SQL注入漏洞过程实例及解决方案

public class JDBCDemo3 {
  public static void demo3_1(){
    boolean flag=login("aaa' OR ' ","1651561");  //若已知用户名，用这种方式便可不用知道密码就可登陆成功
    if (flag){
      System.out.println("登陆成功");
    }else{
      System.out.println("登陆失败");
    }

  }
  public static boolean login(String username,String password){
    Connection conn=null;
    Statement stat=null;
    ResultSet rs=null;
    boolean flag=false;
    try {
      conn=JDBCUtils.getConnection();
      String sql="SELECT * FROM user WHERE username='"+username+"'AND password='"+password+"'"; //此处是SQL注入漏洞的关键，因为是字符串的拼接，会使查询语句变为：SELECT * FROM user WHERE username='aaa' OR '' AND password='1651561',此查询语句是可得到结果集的，便出现此漏洞
      stat=conn.createStatement();
      rs=stat.executeQuery(sql);
      if(rs.next()){
        flag=true;
      }else{
        flag=false;
      }
    } catch (SQLException e) {
      e.printStackTrace();
    }
    return flag;
  }

public static void demo3_1(){
    boolean flag=login1("aaa' OR ' ","1651561");
    if (flag){
      System.out.println("登陆成功");
    }else{
      System.out.println("登陆失败");
    }

  }
  public static boolean login1(String username,String password){
    Connection conn=null;
    PreparedStatement pstat=null;
    ResultSet rs=null;
    boolean flag=false;

    try {
      conn=JDBCUtils.getConnection();
      String sql="SELECT * FROM user WHERE username="; //使用?代替参数，预先设置好sql格式，就算在输入sql关键字也不会被sql识别
      pstat=conn.prepareStatement(sql);
      pstat.setString(1,username); //设置问号的值
      pstat.setString(2,password);
      rs=pstat.executeQuery();
      if(rs.next()){
        flag=true;
      }else{
        flag=false;
      }
    } catch (SQLException e) {
      e.printStackTrace();
    }
    return flag;
  }
}