CODEWITHSUNDEEP
phpmysql
Leuven Wang
Leuven Wang

Reputation: 151

Trouble adding data to MySQL Table With PHP

I have a simple code that is supposed to add data to a MySQL table through PHP. I have code that checks if the machine manages to connect to the database and everything passes successfully however it's not working. Can anyone help me found the problem?

<!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 http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<?php
$mysqlhost="host";
$mysqldatabase="database";
$mysqlusername="username";
$mysqlpassword="password";
$connect=new mysqli($mysqlhost,$mysqlusername,$mysqlpassword);
if (!$connect) {
    die("Connection failed: " . $connect->connect_error);
}
echo "Connected Successfully";
mysqli_select_db($connect,"b33_15887129_Accounts");
$username=$_POST['username'];
$password=$_POST['Password'];
$email=$_POST['Email'];
$gender=$_POST['Gender'];
$tutorgroup=$_POST['tutorgroup'];
	$newaccount="INSERT INTO Users (Username,Password,Gender,Email,Tutor Group)
	VALUES('$username','$password','$email','$gender','$tutorgroup')";
if(!$newaccount){
	die("Failed to create new account.");
	
	}else{
		echo "New Account Successfully Created!";
		
		}
?>
</body>
</html>

Upvotes: 0

Views: 48

Answers (2)

ksbg
ksbg

Reputation: 3284

The problem seems to be that you are not executing your query. After storing the query in $newaccount, you should replace the if-statement that follows with that one:

if(!$connect->query($newaccount))
    die("Failed to create new account. DB Error: [" . $connect->error . "]");
else
    echo "New Account Successfully Created!";

Additionally, if you do it like I described, any error message that results from an unsuccessful query will be displayed.

Check here for an easy tutorial on how to use mysqli properly.

Upvotes: 2

Vineet
Vineet

Reputation: 4635

PLease execute your query like

mysqli_query($newaccount) or die(mysqli_error());

Upvotes: 1

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