PHP页面显示第一个else$errormessage而不检查代码
我一直在尝试创建一个会员登录页面来链接到我的网站。我的PHP页面中有许多嵌套的If，但是无论我怎么做，我都会从第一条else语句中收到$errormsg 我唯一能够改变这一点的方法是在代码底部（表格之前）添加一个额外的空else，当我单击reigisterbtn时，我要么收到第一个else结果，要么收到一个空白页，没有显示任何内容 我是不是错过了一些很明显的东西PHP页面显示第一个else$errormessage而不检查代码,php,html,mysql,Php,Html,Mysql,我一直在尝试创建一个会员登录页面来链接到我的网站。我的PHP页面中有许多嵌套的If，但是无论我怎么做，我都会从第一条else语句中收到$errormsg 我唯一能够改变这一点的方法是在代码底部（表格之前）添加一个额外的空else，当我单击reigisterbtn时，我要么收到第一个else结果，要么收到一个空白页，没有显示任何内容 我是不是错过了一些很明显的东西 <?php error_reporting (E_ALL ^ E_NOTICE); ?> <!DOCTYPE
<?php
error_reporting (E_ALL ^ E_NOTICE);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta content="text/html; charset=utf-8" http-equiv="Content-Type" />
<title>Register Page</title>
</head>
<body>
<?php
if ( $_POST ['registerbtn']) {
$getuser =$_post ['user'];
$getemail =$_post ['email'];
$getpass =$_post ['pass'];
$getretypepass =$_post ['retypepass'];
if($getuser){
if ($getemail){
if ($getpass){
if ($getretypepass){
if ($getpass === $getretypepass){
if ((strlen ($getemail) >=7 ) && (strstr ($getemail, "@")) && (strstr ($getemail, "."))) {
require ("./connect.php");
$query = mysql_query ("SELECT * FROM users WHERE username ='$getuser'");
$numrows = mysql_num_rows ($query);
if ($numrows == 0){
$query = mysql_query ("SELECT * FROM users WHERE email ='$getemail'");
$numrows = mysql_num_rows ($query);
if ($numrows == 0){
$password =md5 (md5 ("kjfiufj".$password."Fj56fj"));
$date("F d, Y");
$code = md5(rand ());
mysql_query ("INSERT INTO users VALUES (
'', '$getuser','$password','$getemail', '0', '$code', '$date')");
$query = mysql_query ("SELECT * FROM users WHERE username='$getuser'");
$numrows = mysql_num_rows ($query);
if ($numrows == 1){
$site = "http://www.andyhoole.co.uk";
$webmaster = "AndyHoole <admin@andyhoole.co.uk>";
$headers = "From: $webmaster";
$subject = "Activate your account";
$message = " Thank you for registering :) . Clink the link below to activate your account.\n ";
$message .="$site/activate.php?user=$getuser&code=$code\n";
$message .= "You must activate your account to log in.";
if ( mail($getemail,$subject, $message, $headers) ){
$errormsg = "You have been registered, you must activate your account form the activation link sent to <b> $getemail </b>";
$getuser = "";
$getemail = "";
}else
$errormsg = "An error has occured. Your activation email was not sent.";
}else
$errormsg = " An error has occured and your account has not been created. ";
}else
$errormsg = " This email address already exsists.";
}else
$errormsg = " This username already exsists.";
mysql_close();
}else
$errormsg = "You must enter a valid email address to register.";
}else
$errormsg = "Your passwords did not match.";
}else
$errormsg = "You must retype your password to register.";
}else
$errormsg = "You must enter your password to register.";
}else
$errormsg = "You must enter your email address to register.";
}else
$errormsg = "You must enter your User name to register.";
}
$form = "<form action='./register.php' method='post'>
<table>
<tr>
<td></td>
<td><font color='red'>$errormsg</font></td>
</tr>
<tr>
<td>Username:</td>
<td><input type='text' name='user' value='$getuser' /></td>
</tr>
<tr>
<td>Email:</td>
<td><input type='text' name='email' value='$getemail' /></td>
</tr>
<tr>
<td>Password:</td>
<td><input type='password' name='pass' value='' /></td>
</tr>
<tr>
<td>Retype:</td>
<td><input type='password' name='retypepass' value='' /></td>
</tr>
<tr>
<td></td>
<td><input type='submit' name='registerbtn' value='Register' /></td>
</tr>
</table>
</form>";
echo $form;
?>
</body>
</html>
注册页
您还应该使用$\u POST
而不是$\u POST
，因为php是区分大小写的 您没有为sql查询设置任何错误处理指令，顺便说一句，您不应该首先使用mysql.*
函数。到底抛出了什么错误？（“第一个else语句”是含糊不清的）关于设计选择，我本可以用一个数组进行错误检查，但不会像代码那样陷入更深的混乱，但这不是quesiton.vjones-在pal上，我已经看了大约12个小时的代码，没有注意到小写。非常感谢。