PHP Parse Error using WAMP Server

dannyabbott89 · 04-19-2010, 07:32 AM

Hi there. I hope somebody can help me with this.

I have a basic HTML file set up which allows my to input some data. I have a MySql database set up behind the scenes with a table for this information to go in. When I click the submit button in my HTML file this PHP file opens and it comes up with

"Parse error: parse error in C:\wamp\www\comp39x\insert_student.php on line 32"

Here is my code I am using in the PHP file:-

<html>
<head>
<title>COMP 39x FInal Year Project - Student Added?</title>
</head>

<body>
<h1>Student Addition Results</h1>
<?php
//create short variable names
$studid=$_POST['studentid'];
$fname=$_POST['firstname'];
$sname=$_POST['surname'];
$projid=$_POST['projectid'];
$mail=$_POST['emailaddress'];

if (!$studid || !$fname || !$sname || !$projid || !$mail)
{
echo 'You have not entered all the required information.<br />'
.'Please go back and try again.';
exit;
}

@ $db = new mysqli('localhost', 'root', '', 'comp39x');

if (mysqli_connect_errno())
{
echo 'Error: Could not connect to database. Please try again later.';
exit;
}

$query = "insert into student values
('".$studid"', '".$fname"', '".$sname"', '".$projid"', '".$mail"')";
$result = $db->query($query);

if ($result)
echo $db->affected_rows. 'student inserted into the database.';
$db->close();
?>
</body>
</html>

I have created short variable names at the top of the code to make things for myself.

Hope somebody can help.

dannyabbott89 · 04-19-2010, 07:43 AM

by the way I am using the textbook and the examples contained within "PHP and MySQL Web Development" as an outline for my code

graemef · 04-19-2010, 09:08 AM

It would help to indicate which is line 32.

Having said that the line(s)

PHP Code:



$query = "insert into student values
('".$studid"', '".$fname"', '".$sname"', '".$projid"', '".$mail"')";

you need to use the string concatenation operator after the variables, hence:

PHP Code:



$query = "insert into student values
('".$studid."', '".$fname."', '".$sname."', '".$projid."', '".$mail."')";

Having said that string literals with double quotes will automatically expand simple variables so you can write it as:

PHP Code:



$query = "insert into student values('$studid', '$fname', '$sname', '$projid', '$mail')";