reload PHP variable as submitted in dropdown menu via AJAX

Question

On a webpage I am trying to create a dropdown menu with some features. After a selection has been made in the dropdown menu the PHP variable "$bin_sec" gets updated to a new value. After $bin_sec has been updated I want a div element (with the id tag of "WHAT") in my webpage to now be reloaded and now take on the new value of $bin_sec.

I'm open to using AJAX, PHP, jQuery, MySQL, etc...

I can't find or come to a working solution.

My code for the dropdown menu:

<form  id="container2" method="post">
    <select name="name" onchange='$("#WHAT").load("testing.php",{bin_sec:x});'>
        <?php $bin_sec=60;?>
        <option selected="selected" <?php if ($_POST['name'] == '60'){  print 'selected'; $bin_sec=60; }?> value="60">1 Minute</option>
        <option <?php if ($_POST['name'] == '120'){ print 'selected'; $bin_sec=120; }?> value="120">2 Minutes</option>
        <option <?php if ($_POST['name'] == '300'){ print 'selected'; $bin_sec=300; }?> value="300">5 Minutes</option>
        <option <?php if ($_POST['name'] == '600'){ print 'selected'; $bin_sec=600; }?> value="600">10 Minutes</option>
    </select> 
</form>

<div id="WHAT">
<?php
echo $bin_sec;
...
?>
</div>

testing.php:

<?php
$bin_sec = $_POST['name'];
?>

<script type="text/javascript">
$("#WHAT").load("index.php #WHAT")
</script>

I think this code should work but it doesn't... Any corrections would be very much appreciated. Thank you in advance.

AS far as i have understood your question is whenever user chooses the dropdown the value of the drop down he chooses is required and to be displayed in the what div am i right please correct me if your requirement is a different one — NaveenThally
– NaveenThally, Commented Aug 1, 2014 at 11:36
@NaveenThally - yes- that is correct. But also bear in mind that because PHP runs 1st and then the Javascript runs second, the div element need to be updated with the new PHP variable. — user3882512
– user3882512, Commented Aug 1, 2014 at 19:02
Better: if(isset($_POST['name']) && $_POST['name'] == 'X') { .. } — Ryan
– Ryan, Commented Aug 1, 2014 at 19:05
@true - so do you mean something like-- <option <?php if ($_POST['name'] == '120'){ print 'selected'; $bin_sec=120; }?> value="120">2 Minutes</option> --to become-- <option <?php if(isset($_POST['name']) && $_POST['name'] == '120'){ print 'selected'; $bin_sec=120; }?> value="120">2 Minutes</option> --is this what you had mind? — user3882512
– user3882512, Commented Aug 1, 2014 at 19:49

NaveenThally · Accepted Answer · 2014-08-02 06:12:43Z

1

There are Many Ways to do it one sample way is given below




        <?php 

        if(isset($_GET['bin_sec']))
        {
            $bin_sec=$_GET['bin_sec'];

        }
        else
        {
            $bin_sec=60;
        }



        ?>
        <select name="bin" onchange="window.location.href='stack.php?bin_sec='+this.value">

            <option <?php if($bin_sec == '60'){  print 'selected';}?> value="60">1 Minute</option>
            <option <?php if ($bin_sec == '120'){ print 'selected'; }?> value="120">2 Minutes</option>
            <option <?php if ($bin_sec == '300'){ print 'selected'; }?> value="300">5 Minutes</option>
            <option <?php if ($bin_sec == '600'){ print 'selected'; }?> value="600">10 Minutes</option>
        </select> 

    <div id="WHAT">
    <?php
    echo $bin_sec;
    ?>
    </div>

I hope it helps and stack.php is your php file where dropdown exists i hope you understand for any doubts revert back

answered Aug 2, 2014 at 6:12

NaveenThally

1,0241 gold badge8 silver badges18 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

user3882512 Over a year ago

this code works! THANK YOU SO MUCH FOR THIS! you have just ended my rainstorm of embarrassment :)

user3882512 Over a year ago

what were the links that you tried to embed in your code above? When I click "There" and "Many Ways" I'm not forwarded to any site.

NaveenThally Over a year ago

hi sorry for that, those are not links i had formatted the whole reply has code which feels like links , cheers

Michael Bearjaws · Accepted Answer · 2014-08-01 11:13:15Z

0

You will need to echo out $bin_sec in PHP in order for it to be sent to the browser.

Additionally you need to change your $_POST var to the same name as you are sending from the browser as $_POST['name'] does is undefined

Example below

<?php
$bin_sec = $_POST['bin_sec'];
echo $bin_sec;
?>

edited Aug 1, 2014 at 11:13

answered Aug 1, 2014 at 11:08

Michael Bearjaws

2011 silver badge4 bronze badges

1 Comment

user3882512 Over a year ago

your reasoning seems to be perfect but when I apply the change I still don't 1) update the PHP variable $bin_sec and 2) the div (id="WHAT") does not reload to reflect the updated value of $bin_sec. my change is the following: testing.php now looks like:

<?php $bin_sec = $_POST['name']; echo $bin_sec;?>  <script type="text/javascript"> $("#WHAT").load("index.php #WHAT")</script>

- does this change reflect what you had in mind?

Eugene Podskal · Accepted Answer · 2014-08-01 11:35:10Z

0

You can use this code:

<option <?php if ($_POST['name'] == '120'){ print 'selected'; =120; }?> value=" <?php $_POST['name'] == '120'? "$bin_sec_value":120 ; ?>">2 Minutes</option>

edited Aug 1, 2014 at 11:35

Eugene Podskal

10.4k5 gold badges34 silver badges56 bronze badges

answered Aug 1, 2014 at 11:11

karan

16410 bronze badges

Collectives™ on Stack Overflow

reload PHP variable as submitted in dropdown menu via AJAX

3 Answers 3

3 Comments

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

3 Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related