# Members - DaVinciCTF
- Category: Web- Points: 66- Solves: 61- Solved by: raff01
## Description
Can you get more information about the members?
`http://challs.dvc.tf:1337/members`
## Analysis
In order to solve this challenge it's necessary to have solved the previous one called "Authentication" that permits to access the following web service:
`http://challs.dvc.tf:1337/members`
Let's have a look to the web site. We have a page where there is a table with some information about members on the right and a form that allows to search members on the left. By analysing the source code of the page we can see that the form uses the GET method to send search-parameters, so all the text we write will be encoded into the url. Once the server has recived our data, it will return information about members. So it seems there's a MySQL Database back the application. Let's inject some malicious code into the text field:
`Leonard" OR 1=1; --`
and the application will return all the records of the table, so the application is vulnerable to SQL Injection!
## Solution
Probably the flag isn't in the same table where are archived all the members. So the best thing to do is to show all the table names and see if there is something interesting. To do this we can use the **UNION** command that permits to add a malicious sub-query to the original one and get other data from the database. In particular the table **information_schema.tables** contains the information about all the tables located in the db. Let's inject the following code:
`Leonard" UNION (SELECT TABLE_NAME,2,3 FROM INFORMATION_SCHEMA.TABLES); --`
and the application will return all the table names that are stored in the database. If we scroll down we can see two interesting tables: **members** that is the table that contains all the members information and another table called **supa_secret_table**, let's analyse it. The table called **information_schema.columns** contains all the information about the columns of all the tables stored. So we can get the field names of **supa_secret_table** by injecting this code:
`Leonard" UNION (SELECT COLUMN_NAME,2,3 FROM INFORMATION_SCHEMA.COLUMNS WHERE TABLE_NAME='supa_secret_table'); --`
the application will return two records with the name of the fields: **id** and **flag**. So now let's get the content of flag:
`Leonard" UNION (SELECT flag,2,3 FROM supa_secret_table); --`
and the application will print the flag!
## Flag
`dvCTF{1_h0p3_u_d1dnt_us3_sqlm4p}`
