如何检查下拉框是否与if语句匹配[PHP]

I have been coding a script, and I am stuck at one small part. I am checking to make sure that the form the user submits is secured. But I'm not sure how to approach the dropdown box, this is the code I have so far:

if ($_POST['type'] != '1' || $_POST['type'] != '2' || $_POST['type'] != '3') // Checks to see if the submitted type matches, prevents exploiting.
{
    $supportErrors[] = "Support type you've chosen was not found, possibly trying to exploit this script?";
}

I am not sure if this is the right way to check its value. This is my HTML side for this part of the script:

<select name="type" id="type" size="1" style="border: 1px solid #000000; background-color: #FFFFFF;">
      <option value="1" selected>Support</option>
      <option value="2">Question</option>
      <option value="3">Complaint</option>
</select>

Now, to my understanding, the value is the data being posted VIA 'type' - so, in my mind, shouldn't it post that value (1 or 2 or 3) and then in the PHP side, it checks if it isn't 1, 2 or 3?

in_array() is a nice solution when checking if 1 thing matches one of multiple options

if (!in_array($_POST['type'],array(1,2,3))){..}

you should be developing with errors turned on:

error_reporting(E_ALL);
ini_set('display_errors', 1);

You need to change the condition logic from OR to AND :

if($_POST['type'] != '1' && $_POST['type'] != '2' && $_POST['type'] != '3')...

Make sure to use ISSET to check if the form was posted

<?php
if ( isset( $_POST['type'] ) and $_POST['type'] != '1' || $_POST['type'] != '2' || $_POST['type'] != '3') // Checks to see if the submitted type matches, prevents exploiting.
{
    $supportErrors[] = ("Support type you've chosen was not found, possibly trying to exploit this script?");
}
?>

//  Assumimg type as a numeric value , I might achieve it like this

    $type = 0;

    if( isset( $_POST['type'] ) ) $type = trim( $_POST['type'] ) * 1;

    if( $type < 1 )
    {
        $supportErrors[] = "Please specify type";
    }
    elseif( $type > 3 )
    {
        $supportErrors[] = "Invalid type";
    }