sql injection "union select null" - CodeProject
Union select null, null, null, null, null, null, null from information_schema.tables. for a small database containing three tables. this instruction is used in sql injection I tried it and it worked but I didn't really know how it works can somebody help me...
Шпаргалка по SQL инъекциям | DefconRU
(M): SELECT CONCAT(login, password) FROM members. 7. Строки без кавычек Есть несколько
Confusing SQL error in SELECT NULL, *, NULL, NULL - Stack Overflow
UNION ALL SELECT NULL, *, NULL, NULL FROM email. I understand what this does and why; the hacker needs to create a query that has the same number
sql - Counting null values as unique value - Stack Overflow
Hours 1 1 2 null null null. The result must be: 3. My query is: select count(distinct hour) from hours; but it returns: 2. I tested also
SQL Инъекции | Page 798 | ANTICHAT - Security online community
http://www.indianamri.com/index.php?page=Indiana_MRI_Bloomington_-_MRI_Services_for_Bloomington_and_Southern_Indiana_Magnetic_Resonance_Imaging&menu_id=1'+limit+0+UNION+SELECT+1,concat_ws(0x3a,user(),version(),database())+
Заметка про NULL / Хабр | select col from ttt where col is not null
Существуют специальные операторы IS NULL и IS NOT NULL, которые позволяют производить сравнения с NULLами.
COUNT, DISTINCT, and NULLs - SQL with Bert
Viewing and COUNTing the NULLs. Sometimes we might have to do the opposite of what the default functionality does when using DISTINCT and COUNT
SQL Tip: COUNTing NULL values – Benjamin's Blog
sqlbenjamin.wordpress.com
· Using SELECT COUNT(*) or SELECT COUNT(1) (which is what I prefer to use) will return the total of all records returned in the result set regardless of NULL values.
SQL NULL Values - IS NULL and IS NOT NULL
A field with a NULL value is a field with no value. If a field in a table is optional, it is possible to insert a new record or update a record without adding a value to this
payloads/Injection at master · xsuperbug/payloads · GitHub
and+false%21=false--+. and(select+1+from(select+count(*),floor(rand(0)*2)from+information_schema.tables+group+by+2)