按人群分组的金牌数
（仅限西欧国家）
+----------+--------------------+-------+
| season | demographic_group | golds |
|----------+--------------------+-------|
| Winter | Male Age 26+ | 13 |
| Winter | Female Age 26+ | 8 |
| Summer | Male Age 13-25 | 7 |
| Summer | Female Age 13-25 | 6 |
| Winter | Male Age 13-25 | 4 |
| Summer | Male Age 26+ | 4 |
| Winter | Female Age 13-25 | 4 |
| Summer | Female Age 26+ | 2 |
+----------+--------------------+-------+
相关表
相关表
选项 A：先 JOIN，后 UNION
选项 A：先 JOIN，后 UNION
步骤 1： 编写上方含 JOIN 的查询
SELECT
athlete_id,
gender,
age,
gold
FROM summer_games AS sg
JOIN athletes AS a
ON sg.athlete_id = a.id;
查询成功运行！
选项 A：先 JOIN，后 UNION
步骤 2： 编写下方查询，并用 UNION 合并两者
SELECT
athlete_id,
gender,
age,
gold
FROM summer_games AS sg
JOIN athletes AS a
ON sg.athlete_id = a.id
UNION ALL
SELECT
athlete_id,
gender,
age,
gold
FROM winter_games AS wg
JOIN athletes AS a
ON wg.athlete_id = a.id;
选项 B：先 UNION，后 JOIN
选项 B：先 UNION，后 JOIN
步骤 1： 先创建初始 UNION
SELECT
athlete_id,
gold
FROM summer_games AS sg
UNION
SELECT
athlete_id,
gold
FROM winter_games AS wg;
选项 B：先 UNION，后 JOIN
步骤 2： 转为子查询并 JOIN
SELECT
athlete_id,
gender,
age,
gold
FROM
(SELECT
athlete_id,
gold
FROM summer_games AS sg
UNION ALL
SELECT athlete_id, gold
FROM winter_games AS wg) AS g
JOIN athletes AS a
ON g.athlete_id = a.id;
对比
选项 A
SELECT
athlete_id,
gender,
age,
gold
FROM summer_games AS sg
JOIN athletes AS a
ON sg.athlete_id = a.id
UNION ALL
SELECT
athlete_id,
gender,
age,
gold
FROM winter_games AS wg
JOIN athletes AS a
ON wg.athlete_id = a.id;
选项 B
SELECT
athlete_id,
gender,
age,
gold
FROM
(SELECT
athlete_id,
gold
FROM summer_games AS sg
UNION ALL
SELECT athlete_id, gold
FROM winter_games AS wg) AS g
JOIN athletes AS a
ON g.athlete_id = a.id;
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