Javascript 如何使用jasmineajax来验证send方法是否被调用?
如果我从未调用Javascript 如何使用jasmineajax来验证send方法是否被调用?,javascript,ajax,jasmine,jasmine-ajax,Javascript,Ajax,Jasmine,Jasmine Ajax,如果我从未调用send方法,那么jasmineajax是否应该调用onreadystatechange并将readyState设置为4 如果上述行为不是预期的行为,我如何使用jasmineajax验证调用了send方法 下面是正在测试的代码: Loader = (function() { var loadNames = function(url, success_callback, error_callback) { var ajax = new XMLH
send
方法,那么jasmineajax
是否应该调用onreadystatechange
并将readyState
设置为4
如果上述行为不是预期的行为,我如何使用jasmineajax
验证调用了send
方法
下面是正在测试的代码:
Loader = (function() {
var loadNames = function(url, success_callback, error_callback) {
var ajax = new XMLHttpRequest();
ajax.open("GET", url);
ajax.onreadystatechange = function () {
console.log("Ready state is " + ajax.readyState);
if (ajax.readyState === 4 && ajax.status === 200) {
success_callback(JSON.parse(ajax.responseText));
} else if (ajax.readyState === 4 && ajax.status !== 200) {
error_callback("There was a problem. Status returned was " + ajax.status);
}
};
ajax.onerror = function () {
error_callback("Unknown error");
};
// Shouldn't removing the call to send prevent
// onredystatechange from being called with readyState 4?
// ajax.send();
};
return {
loadNames: loadNames
};
})();
以下是测试:
describe("Loader", function () {
var successFunction, failFunction;
beforeEach(function () {
jasmine.Ajax.install();
successFunction = jasmine.createSpy("successFunction");
failFunction = jasmine.createSpy("failFunction");
});
afterEach(function () {
jasmine.Ajax.uninstall();
});
describe("#loadNames", function () {
it("Makes a success callback with the data when successful", function () {
Loader.loadNames("someURL", successFunction, failFunction);
jasmine.Ajax.requests.mostRecent().respondWith({
"status": 200,
"contentType": 'application/json',
"responseText": '[1, 2, 4, 3, 5]'
});
// Shouldn't this fail since I never called send?
expect(successFunction).toHaveBeenCalledWith([1, 2, 4, 3, 5]);
});
});
});
我很惊讶地看到调用了
successFunction
,因为测试中的代码从未调用ajax.send()
。如果这是库的预期行为,那么我如何spyOn
底层的ajax
对象,以便验证测试中的代码是否调用send
?是的,您没有调用ajax.send()
,但是由于这段代码,您正在触发ajax.onreadystatechange
事件:
jasmine.Ajax.requests.mostRecent().respondWith({
"status": 200,
"contentType": 'application/json',
"responseText": '[1, 2, 4, 3, 5]'
});
它更改readystate并将readystate设置为“完成”。事实上,这正是文件中所说的:
至于如何检查是否确实调用了xhr.send,这说明您可以在每次调用之前执行以下操作来监视它:
spyOn(XMLHttpRequest.prototype, 'send');
在取消加载程序中xhr.send()部分的注释后,可以检查以下方法调用:
describe("#loadNames", function () {
it("Makes a success callback with the data when successful", function () {
Loader.loadNames("someURL", successFunction, failFunction);
jasmine.Ajax.requests.mostRecent().respondWith({
"status": 200,
"contentType": 'application/json',
"responseText": '[1, 2, 4, 3, 5]'
});
expect(XMLHttpRequest.prototype.open).toHaveBeenCalled();
});
});
从该页面我不清楚
respondWith
是否会在未调用send
的情况下触发回调;但是,知道我正在观察预期的行为是有帮助的。值得一提的是,使用stubRequest
也会产生我需要的行为。@Zack是的,这部分没有解释得那么好,它有点隐含了:jasmine.Ajax.requests.mostRecent().respondWith({“status”:200,“contentType”:“text/plain”,“responseText”:“awesponse”});期望(doneFn)。得到“令人敬畏的响应”代码>。这似乎会触发文档中的doneFn
回调,这相当于您的successFunction