【发布时间】:2015-01-04 14:38:41
【问题描述】:
我的连接语句有问题
String q1="select e.employee_id,e.manager_id,e.first_name,e.last_name,e.salary,e.commission_pct,d.manager_id,d.employee_id from employees as e ,employees as d where e.manager_id=d.employee_id and e.employee_id="+jComboBox1.getSelectedItem();
try{
OracleDataSource ods=new OracleDataSource();
ods.setURL("jdbc:oracle:thin:hr/hr@localhost:1521/XE");
Connection con=ods.getConnection();
Statement s=con.createStatement();
s.execute(q1);
ResultSet rs=s.getResultSet();
String x=(String)jComboBox1.getSelectedItem();
while(rs.next()){
if (x.equals(rs.getString("e.employee_id"))){
jTextField1.setText(rs.getString("e.first_name"));
jTextField2.setText(rs.getString("e.last_name"));
jTextField3.setText(rs.getString("e.salary"));
jTextField4.setText(rs.getString("e.commission_pct"));
jTextField5.setText(rs.getString("d.first_name"));
}
}
con.close();
}catch(Exception e){e.printStackTrace();}
我该如何解决这个问题?! ..................................................... ..................................................... ..................................................... ..................................................... .....................
【问题讨论】:
-
到底是什么问题?
-
employee_id 的类型是什么?
-
可爱的sql injection attack。高枕无忧。很快有人会 pwn 你的服务器,让你的问题没有实际意义。
-
如果遇到异常,能否发布堆栈跟踪信息?
标签: java javascript sql command spring-jdbc