Question

I’m trying to create a POST request, unfortunately the body of the POST never seems to be sent.

Below is the code that I’m using. The code is invoked when a user clicks on a link, not a form ‘submit’ button. It runs without error, invokes the servlet that is being called but, as I mentioned earlier, the body of the POST never seems to be sent.

I can validate that the request body is never sent since I have access to the servlet being called.

I’ve tried using ‘parameters’ in replace of ‘requestBody.’ I’ve also tried using a parameter string (x=a?y=b). I’ve also validated that ‘ckULK’ does contain a valid value.

Any ideas?

new Ajax.Request(sURL, {     method: 'POST'     , contentType: 'text/x-json'     , requestBody: {ulk:ckULK}     , onFailure:         function(transport)         {             vJSONResp = transport.responseText;             var JSON = eval( '(' + vJSONResp + ')' );             updateStatus(JSON.code + ': ' + JSON.message);         } // End onFailure     , onSuccess:          function(transport)          {             if (200 == transport.status)             {                 vJSONResp = transport.responseText;             }             else             {                 log.value += '\n' + transport.status;             }          } // End onSuccess }); // End Ajax.request 

Answer 1

These are the kind of situations where Firebug and Firefox are really helpful. I suggest you install Firebug if you don’t have it and check the request that is being sent.

You also definitely need to stick to parameters instead of requestBody.

This:

new Ajax.Request(sURL, {     method: 'POST',     parameters: 'hello=world&test=yes',     onFailure: function(transport) {         vJSONResp = transport.responseText;         var JSON = eval( '(' + vJSONResp + ')' );         updateStatus(JSON.code + ': ' + JSON.message);     },     onSuccess: function(transport) {         if (200 == transport.status) {             vJSONResp = transport.responseText;         } else {             log.value += '\n' + transport.status;         }      } }); 

Should definitely work.