我创建了用户表
CREATE TABLE `user` (
`id` BIGINT UNSIGNED NOT NULL AUTO_INCREMENT ,
`first_name` VARBINARY(100) NULL ,
`address` VARBINARY(200) NOT NULL ,
PRIMARY KEY (`id`))
ENGINE = InnoDB
DEFAULT CHARACTER SET = utf8
COLLATE = utf8_general_ci;
我插入了一行:
INSERT into user (first_name, address) VALUES (AES_ENCRYPT('Obama', 'usa2010'),AES_ENCRYPT('Obama', 'usa2010'));
要选择我使用的行,请执行以下操作:
SELECT AES_DECRYPT(first_name, 'usa2010'), AES_DECRYPT(address, 'usa2010') from user;
我得到了下面的result.What,我需要确保我的data.No数据对我是可见的。
发布于 2015-07-10 19:45:39
在mysql命令行客户端,不需要使用CAST
mysql> SELECT AES_DECRYPT(AES_ENCRYPT('admin','abc'),'abc');
+-----------------------------------------------+
| AES_DECRYPT(AES_ENCRYPT('admin','abc'),'abc') |
+-----------------------------------------------+
| admin |
+-----------------------------------------------+
1 row in set (0.00 sec)
mysql> SELECT CAST(AES_DECRYPT(AES_ENCRYPT('admin','abc'),'abc') AS CHAR (50));
+------------------------------------------------------------------+
| CAST(AES_DECRYPT(AES_ENCRYPT('admin','abc'),'abc') AS CHAR (50)) |
+------------------------------------------------------------------+
| admin |
+------------------------------------------------------------------+
1 row in set (0.02 sec)
如你所见,在命令行中使用cast会稍微慢一点。但我注意到,如果你使用一些工具,如phpmyadmin,那么你需要使用CAST
,否则结果将是错误的。
发布于 2018-05-23 18:11:38
if (isset($_POST['user_name']) and isset($_POST['user_password'])){
$user_name = $_POST['user_name'];
$user_password = $_POST['user_password'];
$query = "SELECT * FROM `user_tbl` WHERE user_name='$user_name' and AES_DECRYPT(user_password , '@ert') = '$user_password'";
$result = mysqli_query($connection, $query) or die(mysqli_error($connection));
$count = mysqli_num_rows($result);
https://stackoverflow.com/questions/16556375
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