Resource id # 4 php - php

Question

everytime i try and add one to the second column of a certain name, it changes the value to 5, if i echo my event it says it is equal to resource id #4. Anyone have any fixes?
<form action="new.php" method="POST">
<input type="text" name="input_value">
<br />
<input name="new_User" type="submit" value="Add to Users">
<input type="submit" name="event_Up" value="Attended Event">
<?php
//Connect to Database
mysql_connect("localhost", "root", "");
//If Add New user butten is clicked execute
if (isset($_POST['new_User']))
{
$username = $_POST['input_value'];
$make = "INSERT INTO `my_db`.`profile` (`Name`, `Events`) VALUES ('$username', '1')";
mysql_query($make);
}
//If Event up is pushed then add one
if (isset($_POST['event_Up']))
{
$username = $_POST['input_value'];
$event = mysql_query("SELECT 'Events' FROM `my_db`.`profile` WHERE Name ='$username'");
$newEvent = $event +1;
$update = "UPDATE `my_db`.`profile` SET Events = '$newEvent' WHERE Name = '$username'";
mysql_query($update);
}
//Print Table
$data = mysql_query("SELECT * FROM `my_db`.`profile`");
Print "<table border cellpadding=4>";
while($info = mysql_fetch_array($data))
{
Print "<tr>";
Print "<th>Name:</th> <td> ".$info['Name'] . "</td>";
Print "<th>Events:</th> <td>".$info['Events'] . " </td>";
}
Print "</table>";
?>

Answer 1

I've cleaned up your code a little bit.
It's still a mess, but should at least work (un-tested though).
<form action="new.php" method="post">
<input type="text" name="input_value">
<br />
<input name="new_User" type="submit" value="Add to Users">
<input type="submit" name="event_Up" value="Attended Event">
</form>
<?php
//Connect to Database
mysql_connect("localhost", "root", "");
//If Add New user butten is clicked execute
if (isset($_POST['new_User']))
{
$username = empty($_POST['input_value']) ? NULL : $_POST['input_value'];
if ( ! empty($username))
{
mysql_query("
INSERT INTO `my_db`.`profile`
(`Name`, `Events`)
VALUES
('". mysql_real_escape_string($username) ."', 1)
");
}
}
//If Event up is pushed then add one
if (isset($_POST['event_Up']))
{
$username = empty($_POST['input_value']) ? NULL : $_POST['input_value'];
if ( ! empty($username))
{
$event = mysql_query("
SELECT
Events
FROM
`my_db`.`profile`
WHERE
Name = '". mysql_real_escape_string($username) ."'
");
$newEvent = (int) (mysql_result($event, 0, 'Events') + 1);
mysql_query("
UPDATE
`my_db`.`profile`
SET
Events = $newEvent
WHERE
Name = '". mysql_real_escape_string($username) ."'
");
}
}
//Print Table
$data = mysql_query("SELECT * FROM `my_db`.`profile`");
Print "<table border cellpadding=4>";
while($info = mysql_fetch_assoc($data))
{
Print "<tr>";
Print "<th>Name:</th> <td> ". htmlentities($info['Name'], ENT_COMPAT, 'UTF-8') . "</td>";
Print "<th>Events:</th> <td>". htmlentities($info['Events'], ENT_COMPAT, 'UTF-8') . " </td>";
}
Print "</table>";
?>
Edit:
Just so you are aware... your issue was $newEvent = $event +1;.
$event is a MySQL resource, not the query's result. You have to use one of the mysql_* functions to get the data (see my code above.)
It seems you are just learning PHP, and I would highly recommend you stop using the mysql_* functions right now and start using PDO.

Answer 2

use mysql_fetch_assoc not mysql_fetch_array
any time you get a resource id rather than data it means you have just a pointer to something and most likely need a function call to get the data out.

Answer 3

You need to fetch the array and then define $event based on the results. You're assigning $events on the mysql query itself.
$result = mysql_query("SELECT 'Events' FROM `my_db`.`profile` WHERE Name ='$username'");
while($row = mysql_fetch_array( $result )) {
$event = $row['Events'];
}