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1 使用/数据库/查找连续两天漏打卡的人.md

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+#### **查找连续两天漏打卡的人**
+
+sql
+
+复制
+
+下载
+
+-- 方法1：使用自连接
+SELECT DISTINCT a.user_id
+FROM attendance a
+LEFT JOIN attendance b ON a.user_id = b.user_id 
+    AND b.attendance_date = DATE_ADD(a.attendance_date, INTERVAL 1 DAY)
+WHERE b.id IS NULL  -- 找不到下一天的记录
+    AND a.attendance_date >= CURDATE() - INTERVAL 30 DAY;
+
+-- 方法2：使用窗口函数（推荐，MySQL 8.0+）
+WITH attendance_gaps AS (
+    SELECT 
+        user_id,
+        attendance_date,
+        LEAD(attendance_date) OVER (
+            PARTITION BY user_id ORDER BY attendance_date
+        ) as next_date
+    FROM attendance
+    WHERE attendance_date >= CURDATE() - INTERVAL 30 DAY
+)
+SELECT DISTINCT user_id
+FROM attendance_gaps
+WHERE DATEDIFF(next_date, attendance_date) >= 2
+   OR (next_date IS NULL AND attendance_date = CURDATE() - INTERVAL 1 DAY);
+
+-- 方法3：使用日期维度表
+SELECT d.user_id
+FROM (
+    -- 生成用户所有应该打卡的日期
+    SELECT DISTINCT user_id, date_range.date
+    FROM attendance
+    CROSS JOIN (
+        SELECT CURDATE() - INTERVAL n DAY as date
+        FROM (SELECT 0 AS n UNION SELECT 1 UNION SELECT 2) days
+    ) date_range
+    WHERE date_range.date BETWEEN CURDATE() - INTERVAL 30 DAY AND CURDATE()
+) d
+LEFT JOIN attendance a ON d.user_id = a.user_id AND d.date = a.attendance_date
+GROUP BY d.user_id
+HAVING SUM(CASE WHEN a.id IS NULL THEN 1 ELSE 0 END) >= 2
+   AND MAX(d.date) - MIN(d.date) = 1;  -- 连续两天