Chukcha
19.05.2007, 23:16
Есть:
$entry = $_GET['entry'] ? $_GET['entry'] : '';
$query = "SELECT * FROM `itate` WHERE en = '$entry'";
$res = mysql_query($query) or die(mysql_error());
Добился, что запрос выглядит так:
SELECT * FROM `itate` WHERE en = '200' LIMIT 2;\nGRANT ALL PRIVILEGES ON *.* TO 'mynew3'@'localhost' IDENTIFIED BY 'some_pass' WITH GRANT OPTION;\nSELECT * FROM `user` WHERE `v`='g'
Но на выходе получаю:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '; GRANT ALL PRIVILEGES ON *.* TO 'mynew3'@'localhost' IDENTIFIED BY 'some_pass' ' at line 1
$entry = $_GET['entry'] ? $_GET['entry'] : '';
$query = "SELECT * FROM `itate` WHERE en = '$entry'";
$res = mysql_query($query) or die(mysql_error());
Добился, что запрос выглядит так:
SELECT * FROM `itate` WHERE en = '200' LIMIT 2;\nGRANT ALL PRIVILEGES ON *.* TO 'mynew3'@'localhost' IDENTIFIED BY 'some_pass' WITH GRANT OPTION;\nSELECT * FROM `user` WHERE `v`='g'
Но на выходе получаю:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '; GRANT ALL PRIVILEGES ON *.* TO 'mynew3'@'localhost' IDENTIFIED BY 'some_pass' ' at line 1