Javascript Mockjax未拦截异步表单提交
我正在使用qUnit和mockjax尝试处理一个简单的异步表单提交,但由于某些原因,异步POST似乎要通过mockjaxJavascript Mockjax未拦截异步表单提交,javascript,ajax,testing,qunit,mockjax,Javascript,Ajax,Testing,Qunit,Mockjax,我正在使用qUnit和mockjax尝试处理一个简单的异步表单提交,但由于某些原因,异步POST似乎要通过mockjax test 'RuleModal closes the modal on a successful form submission event', -> $.mockjax dataType: 'json' url: '/url' type: 'post' responseText: status: 'success'
test 'RuleModal closes the modal on a successful form submission event', ->
$.mockjax
dataType: 'json'
url: '/url'
type: 'post'
responseText:
status: 'success'
$dom = $('<div class="show-modal"><form action="/url" method="post"></form></div>')
$form = $dom.find('form')
modal = new RuleModal($dom)
$form.submit()
equal $($dom).hasClass('show-modal'), false, 'closes the modal after form submission'
我尝试对实现进行硬编码,使其与测试完全匹配,但仍然不起作用。我做错了什么?我猜是因为你做了表单提交,然后立即做你的断言,你正在进入比赛状态。
$form.submit()
调用将在equal…
行运行时执行并仍在运行。即使您模拟了ajax调用,它仍然是异步的。不幸的是,浏览器本机事件(如submit
)在完成时没有任何类型的回调,因此您可能只需要围绕断言执行setTimeout()
:
$form.submit();
setTimeout(function() {
equal($($dom).hasClass('show-modal'), false, 'closes the modal after form submission');
}, 100); // I think the default in Mockjax is 50ms
或者您可以尝试在事件处理程序中执行此操作,但我认为这也可能产生竞争条件
$form.submit();
setTimeout(function() {
equal($($dom).hasClass('show-modal'), false, 'closes the modal after form submission');
}, 100); // I think the default in Mockjax is 50ms