Question

function addrewarddb()
{
    var rval="98";
    $.post("db/mkreward.php",{
        proid:getURLParameter("id")
    },function(data){


         rval="99";
         alert(rval);
    });
    alert(rval);
     return rval;

}

when this function executes the function returns 98 instead of 99.the two alerts showing 98,then 99.how to make my function returns 99.i think the problem here is the function returnig before the response from the server page mkreward.php.how i can make the function return after the response from server

Answer 1

The $.post call is asynchronous.

This means that the function returns before the HTTP request has been completed.

Then, when the $.post() call finishes, it then updates the variable and alerts 99.

From the jquery docs:

async Boolean Default: true By default, all requests are sent
asynchronously (i.e. this is set to true by default). If you need
synchronous requests, set this option to false. Cross-domain requests
and dataType: “jsonp” requests do not support synchronous operation.
Note that synchronous requests may temporarily lock the browser,
disabling any actions while the request is active. As of jQuery 1.8,
the use of async: false with jqXHR ($.Deferred) is deprecated; you
must use the complete/success/error callbacks.