How To create a Dependable Dropdown From Two Tables - php

Question

I have two tables "oc_t_item_car_make_attr" , "oc_t_item_car_model_attr"
So how can i make a Dependable Dropdown From these Two Tables?

Answer 1

You can only do this by Ajax.
Following is city country reference site, you can get idea from that.
Referance Site
if you stuck any where feel free to ask me.

Answer 2

Actually didn't understand question very well, but you probably want something like this:
SELECT * FROM oc_t_item_car_make_attr
INNER JOIN oc_t_item_car_model_attr
ON oc_t_item_car_model_attr.fk_i_make_id = oc_t_item_car_make_attr.pk_i_id
WHERE oc_t_item_car_make_attr.pk_i_id = ?
You need to insert id in the place of question mark.

Answer 3

Here is the entire code what your looking for:
Below code will display all car names from database dynamically:
<select name="cars" id="carmodels">
<option value="">select</option>
<?php
$qry = mysqli_query("select * from oc_t_item_car_make_attr");
while($row=mysqli_fetch_array($qry)){
echo "<option value=".$row['pk_i_id'].">".$row['s_name']."</option>";
}
?>
</select>
Below result will display car models dependent on the names selected above:
html:
<select name="models" id="models">
</select>
you need to use ajax to get the car models from database:
$(document).on('change','#carmodels',function(){
var value = $(this).val();
$.ajax({
url: "demo.php",
type: "GET",
data: { id : value},
success: function(data){
$("#models").html(data);
}
});
demo.php:
<?php
include "connection.php"; //database connection code
$qry = mysql_query("SELECT * FROM oc_t_item_car_make_attr
INNER JOIN oc_t_item_car_model_attr
ON oc_t_item_car_model_attr.pk_i_id = oc_t_item_car_make_attr.pk_i_id
WHERE oc_t_item_car_make_attr.pk_i_id =".$_GET['id'].");
while($row=mysqli_fetch_array($qry)){
echo "<option value=".$row['pk_i_id'].">".$row['s_name']."</option>";
}
?>

Answer 4

Hey you can do this by ajax
Reference Link 1
Reference Link 2
also study inner join