明确题意：

当有0级用户未完成试卷数大于2时输出每个0级用户的试卷未完成数和未完成率（保留3位小数）； 若不存在这样的用户，则输出所有有作答记录的用户的这两个指标。

结果按未完成率升序排序。

问题分解：

• 统计每个用户的等级、未完成数、未完成率和总作答数，生成临时表 t_tag_count:
  • 右连接试卷作答表和用户信息表：exam_record RIGHT JOIN user_info USING(uid)
  • 按用户分组：GROUP BY uid
  • 统计未完成数：COUNT(start_time) - COUNT(submit_time) as incomplete_cnt
  • 统计未完成率：IFNULL(1 - COUNT(submit_time) / COUNT(start_time), 0) as incomplete_rate
    • 保留3位小数：ROUND(x, 3)
  • 统计总作答数：COUNT(start_time) as total_cnt
• 当存在0级用户未完成试卷数大于2时：
  • 筛选存在性条件：WHERE EXISTS (SELECT uid FROM t_tag_count WHERE level = 0 AND incomplete_cnt > 2)
  • 输出每个0级用户的试卷未完成数和未完成率：
    • 筛选0级用户：level = 0
    • SELECT uid, incomplete_cnt, incomplete_rate
• 当不存在0级用户未完成试卷数大于2时：
  • 筛选存在性条件：WHERE NOT EXISTS (SELECT uid FROM t_tag_count WHERE level = 0 AND incomplete_cnt > 2)
  • 输出所有有作答记录的用户的这两个指标：
    • 筛选有作答记录的用户：total_cnt > 0
    • SELECT uid, incomplete_cnt, incomplete_rate
• 合并上述结果，条件互斥，so只可能有一个结果集：UNION ALL

细节问题：

• 表头重命名：as
• 按未完成率升序排序：ORDER BY incomplete_rate

完整代码：

WITH t_tag_count as (
    SELECT uid, `level`,
        COUNT(start_time) - COUNT(submit_time) as incomplete_cnt, -- 未完成数
        ROUND(
            IFNULL(1 - COUNT(submit_time) / COUNT(start_time), 0),
            3) as incomplete_rate, -- 此人未完成率
        COUNT(start_time) as total_cnt -- 总作答数
    FROM exam_record
    RIGHT JOIN user_info USING(uid)
    GROUP BY uid
)

SELECT uid, incomplete_cnt, incomplete_rate
FROM t_tag_count
WHERE EXISTS (
    SELECT uid FROM t_tag_count WHERE `level` = 0 AND incomplete_cnt > 2
) AND `level` = 0
UNION ALL
SELECT uid, incomplete_cnt, incomplete_rate
FROM t_tag_count
WHERE NOT EXISTS (
    SELECT uid FROM t_tag_count WHERE `level` = 0 AND incomplete_cnt > 2
) AND total_cnt > 0
ORDER BY incomplete_rate;