php radio button auto update db table

Question

I have two radio buttons

<div align="center" class="radio_group">
    <input type="radio" id="gallerymenustyle1" class="element radio" name="gallerymenustyle[]" value="1" /> Gallery Link - In the navigation of my website, display one "gallery" link<br />
    <input type="radio" id="gallerymenustyle2" class="element radio" name="gallerymenustyle[]" value="2" /> Category Links - In the navigation of my website, display a separate link to each category.
</div>

how to update one field from a row, once I clicked one of those radio buttons without hitting a submit button ?

here's what i have so far

$(function(){
           $(':radio').click(function(){
             if($(this).is(':checked')) 
              {
                   var galleryMenuStyleVal = $(this).val();
                   $.ajax({
                  type: 'POST',
                  url: '<?php echo BASE_URL; ?>ajax/ajax_methods_gallery.php',
                        data: 'showGalleryMenuStyleFromEditPage=' + galleryMenuStyleVal
                    });
              }
           });
        }); 


if($_POST['showGalleryMenuStyleFromEditPage'] !== null || $_POST['shoGalleryMenuStyleFromEditPage'] !== 0){
    $userObj = new User();
    if($_POST['showGalleryMenuStyleFromEditPage'] == 1){
        $query = "UPDATE users SET gallery_menu_style = 1";
        $userObj->updateUserGalleryMenuStyle($query);
    } else if ($_POST['showGalleryMenuStyleFromEditPage'] == 2){
        $query = "UPDATE users SET gallery_menu_style = 2";
        $userObj->updateUserGalleryMenuStyle($query);
    } 
}

is my jquery correct including some portions of PHP ? ..because it's not working

hohner · Accepted Answer · 2011-07-31 08:29:58Z

1

On this line:

if($_POST['showGalleryMenuStyleFromEditPage'] !== null || $_POST['shoGalleryMenuStyleFromEditPage'] !== 0){

You used 'shoGalleryMenuStyleFromEditPage' instead of 'showGalleryMenuStyleFromEditPage' on the second $_POST value. It's therefore not allowing anything to get through that initial if statement.

You've also used !== 0 instead of != 0.

This is why using short keys/names isn't such a bad idea...

If it's still not working, use Firebug to ascertain whether your jQuery's actually sending anything.

answered Jul 31, 2011 at 8:29

hohner

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7 Comments

sasori Over a year ago

didn't helped, it still doesn't update the gallery_menu_style field from the table, no matter which of those 2 radio buttons i ticked :(

hohner Over a year ago

Replace the $_POST variables with defined ones (like $post1 = 1; $post2 = 2) and see if the PHP is behaving correctly. If it's inputting data, then it must be a problem with your jQuery. Another little test you can do is to check your JavaScript: remove everything inside the .click(function(){}) and replace with $(this).remove(); If the radio button disappears when clicked, the jQuery is recognising the click event but not the AJAX call

sasori Over a year ago

i checked the console from firebug..the post tab showed this showGalleryMenuStyleFromEditPage=2 ..or when i selected the first radio buttn showGalleryMenuStyleFromEditPage=1

hohner Over a year ago

Remove the first if statement, it's unnecessary - you're already checking the values with the two additional statements. As I said, define some variables like ($post1 = 1; $post2 = 2) and call the PHP with the AJAX function. You're replacing the $_POST variables with your own defined ones just to see if it's the User() class that's screwing things up...

sasori Over a year ago

ok, first i did was, I removed the if statement in the jquery, and tried the $(this)remove() , the button dissappeared..so I returned back the ajax, still not working... i tried replacing the value of each radio button to a predefined variable ..didn't work

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