Reputation: 26301
How can one mock an ajax response?
https://stackoverflow.com/a/13915248/1032531 provides a link to an answer but the link is broken.
https://stackoverflow.com/a/29680013/1032531 provides the following answer but it errors with $.ajax is not a function
. https://jsfiddle.net/bdw0gfan/
function ajax_response(response) {
var deferred = $.Deferred().resolve(response);
return deferred.promise();
}
$(function() {
$.ajax = ajax_response([1, 2, 3]);
$.ajax('GET', 'some/url/i/fancy').done(function(data) {
console.log(data); // [1, 2, 3]
});
});
Upvotes: 1
Views: 2436
Reputation: 5584
$.ajax
needs to be a function or else it won't return the promise.
function ajax_response(response) {
var deferred = $.Deferred().resolve(response);
return deferred.promise();
}
$.ajax = function() {
return ajax_response([1, 2, 3]);
};
$.ajax('GET', 'some/url/i/fancy').done(function(data) {
console.log(data); // [1, 2, 3]
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
Upvotes: 2