Resource id #6 error, Not sure how to fix it - php

Question

I keep getting a 'Resource id # 6' failure when submitting a script on my website. The code I'm using is the same type of code I use for registering for the website and that works but this script doesn't work at all. What my code does is send a booking request with the fields as shown to the database. I keep getting a Resource id#6 error , and I've googled what that is but I can't seem to figure out whats wrong. I am a beginner at php , so any tips on whats to look for to avoid a resource id # 6 error would be a lot of help
<?php
//$pattern="/^.+#.+/.com/";
//error_reporting(0);
if(isset($_POST["submit"])){
$Name_of_Person = $_POST['Name_of_Person'];
$Name_of_Group = $_POST['Name_of_Group'];
$room = $_POST['room'];
$How_Many_People = $_POST['How_Many_People'];
$Date_of_Booking = $_POST['Date_of_Booking'];
$End_time = $_POST['End_time'];
$Purpose = $_POST['Purpose'];
$Contact_Number = $_POST['Contact_Number'];
$Contact_Email = $_POST['Contact_Email'];
$Alcohol = $_POST['Alcohol'];
$Security = $_POST['Security'];
$Projector = $_POST['Projector'];
$Extra_Chairs = $_POST['Extra_Chairs'];
$Extra_Info = $_POST['Extra_Info'];
$Activated = '0';
$con = mysql_connect('localhost','root','test123') or die("couldn't connect");
mysql_select_db('bookerdb') or die("couldn't connect to DB");
//if(filter_var($email, FILTER_VALIDATE_EMAIL)){//(preg_match($pattern, $_POST['Contact_Email'])){
$query = mysql_query("SELECT * FROM `booking_table` WHERE Date_of_Booking='".$Date_of_Booking."' AND room='".$room."'");
$numrows = mysql_num_rows($query);
echo $query;
if($numrows==0){
$sql="INSERT INTO `booking_table` (Name_of_Person,Name_of_Group,room,How_Many_People,Date_of_Booking,End_time,Purpose,Contact_Number,Contact_Email,Alcohol,Security,Projector,Extra_Chairs,Extra_Info, Activated) VALUES ('$Name_of_Person','$Name_of_Group','$room','$How_Many_People','$Date_of_Booking','$End_time','$Purpose','$Contact_Number','$Alcohol','$Security','$Projector','$Extra_Chairs','$Extra_Info',$Activated)";
$result = mysql_query($sql);
if($result){
echo "Sent to be approved";
$redirect_page = '../ASC.php';
$redirect = true;
if($redirect==true){
header('Location: ' .$redirect_page);
}
}else{
echo "Failed";
}
}else{
echo"There is already a requested booking on that date & time";
$redirect_page = '../EAR.php';
$redirect = true;
if($redirect==true){
header('Location: ' .$redirect_page);
}
}
/*}else{
echo "error";
$redirect_page = '../EWF.php';
$redirect = true;
if($redirect==true){
header('Location: ' .$redirect_page);
}
}*/
}
?>

Answer 1

You have error in your second SQL query. You try to insert 14 values into 15 columns (in values you forgot $Contact_Email).
$sql="INSERT INTO `booking_table` (Name_of_Person,Name_of_Group,room,How_Many_People,Date_of_Booking,End_time,Purpose,Contact_Number,Contact_Email,Alcohol,Security,Projector,Extra_Chairs,Extra_Info, Activated) VALUES ('$Name_of_Person','$Name_of_Group','$room','$How_Many_People','$Date_of_Booking','$End_time','$Purpose','$Contact_Number','$Contact_Email','$Alcohol','$Security','$Projector','$Extra_Chairs','$Extra_Info',$Activated)";
Than remove echo $query from your code, line 30.
In $query isn't query, but mysql result object. You can't work with that by this way, you can't echo it.