I'm having this select, let's call it "X", that is populated with car brands as options from the database via a SELECT.
<select name="X">
<?php
$row = mysqli_query("SELECT * FROM brands");
while($row2 = mysqli_fetch_array($row))
echo '<option value="' . $row2['brandID'] . '">' . $row2['brandName'] . '</option>
?>
</select>
Now i have to populate the second select, called "Y", with the models specifics to a brand selected.
For example, if the option that's selected in the first select box (X) is Audi i should have as options in the second select (Y) the following: A4,A6,TT,TTs
To populate the second select box manually is easy, basicaly the same thing as for the first just with a different SQL request.
$row = mysqli_query("SELECT modelName from modele WHERE brandName = '$brand'");
Where $brand would have a value according to the first select's selection.
Thanks
Put an id in your <select name="X"> code like <select name="X" id ="X">
Put another select like <select name="Y" id="Y">. Which will be blank.
put this jquery in your page.
$("X").on("change",function(){
var x_value=$("X").val();
$.ajax({
url:'ajax.php',
data:{brand:x_value},
type: 'post',
success : function(resp){
$("#Y").html(resp);
},
error : function(resp){}
});
});
in your ajax.php add the query.
<?php
$row = mysqli_query("SELECT modelName from modele WHERE brandName =".$_POST['brand']);
while($row2 = mysqli_fetch_array($row))
echo '<option value="' . $row2['modelId'] . '">' . $row2['modelName'] . '</option>
?>
Hopefully it will work. Please tell me if you need anything.
Use ajax and on change event together,
method:
1- build a page where you post your queries to.
2- that page should react to the query and sends a respond with the desired list of options.
example :
if ($_POST["vehicle"] == car) // send response with array of car names as json obj for example
else // send response with array of motorbikes names
your ajax should receive it and populate the filed on-complete.
If you need more details, I'll be happy to provide it