mySQL在表中添加数据出错
发布网友
发布时间：2022-04-30 03:24
我来回答
共2个回答
懂视网
时间：2022-04-30 07:45
= t0.commentId
LEFT JOIN wxcommunity.t_ask t2 ON t2.askId = t1.askId
LEFT JOIN wxcommunity.t_school t9 ON t9.schoolId = t2.schoolId
LEFT JOIN wxcommunity.t_answer t3 ON t1.answer_id = t3.answer_id
LEFT JOIN wxcommunity.t_ask t11 on t11.askId = t3.askId
LEFT JOIN wxcommunity.t_school t12 ON t12.schoolId = t11.schoolId
LEFT JOIN wxcommunity.t_member t4 ON t4.userid = t0.userid
left join wxcommunity.t_answer t5 on t5.answer_id = t0.answer_id
LEFT JOIN wxcommunity.t_ask t13 on t13.askId = t5.askId
LEFT JOIN wxcommunity.t_school t14 ON t14.schoolId = t13.schoolId
LEFT JOIN wxcommunity.t_ask t6 on t6.askId = t0.askId
LEFT JOIN wxcommunity.t_school t10 ON t10.schoolId = t6.schoolId
LEFT JOIN wxcommunity.t_handle_state t7 on t7.stateId = t0.stateId
LEFT JOIN wxcommunity.t_accusation_item t8 on t8.accusation_item_id = t0.accusation_item_id
WHERE (t0.askId > 0 OR t0.answer_id > 0 OR t1.askId > 0 OR t1.answer_id > 0)
总结：1，带有or逻辑关系的sql语句jpa查询出错，初步判断jpa本身bug。
2，上述语句在数据库中直接查询（navicat执行sql语句）没问题，但是用在java代码中执行 executeNativeSql（）出错，开始时因mysql-connector-java查询出错查不出数据，换了一个mysql-connector-java5.1.30的能查出数据但是还是有错（查询记录老是少了一条）。初步判断java代码中mysql-connector-java驱动本身缺陷。
mysql存入数据出错总结
标签：
热心网友
时间：2022-04-30 04:53
摘要MySQL数据库导入SQL server数据库不成功，可能是SQL server导出数据库的时候，字符集或者SQL语句出现了不兼容的情况，你可以尝试改变一下字符集，重新导出数据库，然后在MySQL中导入。更简单的方法是，你从SQL server建立一个到MySQL的连接，然后将数据从SQL server直接导出到MySQL上面。咨询记录 · 回答于2021-12-31mySQL在表中添加数据出错MySQL数据库导入SQL server数据库不成功，可能是SQL server导出数据库的时候，字符集或者SQL语句出现了不兼容的情况，你可以尝试改变一下字符集，重新导出数据库，然后在MySQL中导入。更简单的方法是，你从SQL server建立一个到MySQL的连接，然后将数据从SQL server直接导出到MySQL上面。insert into zhanghao（`zhanghao_username`，`zhanghao_password`，`zhanghao_status`） -> values（‘stud’,‘stud’,‘stud’）;ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '`zhanghao_username`，`zhanghao_password`，`zhanghao_status`）values（‘stu' at line 1如果是只用php的话用php的字符串截取因为php是弱类型语言，所有变量都可以视为字符串使用$time='2011-1-1 13:15:20';echo trim(substr($time,0,9));取到9和用trim是为了防止你的日、月出现两位数
