示例
示例
SELECT * FROM users WHERE phone IS NULL;
SELECT * FROM users WHERE phone IS NOT NULL;
SELECT COUNT(*) AS all_rows,
COUNT(phone) AS has_phone,
SUM(phone IS NULL) AS null_phone -- MySQL：布尔转 0/1
FROM users;| 判空 | IS NULL / IS NOT NULL |
|---|---|
| 错误写法 | = NULL、<> NULL 永远不成立 |
| 聚合 | COUNT(col) 不统计 NULL；AVG/SUM 忽略 NULL |
PostgreSQL 中可写作 SUM((phone IS NULL)::int)；MySQL 里布尔表达式本身即 0/1。