SQL一对多关系过滤:筛选从未有status=1的id
筛选从未有status=1的ID
给定数据表如下:
| id | status |
|---|---|
| 1 | -1 |
| 2 | 1 |
| 3 | -1 |
| 3 | 1 |
需要筛选出从未出现过status=1的ID,示例中仅返回id=1(id=3因存在status=1的记录,故排除)。以下是几种可行的实现方式:
方法1:NOT EXISTS子查询
通过子查询判断当前ID是否不存在status=1的记录,逻辑直观且性能优异(尤其当id和status字段有索引时):
SELECT DISTINCT t.id FROM your_table t WHERE NOT EXISTS ( SELECT 1 FROM your_table t2 WHERE t2.id = t.id AND t2.status = 1 );
方法2:GROUP BY + HAVING分组过滤
对ID分组后,通过聚合函数判断分组内是否存在status=1的记录:
SELECT id FROM your_table GROUP BY id HAVING MAX(CASE WHEN status = 1 THEN 1 ELSE 0 END) = 0;
如果status只有-1和1两种取值,也可以用更简洁的写法:
SELECT id FROM your_table GROUP BY id HAVING SUM(CASE WHEN status = 1 THEN 1 ELSE 0 END) = 0;
方法3:LEFT JOIN + IS NULL
通过左连接匹配status=1的记录,筛选未匹配成功的ID:
SELECT DISTINCT t.id FROM your_table t LEFT JOIN your_table t2 ON t.id = t2.id AND t2.status = 1 WHERE t2.id IS NULL;
内容的提问来源于stack exchange,提问作者Casper Lindberg
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