我是Jasmine的新手,并遇到了我希望String或null的情况。我尝试做一个或在toEqual之内,但我看到一些奇怪的结果让我相信我会以错误的方式进行。这种情况的最佳方法是什么?
也许我只是在考试错误。我是否应该放弃一个测试以测试这两种情况的想法?
describe("Jasmine", function () {
//1
it("should be able to handle any(String) || null within toEqual for string", function () {
expect("aString").toEqual(jasmine.any(String) || null);
});
//2
it("should be able to handle any(String) || null within toEqual for null", function () {
expect(null).toEqual(jasmine.any(String) || null);
});
//3
it("should be able to handle null || any(String) within toEqual for string", function () {
expect("aString").toEqual(null || jasmine.any(String));
});
//4
it("should be able to handle null || any(String) within toEqual for null", function () {
expect(null).toEqual(null || jasmine.any(String));
});
});
Expected null to equal <jasmine.any(function String() { [native code] })>.
Expected null to equal <jasmine.any(function String() { [native code] })>.
我意识到还有一个toBeNull(),这可能是为什么结果如此不稳定,但没有“或”链接我不知道如何合并它。
(运行Jasmine 1.3.1修订版1354556913)
解决了!如果有人有兴趣,请在下方提供完整的解决方案
describe("Jasmine", function () {
beforeEach(function () {
this.addMatchers({
toBeStringOrNull: function () {
var actual = this.actual;
this.message = function () {
return "Expected " + actual + " to be either string or null";
};
return typeof actual === 'string' || actual instanceof String || actual === null;
}
});
});
//1
it("should be able to handle any(String) || null within toEqual for string", function () {
expect("aString").toBeStringOrNull();
});
//2
it("should be able to handle any(String) || null within toEqual for null", function () {
expect(null).toBeStringOrNull();
});
//3
it("should be able to handle null || any(String) within toEqual for string", function () {
expect("aString").toBeStringOrNull();
});
//4
it("should be able to handle null || any(String) within toEqual for null", function () {
expect(null).toBeStringOrNull();
});
});
答案 0 :(得分:12)
编写自己的自定义匹配器:
toBeStringOrNull: function() {
var actual = this.actual;
this.message = function () {
return "Expected " + actual + " to be either string or null";
}
return typeof actual === 'string' || actual instanceof String || actual === null;
}
答案 1 :(得分:1)
如果您确实希望这些案例能够在同一个测试中工作,那么您可以编写自己的matcher。像
这样的东西toBeStringOrNull: function() {
var actual = this.actual;
this.message = function () {
return "Expected " + actual + " to be either string or null";
}
return jasmine.any(String) || actual === null;
}