0

我有以下代码

async function searchCharacter(){
   
  const response = await fetch("https://rickandmortyapi.com/api/character/?name=")
  .then((response) => {
      if(response.status !== 200) throw `HTTP error: ${response.status}`;
      const data = response.json();
     return data
    })
  .then( (body) => {
      const [first] = body.results;
      console.log(first)
  })
  .catch( (error) => {
      console.log(`Error: ${error}`)
  })
}

有了这个笑话测试

describe('searchCharacter', () => {

  it('gets the character data', async () => {

    const responses = {
      status: "200",
      json() => {return {name: "ricky"}}
    }

    fetch.mockResolvedValueOnce(responses);

    const char = await searchCharacter();
    expect(char.name).toBe("ricky")

  });
  
  });

这就是我遇到运行纱线测试的后续输出的问题

TypeError: Cannot read property 'name' of undefined

      32 |
      33 |     const char = await searchCharacter();
    > 34 |     expect(char.name).toBe("ricky")
         |                 ^
      35 |
      36 |   });
      37 |   /*

      at Object.<anonymous> (tests/searchCharacter.test.js:34:17)

当我删除 .then() 并降低异步函数的复杂性时,这是开玩笑的测试,但我需要代码保持这样,我想当我通过第一个 .then() 时会出现一些错误或冲突,但我不知道怎么解决。

4

0 回答 0