Parse error: syntax error, unexpected '"', expecting identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING) in C:... on line 22 - sql

Question

I want to update a row in a table for my project, I'm copying a syntax I saw somewhere else here however, I think my problem comes when I try updating where ApplicantID is equal to $_SESSION["ID"].
I get this error
Parse error: syntax error, unexpected '"', expecting identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING) in C:\xampp\...\InsertPData.php on line 22
here is the php along side the SQL:
<?php
include_once'dbconnect.php';
session_start();
function INSERT()
{
$Name=$_POST['name'];
$Relation=$_POST['Relation'];
$Email=$_POST['Email'];
$Address=$_POST['Address'];
$Postcode=$_POST['Postcode'];
$Mobile_Number=$_POST['Mobile_Number'];
$Home_Number=$_POST['Home_Number'];
$INSERT="UPDATE Applicants
SET ParentName='$Name',
Relationtoapplicant='$Relation',
ParentEmail='$Email',
ParentAddress='$Address',
ParentPostcode='$Postcode',
ParentMobile='$Mobile_Number',
ParentHome='$Home_Number',
WHERE ApplicantID=$_SESSION["ID"] "; #THIS IS LINE 22
$data=mysql_query($INSERT) or die(mysql_error());
if($data)
{
echo "Parents/Gauridan details hav been entered";
}
else print "error";
}
INSERT()
?>
I've already searched for a solution to this but haven't found something where the user is using a session thing. Thank you.

Answer 1

This is why an IDE with syntax highlighting is helpful. StackOverflow uses syntax highlighting on code blocks as well and actually already gives you the answer based on your code:
$INSERT="UPDATE Applicants
WHERE ApplicantID=$_SESSION["ID"] ";
See how ID is suddenly black instead of dark red? That's because you are terminating the string there. The double quotes should either be escaped or replaced with single quotes, like:
$INSERT="UPDATE Applicants
WHERE ApplicantID=$_SESSION[\"ID\"] ";
Or
$INSERT="UPDATE Applicants
WHERE ApplicantID=$_SESSION['ID'] ";
See how the ID bit stays dark red? This is because now your string is not suddenly terminated.
Also, please do not use mysql_ functions anymore. They have been deprecated since 2013 and are currently not even a part of PHP anymore. So if you'd update your PHP to the latest version, this code would not work. On top of that, this code is vulnerable to SQL injection attacks.
Also see Why shouldn't I use mysql_* functions in PHP? and How can I prevent SQL-injection in PHP?.