开发者

php radio button auto update db table

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'开发者_开发百科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


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.

0

上一篇:

下一篇:

精彩评论

暂无评论...
验证码 换一张
取 消

最新问答

问答排行榜