Retrieve name ID from php json file - php

Question

I have this php file for country/state/town list using json method.
PHP file:
sleep(1);
$stateID = $_GET['stateID'];
$countyID = $_GET['countyID'];
$townID = $_GET['townID'];
$html = $_GET['html'];
$states = array();
$states['MA'] = "Massachusetts";
$states['VT'] = "Vermont";
$states['SC'] = "South Carolina";
$counties = array();
$counties['MA']['BARN'] = 'Barnstable';
$counties['MA']['PLYM'] = 'Plymouth';
$counties['VT']['CHIT'] = 'Chittenden';
$counties['SC']['ANDE'] = 'Anderson';
$towns = array();
$towns['MA']['BARN']['CHA'] = "Chatham";
$towns['MA']['BARN']['DEN'] = "Dennis";
$towns['MA']['BARN']['YAR'] = "Yarmouth";
$towns['MA']['PLYM']['BRI'] = "Bridgewater";
$towns['MA']['PLYM']['MAR'] = "Marshfield";
$towns['MA']['PLYM']['WAR'] = "Wareham";
$towns['VT']['CHIT']['BUR'] = "Burlington";
$towns['VT']['CHIT']['ESS'] = "Essex";
if($stateID && !$countyID && !$townID){
echo json_encode( $counties[$stateID] );
} elseif( $stateID && $countyID && !$townID ) {
echo json_encode( $towns[$stateID][$countyID] );
} elseif( isset($villages[$stateID][$countyID][$townID]) ) {
echo json_encode( $villages[$stateID][$countyID][$townID] );
} else {
echo '{}';
}
I retrieve value from MySql database(example:MA for country).
Now I need to print country/state/town name into select box dropdown like this.
<select id="country" class="validate[required]" name="country">
<option value="0">Choose ...</option>
<option selected="selected" value="'.$country['selector'].'">Show Country Name From php File</option>
</select>
<select id="state" class="validate[required]" name="state">
<option value="0">Choose ...</option>
<option selected="selected" value="'.$state['selector'].'">Show state Name From php File</option>
</select>
<select id="town" class="validate[required]" name="town">
<option value="0">Choose ...</option>
<option selected="selected" value="'.$town['selector'].'">Show town Name From php File</option>
</select>
How can I print this?

Answer 1

Create a php file that makes the call to the database. The results are, normally, stored in a php array. You can turn this array into Json using json_encode()(http://www.php.net/manual/en/function.json-encode.php).
Simply echo this in the php-file.
Now call that script with jQuery. There is a simple jQuery method called getJson(). - Read more about it here: http://api.jquery.com/jquery.getjson/
This will bring the MySQL results to you - in Json form - and you could now use JavaScript to loop the results, and display the form accordingly. A simple example would be (I haven't tested this but I hope you get the idea):
$.each(data, function() {
$('#mySelect')
.append($("<option></option>")
.attr("value",data[row].NameOfColumn)
.text(value));
row ++;
});
Note the name of column - that's the name of the key for the value you are looking for. In your case you might want to loop the keys as well - with an outer loop - since the data seems to be presented in such a matter.