我正在尝试为同时具有反应性元素和非反应性元素的对象编写测试。我不知道如何写大理石图,所以测试是清晰的。
在下面的测试中,我有一个我正在测试的对象,它既存储一个值又将它发布给一个主题。我想编写测试,以便可以发出一些值、停止虚拟时间并检查一些断言。然后,我想恢复虚拟时间。所以,我想我想使用flush
.
下面的测试成功,但不清晰:大理石图source2
和expected2
不对齐,所以我不能真正“看到”测试是否正确编写。如果我source2
在第一次调用上面定义expectObservable
,那么第二次调用expectObservable
永远不会看到任何值。
class StoresLatestValue {
constructor() {
this.subject = new Subject();
}
emit(value) {
this.latest = value;
this.subject.next(value);
}
}
test("StoresLatestValue works", () => {
new TestScheduler((actual, expected) => expect(actual).toEqual(expected)).run(
(helpers) => {
const { flush, hot, expectObservable } = helpers;
const obj = new StoresLatestValue();
// First half of the test
const source1 = hot("a-b-c");
const expected1 = " a-b-c";
source1.subscribe((val) => obj.emit(val));
expectObservable(obj.subject).toBe(expected1);
flush();
// Check that the latest value is what we expect
expect(obj.latest).toBe("c");
// These next two marble diagrams work, but since they don't line up,
// it's hard to know that the test is written correctly
const source2 = hot("d-e--");
const expected2 = " ----d-e--";
source2.subscribe((val) => obj.emit(val));
expectObservable(obj.subject).toBe(expected2);
flush();
// Check that the latest value is what we expect
expect(obj.latest).toBe("e");
}
);
});
我尝试将订阅运算符添加^
到source
,但似乎没有帮助。我还尝试使用冷的可观察对象,但我仍然无法编写弹珠,所以一切都很好地排列:
// Making the second observable cold works, but still doesn't line up
const source1 = hot(" a-b-c");
const expected1 = " a-b-c";
const source2 = cold(" -d-e--");
const expected2 = " -----d-e--";
// Making them both cold doesn't seem to work at all, nothing is observed
// by the second expectObservable
const source1 = cold("a-b-c");
const expected1 = " a-b-c";
const source2 = cold("-----d-e--");
const expected2 = " -----d-e--";
有没有办法编写此测试以使其看起来正确?