submit a form via Ajax using prototype and update

2019-02-06 22:09发布

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 :

<form id="myForm" action="/getResults">
    [...]
    <input type="submit" value="submit" />
</form>
<div id="result"></div>

I tried to attach a javascript function (which uses Ajax.Updater) to "onsubmit" (on the form) and "onclick" (on the input) but the form is still "non-Ajax" submitted after the function ends (so the whole page is replaced by the results).

4条回答
萌系小妹纸
2楼-- · 2019-02-06 22:36

You need to stop the default action of the onsubmit event.

For example:

function submit_handler(e){
    Event.stop(e)
}

More info on stopping default event behavior in Prototype »

查看更多
做自己的国王
3楼-- · 2019-02-06 22:39

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>

Your 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
});
查看更多
叛逆
4楼-- · 2019-02-06 22:57

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.

查看更多
一纸荒年 Trace。
5楼-- · 2019-02-06 22:59

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">


function myfunc(){
   ... do prototype ajax stuff...
  return false;

}

Using onsubmit on the form also captures users who submit with the enter key.

查看更多
登录 后发表回答