Jquery drop down option on based previous selction

Question

Is it possible to get data from the database based on the previous drop down selection?

For example:

<select id="paramscountries" class="chzn-done" style="display: none;" name="params[countries][]" multiple="true" size="10">
    <option value="1">Afghanistan</option>
    <option value="2">Albania</option>
    <option value="3">Algeria</option>
    <option value="4">USA</option>
</select>

If I select USA from the above list it will make a query & get all the states name from the database. For example from states.php & will show all the states name in another drop down selection without reloading that page.

Answer 1

You would need to use Javascript or jQuery for this and use it to post to a page that will return your results.

<script src="//ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>

<select id="paramscountries" class="chzn-done" style="display: block;" onchange="getNewList($(this).val());" name="params[countries][]" multiple="true" size="10">
    <option value="1">Afghanistan</option>
    <option value="2">Albania</option>
    <option value="3">Algeria</option>
    <option value="4">USA</option>
</select>

<select name="states" id="states">
</select>

<script type="text/javascript">
function getNewList(country) {
    $.post('post.api.php', {'api': 'getStates', 'countries': country}, function(response) {
        var obj = $.pareseJSON(response);
        $('#states').html(obj.states);
    });
}

post.api.php page

<?php

if(isset($_POST['api'])){

    if($_POST['api'] == 'getStates'){

        // sdet the countries array to a variable
        $countries = $_POST['countries'];

        // set a variable to hold the results
        $results = array();

        // query your db
        $stm = $db->prepare('SELECT * FROM `your_database` WHERE `country` IN("'.implode('","', $countries).'")');
        $stm->execute();
        if($stm->rowCount() > 0){
            foreach($stm->fetchAll(PDO::FETCH_ASSOC) as $row){
                $results['states'] .= '<option value="'.$row['stateName'].'">'.$row['stateName'].'</option>';
            }
        }            
        echo json_encode($results);
    }

}

I would do something like that. Hopefully there is enough here to get you rolling.

Answer 2

Try using jquery ajax request. something like this

jQuery

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8/jquery.min.js"></script>
<script>
$(document).ready(function(){
    $('#paramscountries').change(function(){
        $.ajax({
            url: 'states.php'
            ,data:{val: $(this).val()}
            ,dataType:'json'
            ,type: 'POST'
            ,success:function(data){
                for(var i = 0; i < data.length; i++){
                    var dropDownValue = '<option value="'+ data[i] +'">'+ data[i] +'</option>'
                    $('#secondDropDownList').append(dropDownValue);
                }
            }
        });
    });
});
</script>

In states.php

$con=mysqli_connect("localhost","user","password","db_Name");
if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if(isset($POST['val'])){
$var = $_POST['val'];
$result = mysqli_query($con,"SELECT * FROM table_name WHERE table_column_name = '$var'");
$arr = array();
while($row = mysqli_fetch_array($result)) {
   $arr[] =  $row['name'];
}
echo json_encode($arr);
}
mysqli_close($con);

In HTML

<select id="secondDropDownList"></select>