描述info--登录校验码在PHP
<?php
$loginpassword = $_POST['password'];
$loginemail = $_POST['email'];
//connect
$my_hostname = "localhost";
$my_user = "root";
$my_password = "root";
$connect = mysql_connect($my_hostname, $my_user, $my_password);
$mysql_database = "myapp";
mysql_select_db($mysql_database, $connect);
//email and password check
$query = "SELECT * FROM hw2 WHERE email = '$loginemail'";
$check= mysql_query($query);
$result = mysql_num_rows($check);
if ($result = 0) {
die("EmailID does not exist");
}
// part 3
if ($loginemail = "[email protected]" and $loginpassword = "12345") {
$sql_sel_query = "SELECT * FROM hw2";
$result = mysql_query($sql_sel_query);
$total_num_rows = mysql_num_rows($result);
$i=0;
while ($i < $total_num_rows) {
$new_row = mysql_fetch_array($result);
$firstname_dis = $new_row['firstname'];
$lastname_dis = $new_row['lastname'];
echo "$firstname_dis";
echo "--";
echo "$lastname_dis";
echo "<BR>";
$i=$i+1;
}
?>
我新的PHP代码,在这里我想实现的功能中附加的想象。但我有点在步骤b中迷失了。任何人都可以帮助我,并在可能的情况下检查其他步骤是否正确。
对不起你们,请点击 “在这里输入图像描述”看到附件的想象。 –
使用mysql_ *函数停止!他们不安全,不推荐,不见了。改用mysqli_ *或PDO。 – Jeff
如果电子邮件地址错误或电子邮件地址正确但密码错误,我不建议给出具体的答复。这使得更容易入侵。相反有一个答复“凭据是错误的” – Jeff