如何在Angular中检索状态代码?
我是Angular的新手,为了练习,我想制作一个小应用程序,在这个应用程序中,用户只需使用自己的用户名即可登录。为此,如果后端返回http状态200,我想存储登录的用户。但是我无法获取请求的http状态。我已经在这里和其他网站上查阅了一些帖子,但所有这些解决方案似乎都不适合我 我的角度版本是:8.2.14 这是我的登录服务:如何在Angular中检索状态代码?,angular,httpresponse,Angular,Httpresponse,我是Angular的新手,为了练习,我想制作一个小应用程序,在这个应用程序中,用户只需使用自己的用户名即可登录。为此,如果后端返回http状态200,我想存储登录的用户。但是我无法获取请求的http状态。我已经在这里和其他网站上查阅了一些帖子,但所有这些解决方案似乎都不适合我 我的角度版本是:8.2.14 这是我的登录服务: import { Injectable } from '@angular/core'; import { HttpClient, HttpHeaders, Ht
import { Injectable } from '@angular/core';
import {
HttpClient,
HttpHeaders,
HttpErrorResponse,
} from '@angular/common/http';
import { Observable, throwError } from 'rxjs';
import { catchError, retry } from 'rxjs/operators';
import { User } from '../model/User';
@Injectable({
providedIn: 'root',
})
export class LoginService {
constructor(private http: HttpClient) {}
loginUrl = 'login';
httpOptions = {
headers: new HttpHeaders({
'Content-Type': 'application/json',
}),
};
login(user: User) {
const request = this.http
.post(this.loginUrl, user, this.httpOptions)
.pipe(catchError(this.handleError));
return request;
}
private handleError(error: HttpErrorResponse) {
console.log('handleError Method');
console.log('Errorcode', error.status);
if (error.error instanceof ErrorEvent) {
// A client-side or network error occurred. Handle it accordingly.
console.error('An error occurred:', error.error.message);
} else {
// The backend returned an unsuccessful response code.
// The response body may contain clues as to what went wrong,
console.error(
`Backend returned code ${error.status}, ` + `body was: ${error.error}`
);
}
// return an observable with a user-facing error message
return throwError('Something bad happened; please try again later.');
}
}
这是调用服务的登录组件:
import { Component, OnInit } from '@angular/core';
import { Router, ActivatedRoute } from '@angular/router';
import { FormBuilder, FormGroup, Validators } from '@angular/forms';
import { first } from 'rxjs/operators';
import { LoginService } from 'src/app/services/login.service';
import { User } from '../../../model/User';
@Component({
selector: 'app-login-component',
templateUrl: './login-component.component.html',
styleUrls: ['./login-component.component.css'],
})
export class LoginComponent implements OnInit {
loginForm: FormGroup;
loading = false;
submitted = false;
returnUrl: string;
user: User;
// loginService: LoginService;
constructor(private loginService: LoginService) {}
ngOnInit() {}
login(username: string, password: string) {
const dummyUser = { username, password };
this.loginService.login(dummyUser).subscribe((data) => {
console.log('data', data);
this.user = data;
console.log('user', this.user);
});
}
}
编辑:
有了Mari Mbiru的回答和这篇文章,我能够解决这个问题。实际上,我以前试过设置observe:'response'
,但我没有把它放在httpOptions
中,而是放在HttpHeaders
中,但它不起作用。
现在,我的工作职位请求如下所示:
login(user: User) {
const request = this.http
.post<User>(
`${this.loginUrl}`,
{ username: user.username, password: user.password },
{
headers: new HttpHeaders({ 'Content-Type': 'application/json' }),
observe: 'response',
}
)
.pipe(catchError(this.handleError));
return request;
}
登录(用户:用户){
const request=this.http
.邮政(
`${this.loginUrl}`,
{用户名:user.username,密码:user.password},
{
headers:newhttpheaders({'Content-Type':'application/json'}),
观察:'回应',
}
)
.pipe(catchError(this.handleError));
返回请求;
}
不要将响应转换为JSON,然后您可以在此处找到它
订阅((res)=>res['status'],(err)=> 不要将响应转换为JSON,然后您可以在此处找到它
订阅((res)=>res['status'],(err)=> 通过向请求选项添加
{observe:'response'}
,HttpClient允许您查看完整响应,而不仅仅是正文。这将返回一个HttpResponse对象,该对象具有主体、标题、状态、url等
因此,http选项应该是:
httpOptions = {
observe:'response'
headers: new HttpHeaders({
'Content-Type': 'application/json',
}),
};
在认购书中:
this.loginService.login(dummyUser).subscribe((res) => {
console.log('response', res);
this.user = res.body;
console.log('user', this.user);
});
HttpClient允许您通过向请求选项添加
{observe:'response'}
来查看完整响应,而不仅仅是正文。这将返回一个HttpResponse对象,该对象具有主体、标题、状态、url等
因此,http选项应该是:
httpOptions = {
observe:'response'
headers: new HttpHeaders({
'Content-Type': 'application/json',
}),
};
在认购书中:
this.loginService.login(dummyUser).subscribe((res) => {
console.log('response', res);
this.user = res.body;
console.log('user', this.user);
});
这回答了你的问题吗?请始终尝试缩小示例代码。这是否回答了您的问题?请始终尝试缩小示例代码。