javascript - "map((respone) => respone.json())"和 "map((respone) =>{ respone.json();})"之间的区别

Question

这个问题已经有答案了:

Curly Brackets in Arrow Functions (3 个回答)

已关闭 5 年前。

这里是源代码:

export class PostComponent implements OnInit {

    posts: any[];

    constructor(private http: Http, private service: PostService) {
    }
    getAll1() {
    const url = 'http://jsonplaceholder.typicode.com/posts';
    this.http.get(url)
    .map((respone) => respone.json())
    .subscribe(posts => this.posts = posts );
    }  
  }

`

它工作正常，但如果我为“respone.json”添加大括号，则可视 stdio 代码会提示错误(见下文)。添加大括号后，源代码如下:

export class PostComponent implements OnInit {

    posts: any[];

    constructor(private http: Http, private service: PostService) {
    }
    getAll1() {
    const url = 'http://jsonplaceholder.typicode.com/posts';
    this.http.get(url)
    .map((respone) => {respone.json();})
    .subscribe(posts => this.posts = posts );
    }  
  }

错误出现在最后一行的单词“this.posts”上，错误是:[ts]类型“void”不能分配给类型“any[]”。

为什么？

Answer 1

.map((respone) => respone.json()) 实际上与此相同:

.map((respone) => { return respone.json() })

虽然 .map((respone) => { respone.json() }) 与此类似:

.map((respone) => { respone.json(); return undefined; })

因此，在第二个代码块中，respone.json() 被忽略。