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...
SQL Инъекции | Page 10 | ANTICHAT - Security online community
...null,null,load_file('/etc/passwd'),null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null
Ошибка СУБД | Форум
process=rphost,p:processName=test,t:clientID=111,t:applicationName=1CV8,t:computerName=ois2,t:connectID=88,SessionID=70,Usr=Калошин А.Г.,Trans=1,dbpid=13446,Sql="SELECT T1.Q_001_F_001RRef
How to check space or null in between | in a string — oracle-tech
On the other hand, there is no relation between null (as used in SQL) and NUL (chr(0)), other than the name similarity. Indeed, regular expression functions have little trouble with the empty string... but they choke on chr(0). I would not bring up chr(0) in a discussion of null; that can only create confusion.
SQL | fkn+antitotal
CREATE TABLE tbl ( id INT(11) NOT NULL, name TINYTEXT NOT NULL, PRIMARY KEY (id) ); INSERT INTO tbl VALUES (1, 'Видеоадаптеры'); INSERT INTO tbl VALUES (1, 'Видеоадаптеры'); Если необходимо, чтобы новые записи с дублирующим ключом отбрасывались без ге- нерации...
3 непонятных запроса | Форум
Левое объединение подставляет NULL в случае, когда в правой таблице нет данных. Соответственно, можно использовать это в WHERE.
Getting both read and unread notifications with a twist
For currently logged in user I wanna load all unread notifications and only 5 most recent read notifications.
HackerRank SQL - Programmer Sought | SELECT ALL
Note: Print NULL when there are no more names corresponding to an occupation.
Oracle / [Из песочницы] Некоторые примеры нестандартных...
false OR null = null. Пока все логично. Теперь рассмотрим, как работает IN в запросе №2
SQL FOR XML EXPLICIT
SELECT 3 AS TAG, 2 AS Parent, NULL AS [Customers!1!CustomerID!Element], NULL, NULL, l.lid AS [cts!Loan!2!LoanNumber], l.lamount AS [cts!Loan!2!LoanAmount] FROM @customer c JOIN @loan l on c.cid=l.cid FOR XML EXPLICIT, ROOT('cts')