有人能告诉我什么是狼人 [英] Can somebody tell me what is worng

问题描述

hi


一直在做这个任务，截止日期到期。它是一个必须在服务器上运行的网站



我试过运行它但是一直打破，没有输出，甚至没有得到php错误，没有我认为这是代码中的原因导致它但是无法肯定地说



我需要第三眼或者第四只眼睛。



我尝试了什么：



这是我到目前为止所尝试的

hi
have been doing this assignment and the deadline is due. it is a website where it have to run on the server

I have tried running it but keep breaking down, no output, am not even getting a php error, nothing I think it is something within the code what is causing it but cant say for sure

I need a third or maybe a fourth eye.

What I have tried:

this is what i have tried so far

<pre><?php

		error_reporting(E_ERROR | E_WARNING | E_PARSE | E_NOTICE);
		
		
		if(isset($_POST['register_user']))
		{
			//capture the variable from the form and store in php variables
			
			
			$title=$_POST['title'];
			$fullname=$_POST['fullname'];
			$lastname=$_POST['lastname'];
			$screenname=$_POST['screenname'];
			$email=$_POST['email'];
			$gender=$_POST['gender'];
			$address=$_POST['address'];
			$mypwd=$_POST['mypwd'];
			
			//connecting to the server
			
			include'db_server.php';
			
			$conn = mysqli_connect($db_host,$db_username,$db_password) or die (mysqli_connect_error());
			
			//select the database you want to query
			
			mysqli_select_db($conn, 'national_wonders') or die (mysqli_error($conn));
			$sql = "SELECT * FROM members WHERE screenname='$screenname'";
			$result = mysqli_query($conn, $sql) or die ("ERROR:" .mysqli_error());
			$rowcount = mysqli_num_rows($result);
			
			if($rowcount >= 1)
			{
				echo"<script type=\"text/javascript\">
					alert('Username already exits!!');
					window.location=\"../php/popup.php\";
					</script>";
			}
			else
			{
				//insert data into table
				
				$sql = "INSERT INTO members
				VALUES('$title', '$fullname', '$lastname', '$screenname', '$address', '$email', '$gender',  'md5('$mypwd))";
			
				if(mysqli_query($conn,$sql))
				{
					echo"<script type=\"text/javascript\">
					alert(Welcome);
					window.location=\"login.php\";
					</script>";
				}
				else
				{
					echo "Error inserting values into database";
				}
				
			
			}
			
		}	
			
?>	

推荐答案

_POST ['register_user']））
{
//从表单中捕获变量存储在php变量中

_POST['register_user'])) { //capture the variable from the form and store in php variables

title =

_POST ['title'];

_POST['title'];

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