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

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Curly Brackets in Arrow Functions
                                
                                    （2个答案）
                                
                        
                2年前关闭。
            
        

这是源代码：

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，但是如果我为“ respone.json”添加大括号，则可视stdio代码会提示错误（请参见下文）。
添加花括号后的源代码如下：

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？

最佳答案

.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，在您的第二个代码块中，respone.json()只是被关闭了。