随笔分类 - Mysql
摘要:SELECT @@sql_mode; = 'ONLY_FULL_GROUP_BY,STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_AUTO_CREATE_USER,NO_ENGINE_SUB
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摘要:如图上面是数据: 正确解法(假设分数一样则名次一样) SELECT * FROM student t WHERE (SELECT COUNT(1)+1 FROM student WHERE class_id=t.class_id AND grade>t.grade)<=3 运行结果: 错误解法:来源
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