do not show duplicate values in my drop down box - php

Question

I need some help with my logic here. I am populating a value from the database in my select box.
If it exists, I echo the value else i display the default options.
Now, for example if the value from the database is New Jersey, I do not want to display New Jersey for the second time in my drop down box. How do I do that?
<select name="location" class="field" >
<option value="<?php if(!empty($get_location)){ echo $get_location; } ?>"><?php if(!empty($get_location)){ echo $get_location; }?></option>
<option value="New Jersey">New Jersey</option>
<option value="New York">New York</option>
<option value="California">California</option>
</select>

Answer 1

You make an if statement in every option field and check if the value from the database matches the value of the option field and if it does so you echo "selected=\"true\"" to the option field.
For a code example see my answer for this question:
retieve data from mysql and display in form

Answer 2

If you only want your database values to be shown:
<?php
// You populate an array with the locations from your database
// As an example:
$options = array(1=>'New Jersey',2=>'Los Angeles');
$html = '<select name="locations">';
foreach($options as $id => $option)
{
$html .= '<option id="'.$id.'">'.$option.'</option>';
}
echo $html.'</select>';
?>
If you want something special to happen to your database values, but still load the default values too:
<?php
// You populate an array with the locations from your database
// As an example:
$options = array(1=>'New Jersey',2=>'Los Angeles');
// Compare against your full list of locations
// As an example:
$locations = array(1=>'New Jersey',2=>'Los Angeles',3=>'California',4=>'London');
$html = '<select name="locations">';
foreach($options as $id => $option)
{
if(array_key_exists($id,$locations))
{
// Enter your magic
}
}
echo $html.'</select>';
?>

Answer 3

I would change your php code slightly to add defaults and then you can filter and remove duplicates with jQuery. If there are lots of options this might not be the best solution tho...
php
<select name="location" class="field">
<?php if(!empty($get_location)): ?>
<option value="<?php echo $get_location; ?>">
<?php echo $get_location; ?>
</option>
<?php else: ?>
// Defaults
<?php endif; ?>
</select>
jQuery
var removeDup = function($select){
var val = '';
$select.find('option').each(function(){
if ($(this).val() === val) { $(this).remove(); }
val = $(this).text();
});
};
removeDup($('#yourSelect'));

Answer 4

Get your locations in one request to the database, add them to an array together with the defaults, if only one location do as below, if several locations parse them into an array and merge them with the defaults, do an array_unique to get rid of duplicates and output the array in a loop.
<select name="location" class="field">
<?php
$output = array(1=>'New Jersey', 2=>'New York', 3=>'California', 4=>$get_location);
$options = array_unique($output);
foreach($options as $key => $value) {
echo '<option value="'.$value.'">'.$value.'</option>';
}
?>
</select>