Question

我想知道，是否有可能获得测试的完整嵌套描述路径？

假设：

describe('Smoke Testing - Ensuring all pages are rendering correctly and free of JS errors', function () {
  describe('app', function () {
    describe('app.home', function () {
      it('should render this page correctly', function (done) {
        //name here should be: Smoke Testing - Ensuring all pages are rendering correctly and free of JS errors app app.home should render this page correctly
        done()
      })
    })

    describe('app.dashboard', function () {
      describe('app.dashboard.foobar', function () {
        it('should render this page correctly', function (done) {
        //name here should be: Smoke Testing - Ensuring all pages are rendering correctly and free of JS errors app app.dashboard app.dashboard.foobar should render this page correctly    
          done()
        })
      })
    })

  })

})

Answer 1

jasmine.Suite 和 jasmine.Spec 都有方法 getFullName（）。像你期望的那样工作：

describe("A spec within suite", function() {
  it("has a full name", function() {
    expect(this.getFullName()).toBe('A spec within suite has a full name.');    
  });
  
  it("also knows parent suite name", function() {
    expect(this.suite.getFullName()).toBe('A spec within suite');    
  });
});

<script src="http://searls.github.io/jasmine-all/jasmine-all-min.js"></script>

注意：此答案现在有点过时，并在示例中使用了Jasmine 1.3.1。

Answer 2

当你进入描述回调函数时，这被设置为＆＃34;套件＆＃34;具有套件描述的对象（您传递给描述的文本）和父套件的属性。

以下示例获取描述嵌套描述调用的串联，我不确定如何访问＆＃34; it＆＃34;的描述。但这会让你在那里分道扬。

var getFullDesc = function(suite){
    var desc = "";
    while(suite.parentSuite){
        desc = suite.description + " " + desc;
        suite = suite.parentSuite;
    }

    return desc;
}

describe('Outer describe', function(){
   describe('Inner describe', function(){
        console.log(getFullDesc(this));
       it('some test', function(){

       });
    });
});

Jasmine Testing获得完整描述/它的名称

2 个答案: