pimchanok leuwisetpaiboon Php injection point construction code

Save the following as Test.asp

Copy the code The code is as follows:

$mysql_server_name = "localhost";
$mysql_username = "root";
$mysq l_password = "password";
$ mysql_database = "phpzr"; //??Play??
$c
mysql_select_db($mysql_database,$conn);
$id=$_GET['id'];
$sql = "select username,password from admin where id=$id";
$result=mysql_db_query($mysql_database, $sql,$conn);
$row=mysql_fetch_row($result);
?>

< ;head>
Copy code The code is as follows:

CREATE DATABASE `phpzr`; //Create database name
CREATE TABLE admin (
id int(10) unsigned NOT NULL auto_increment,
username char(10) NOT NULL default '',
password char(10) NOT NULL default '',
useremail char(20) NOT NULL default '',
groupid int(11) NOT NULL default '0',
PRIMARY KEY (id)
) TYPE=MyISAM;
INSERT INTO admin VALUES (1, 'admin', 'itpro.blog.163.com','itpro@163.com', 1);
INSERT INTO admin VALUES (2, 'admin1', 'itpro.blog.163.com','itpro@163.com', 2);
INSERT INTO admin VALUES (3, 'admin2', 'itpro.blog.163.com', 'itpro@163.com', 3);
INSERT INTO admin VALUES (4, 'admin3', 'itpro.blog.163.com','itpro@163.com', 4);
INSERT INTO admin VALUES ( 5, 'admin4', 'itpro.blog.163.com','itpro@163.com', 5);
CREATE TABLE admin1 (
id int(10) unsigned NOT NULL auto_increment,
username char(10) NOT NULL default '',
password char(10) NOT NULL default '',
useremail char(20) NOT NULL default '',
groupid int(11) NOT NULL default '0',
PRIMARY KEY (id)
) TYPE=MyISAM;

INSERT INTO admin1 VALUES (1, 'admin', 'itpro.blog.163.com','itpro@163.com', 1);
INSERT INTO admin1 VALUES (2, 'admin1', 'itpro.blog.163.com','itpro@163.com', 2);
INSERT INTO admin1 VALUES (3, 'admin2', 'itpro.blog.163.com','itpro@163.com' , 3);
INSERT INTO admin1 VALUES (4, 'admin3', 'itpro.blog.163.com','itpro@163.com', 4);
INSERT INTO admin1 VALUES (5, 'admin4', ' itpro.blog.163.com','itpro@163.com', 5);

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