SQL注入漏洞过程实例及解决方案

代码示例：

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=? AND password=?"; //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;  }}

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