一個單一的UPDATE聲明與一些ALTER TABLE命令共同完成。

Answer 2

試試這個：

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.COUNTRY