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Updating the DOM to recognize AJAX inserted inputs

开发者 https://www.devze.com 2023-04-08 18:15 出处:网络
I\'m pretty much an idiot when it comes to AJAX, so if this problem is really simple, please forgive me.

I'm pretty much an idiot when it comes to AJAX, so if this problem is really simple, please forgive me.

I have this little form:

<form id="location_ajax_request">
  <label for="location">Enter Your Location:<开发者_如何学Python;/label>
  <input name="ajax_location" id="ajax_location" type="text" value="Irvine, CA, USA" />
  <input id="requestLocation" type="button" value="Click to Submit" />
  <p id="output"></p>
</form>

When requestLocation is clicked, a GET call to a php script returns something like:

<input type="radio" name="location_selected" value="0" />
<input type="hidden" name="location_x_0" value="-117.8253403" />
<input type="hidden" name="location_y_0" value="33.6868782" />
<input type="hidden" name="location_name_0" value="Irvine, CA, USA" />
[...]
<input type="button" id="confirmAddress" value="Confirm Address" />

Where the _0 is a count of items. If, for instance, someone had entered London, USA, they'd receive some 5 responses.

With jQuery, I grab the click of $('#confirmAddress') successfully using live() and attempt to grab the values of the inputs. I assume they somehow need to be checked for since inserted elements aren't registered with the DOM. Say I'm trying to grab:

document.forms['location_ajax_request']['location_name_0'].value;

How do I first register it with the DOM as a valid object so it stops returning undefined?


Well if you are using jquery as the OP tags say:

$('input[name="location_name_0"]', '#location_ajax_request').val();
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