72 lines
1.9 KiB
SQL
72 lines
1.9 KiB
SQL
-- ========================================
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-- 修复学生多班级问题
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-- ========================================
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-- 第一步：查看有多个班级的学生
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SELECT
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student_id,
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COUNT(*) as class_count,
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GROUP_CONCAT(class_id ORDER BY join_time DESC) as class_ids,
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GROUP_CONCAT(join_time ORDER BY join_time DESC) as join_times
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FROM student_class
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WHERE status = 1
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GROUP BY student_id
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HAVING COUNT(*) > 1;
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-- 第二步：查看编号206的详细记录
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SELECT
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sc.id,
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sc.student_id,
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sc.class_id,
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c.class_name,
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sc.status,
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sc.join_time,
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sc.create_time
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FROM student_class sc
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LEFT JOIN class c ON sc.class_id = c.id
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WHERE sc.student_id = 206
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ORDER BY sc.join_time DESC, sc.create_time DESC;
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-- 第三步：修复数据（保留最新的班级记录，其他设为status=0）
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-- ⚠️ 请先执行上面的查询确认数据后，再执行此更新！
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-- 方法：为每个学生只保留join_time最新的那条记录
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UPDATE student_class sc1
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SET status = 0
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WHERE status = 1
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AND student_id IN (
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-- 找出有多个班级的学生
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SELECT student_id
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FROM (
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SELECT student_id
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FROM student_class
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WHERE status = 1
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GROUP BY student_id
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HAVING COUNT(*) > 1
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) AS multi_class_students
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)
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AND id NOT IN (
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-- 为每个学生保留join_time最新的记录
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SELECT id
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FROM (
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SELECT id
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FROM student_class sc2
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WHERE sc2.status = 1
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AND sc2.student_id = sc1.student_id
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ORDER BY sc2.join_time DESC, sc2.create_time DESC
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LIMIT 1
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) AS keep_records
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);
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-- 第四步：验证修复结果
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SELECT
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student_id,
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COUNT(*) as active_class_count,
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GROUP_CONCAT(class_id) as class_ids
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FROM student_class
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WHERE status = 1
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GROUP BY student_id
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HAVING COUNT(*) > 1;
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-- 应该返回空结果，说明每个学生只有一个活跃班级
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