Altering PHP Variable Based Off Selection - php

Question

I have a html select that holds values for all four of our offices, or the text All. If the user selects All, I want to echo "All Offices", but if the user selects a specific office, I want to echo that number. My problem is that when I run the syntax below, All remains All instead of All Offices.
Did I set this up the incorrect way?
Display Data For Which Office:
<select name="office">
<option value="All">All...</option>
<option value="one">One Quarter</option>
<option value="two">Two Quarter</option>
<option value="three">Three Quarter</option>
<option value="four">Four Quarter</option>
</select>
<?php
if ($officename != 'All') {
$officename = $_POST['officename'];
} else {
$officename = "All Offices";
}
echo $officename;
?>

Answer 1

You never initialized the $officename variable, so it should be null. As a result, won't $officename != 'All' always be true, so $officename = $_POST['officename']; will always be executed?
I think what you want instead is something like:
$officename = $_POST['officename'];
if ($officename == 'All') {
$officename = "All Offices";
}
echo $officename;

Answer 2

I assume you're wanting "All" to come from the value of the <select>, which is named "office". Therefore you need to assign that value to a variable before you can use it for comparison. Note the select is named "office" not "officename", so it's "office" you need to look for in $_POST.
$officeValue = $_POST['office'];
if ($officeValue == 'All') {
$officename = "All Offices";
}
else {
$officename = $_POST["officename"]; //I assume this is some other field in the form which you haven't shown, but which contains a human-readable office name.
}
echo $officename;