在yii中更改所选项目时更改下拉列表的内容[关闭]

Question

i have a dropdown and on selecting items item i want to update the content of this dropdown list using ajax call,everything is fine but i don't know how to put the array returned by ajax call in callback function as data should be used to update my dropdown list. thankx in advance

Answer 1

If there's a lot of data that needs to be added after the ajax call, what I've done in the past is set the html in the controller (yii) and pass the fully updated list as the return for the ajax call. It's much easier to replace the options like this in your ajax success callback than having to convert a ton of variables into javascript. Hopefully that helps.

in your Yii controller I would do something like this:

$response = [];
$response["newDropdownValues"] = "<option>Value 1</option><option>Value 2</option>";
$this->controller->sendResponse(200, $response);

And in your ajax callback you'd write something like:

success: function(output) {
            var reply = jQuery.parseJSON(output);
            $(this).find('select').html(reply.newDropdownValues);
    },