我有以下脚本读取员工ID从Employee表中分配给员工的每一个数据库,并改变了EmployeeLedger_ $ EMPID表中相应的数据库在shell脚本在MySQL使用concat函数
cls_ip="localhost";
mysql="mysql -h $cls_ip -u root"
list=`echo "show databases like '$dbPattern'" | $mysql| grep -v Database`
for db in $list
do
echo "altering EmployeeLedger table for database $db";
${mysql} ${db} -e "use $db";
empId=`${mysql} ${db} -e "select EMPID from Employee"`;
echo "$empId";
${mysql} ${db} -e "alter table concat('EmpTimeLedger',$empId) add column HOLIDAY tinyint(1) not null default 1;";
done
这里我没有成功将雇员表与EmplyeeLedger进行连接,以构成EmployeeLedger_ $ empId表。我该怎么做?
为什么不'你只是这样做'empId =“EmpTimeLedger_ $ empId”;'然后'“ALTER TABLE $ empId ADD COLUMN ...”'? – Cyclonecode 2015-02-23 14:03:32
@Cyclone这是一个伎俩!谢谢 – 2015-02-23 17:00:52
不客气,太糟糕了我没有添加它作为答案=) – Cyclonecode 2015-02-23 17:05:38