thank you both for your answer.
i still don't get any activity in firebug but i guess my php stuff is
just wrong.
On 12 Feb., 19:46, James <james.gp....@gmail.com> wrote:
> The "data" part of ajax should be some pair, for example:
> data: {myVar:value}
> myVar will then become your POST variable with the value 'value' (not
> the literal string) in your PHP script.
>
> To extend on the comment using Firebug, you should use it for Firefox.
> The Firebug console will show you requests that are done through AJAX
> including parameters that are GET/POSTed and the response it returns.
> Very useful for debugging.
>
> On Feb 12, 5:16 am, Liam Potter <radioactiv...@gmail.com> wrote:
>
> > you can test if it would work using firebug, just check the net tab for
> > activity from the page to "some.php"
>
> > weidc wrote:
> > > Hi,
>
> > > i got a radio button. if this radio button is checked it shell send
> > > the value of the button to a .php document. didn't used ajax at all so
> > > i don't know how to do this.
>
> > > at the moment i tried something like this:
>
> > > $(":radio").click(function()
> > > {
> > > $("#infoding").css("display","block");
> > > var value = $(this).val();
> > > $.ajax({
> > > type: "POST",
> > > url: "some.php",
> > > data: value,
> > > success: function(msg){
> > > alert( "Data Saved: " + value );
> > > }
> > > });
>
> > > });
>
> > > got it out of some documentation but don't know if it would work
> > > 'cause i'm not able to test it at the moment.
>
> > > i'd like to know if this would work and if not how it could work.
> > > i'd be happy about any help.
>
> > > thanks
> > > -weidc
