mysql無法修改密碼的兩個解決辦法方法一:# /etc/init.d/mysql stop# mysqld_safe --user=mysql --skip-grant-tables --skip-networking &# mysql -u root mysqlmysql> UPDATE user SET Password=PASSWORD(’newpassword’) where USER=’root’;mysql>
環境介紹:使用yum安裝了一個mysql,因版本不符合需求,使用rpm卸載,又用rpm安裝了一個版本的Mysql。報錯提示:Mysql安裝完畢後,錯誤記錄檔中報錯:[ERROR] Native table 'performance_schema'.'xxxxx' hasthe wrong structure解決方案:在作業系統層級,執行mysql_upgrade報錯原因:當前performance_schema的資訊與mysql實際的情況不一致。參考:If you install MySQL
【現象】匯入MySQL的mysqldump檔案時報錯,提示ERROR 2013 (HY000) at line xxoo: Lost connection to MySQL serverduring query,匯入操作中斷退出。【原因】要匯入的mysqldump檔案中insert的values值太多,超過了MySQL參數max_allowed_packet的值,進而導致匯入操作中斷退出。【解決方案】調高max_allowed_packet的值:SQL> set global
1.1.1. mysql5.6.14的datadir遷移時遇到報錯【環境描述】在機器A上安裝了perconamysql 5.6.14,資料庫停啟正常,datadir路徑為pathA,並且已經做了應用程式資料庫的初始化工作,然後關閉了這個資料庫執行個體,把它的datadir和/etc/my.cnf遷移到另外一台機器B上的percona mysql 5.6.14,遷移後的datadir路徑修改成路徑pathB,在啟動mysql服務的時候遇到問題,啟動失敗。操作步驟:1)
mysql案例:Java程式訪問Mysql報錯'OPTION SQL_SELECT_LIMIT=XXX'【環境描述】mysql版本: 5.6.14-56-log Percona Server (GPL)程式: 使用JDBC驅動mysql-connector-java訪問mysql資料庫【報錯資訊】程式訪問mysql執行SQL語句的時候報錯:You have an error in your SQL syntax; check the manual that corresponds to
【MySQL案例】mysql-libs-5.1.73-3.el6_5.x86_64 conflicts with file from package Percona-Server-server如果遇到mysql-libs-5.1.73-3.el6_5.x86_64 conflicts with file from package Percona-Server-server報錯,有兩種情況導致這個問題:1)卸載作業系統預裝的mysql-libs包rpm -qa | grep -i libs |
1.1.1. ERROR 1665 (HY000)【環境描述】msyql5.6.14【報錯資訊】執行SQL語句的時候報錯:ERROR 1665 (HY000): Cannot executestatement: impossible to write to binary log since BINLOG_FORMAT = STATEMENTand at least one table uses a storage engine limited to row-based
1.1.1. HA: GTID_MODE配置不一致【環境描述】msyql5.6.14【報錯資訊】初始狀態Master和Slave都開啟了enforce-gtid-consistency和gtid-mode,然後在Master上把它倆都修改成了off關閉狀態,這時,Slave發生報錯:Last_IO_Error: The slave IO thread stopsbecause the master has @@GLOBAL.GTID_MODE OFF and this server
MySql的like語句中的萬用字元:百分比符號、底線和escape%代表任意多個字元Sql代碼 http://blog.csdn.net/yc7369/select * from user where username like '%huxiao';select * from user where username like 'huxiao%';select * from user where username like '%huxiao%';_代表一個字元Sql代碼select *