我有以下PHP代码,用于创建数据库,用户和授予用户权限：
$con = MysqL_connect("IP.ADDRESS","user","pass"); MysqL_query("CREATE DATABASE ".$dbuser."",$con)or die(MysqL_error()); MysqL_query("grant all on ".$dbuser.".* to ".$dbname." identified by '".$dbpass."'",$con) or die(MysqL_error());
我想在shell脚本中执行这些相同的操作.它只是这样的：
MyUSER="user" MyPASS="pass" MysqL -u $MyUSER -h -p$MyPASS -Bse "CREATE DATABASE $dbuser;" MysqL -u $MyUSER -h -p$MyPASS -Bse "GRANT ALL ON $DBUSER.* to $DBNAME identified by $DBPASS;"
在postwwwacct(一个cPanel帖子帐户创建钩子脚本)中需要编辑,理想情况下它将是完全独立的
您需要使用小写“MysqL”并在-h之后添加主机名,并且您混合使用单引号和双引号.此外,您需要设置dbname,dbuser和dbpass的值并使用一致的大小写：
MyUSER="user" MyPASS="pass" HostName="host" dbName="dbname" dbUser="dbuser" dbPass="dbpass" MysqL -u $MyUSER -h $HostName -p$MyPASS -Bse "CREATE DATABASE $dbUser;" MysqL -u $MyUSER -h $HostName -p$MyPASS -Bse "GRANT ALL ON ${dbUser}.* to $dbName identified by $dbPass;"
但我对你的sql语法没有百分之百的信心.我认为它看起来更像是这样的：
MysqL -u $MyUSER -h $HostName -p$MyPASS -Bse "CREATE DATABASE $dbName;" MysqL -u $MyUSER -h $HostName -p$MyPASS -Bse "GRANT ALL ON ${dbName}.* to $dbUser identified by $dbPass;"
