$name = isset($_POST['username'])?$_POST['username']:'';
// 验证密码是否填写
$pwd = isset($_POST['password'])?(int)($_POST['password']):'';
// 如果用户没有填写用户名和密码,重定向到登录页面
if($name == '' || $pwd == ''){
header("Location:login.php");
}// 验证用户信息是否正确
$conn = mysql_connect('localhost','root','');
mysql_select_db('ena');
$sql = "select * from student where name='$name' and password=$pwd";
$selcSql = mysql_query($sql); //发送一条sql
$num = mysql_num_rows($selcSql); //返回结果行数
var_dump($num);
最后这句获取不了值,获取的是int(0),输入账号密码数据库有这个人,应该返回int(1)才对啊~ 怎么回事
// 验证密码是否填写
$pwd = isset($_POST['password'])?(int)($_POST['password']):'';
// 如果用户没有填写用户名和密码,重定向到登录页面
if($name == '' || $pwd == ''){
header("Location:login.php");
}// 验证用户信息是否正确
$conn = mysql_connect('localhost','root','');
mysql_select_db('ena');
$sql = "select * from student where name='$name' and password=$pwd";
$selcSql = mysql_query($sql); //发送一条sql
$num = mysql_num_rows($selcSql); //返回结果行数
var_dump($num);
最后这句获取不了值,获取的是int(0),输入账号密码数据库有这个人,应该返回int(1)才对啊~ 怎么回事
不会把sql输出到数据库执行看看?$sql = "select * from student where name='$name' and password='$pwd'";
你用你的输出的试看看?
密码是varchar的你告诉我你不用引号能执行?
银行ATM都不用INT你用INT?有点厉害。 你这明文存储,是我公司的第一个开除走人
Navicat MySQL Data TransferSource Server : 127.0.0.1
Source Server Version : 50540
Source Host : localhost:3306
Source Database : testTarget Server Type : MYSQL
Target Server Version : 50540
File Encoding : 65001Date: 2015-11-25 21:25:46
*/SET FOREIGN_KEY_CHECKS=0;-- ----------------------------
-- Table structure for users
-- ----------------------------
DROP TABLE IF EXISTS `users`;
CREATE TABLE `users` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`username` varchar(16) NOT NULL,
`password` varchar(64) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=3 DEFAULT CHARSET=utf8;-- ----------------------------
-- Records of users
-- ----------------------------
INSERT INTO `users` VALUES ('1', 'ty0716', 'qaz123');
INSERT INTO `users` VALUES ('2', 'wsy', 'qaz123');
<?php
header('content-type:text/html;charset=utf-8');
$con = mysql_connect("localhost","root","root");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}mysql_select_db("test", $con);
$username='ty0716';
$password='qaz123';
$sql="SELECT * FROM users where username='$username' and password='$password' limit 1";
$result = mysql_query($sql,$con);
$temp=mysql_fetch_row($result);if(!$temp){
exit('用户名或密码错误');
}echo '登陆成功,用户名是:'.$temp[1];mysql_close($con);
?>