Submit A Form Via Ajax Using Prototype And Update A Result Div May 30, 2024 Post a Comment I'm wondering how can I submit a form via Ajax (using prototype framework) and display the server response in a 'result' div. The html looks like this : Solution 1: Check out Prototype API's pages on Form.Request and Event handling.Basically, if you have this:<form id='myForm'> .... fields .... <input type='submit' value='Go'> </form> <div id='result'></div> CopyYour js would be, more or less:Event.observe('myForm', 'submit', function(event) { $('myForm').request({ onFailure: function() { .... }, onSuccess: function(t) { $('result').update(t.responseText); } }); Event.stop(event); // stop the form from submitting }); CopySolution 2: You first need to serialize your form, then call an Ajax Updater, using POST options and pass it the serialized form data. The result will then appear in the element you sepcified.Solution 3: You need to return the value false from the ajax function, which blocks the standard form submit.<form id="myForm" onsubmit="return myfunc()" action="/getResults"> functionmyfunc(){ ... do prototype ajax stuff... returnfalse; Copy}Using onsubmit on the form also captures users who submit with the enter key.Solution 4: You need to stop the default action of the onsubmit event.For example:function submit_handler(e){ Event.stop(e) } CopyMore info on stopping default event behavior in Prototype » Share Post a Comment for "Submit A Form Via Ajax Using Prototype And Update A Result Div"
Post a Comment for "Submit A Form Via Ajax Using Prototype And Update A Result Div"