为啥我的 POST 请求仅对我的 API 不起作用？

Posted 2023-02-23

【中文标题】为啥我的 POST 请求仅对我的 API 不起作用？

【英文标题】：Why is my POST request not working in flutter for my API only?为什么我的 POST 请求仅对我的 API 不起作用？

【发布时间】：2020-11-28 05:23:01

【问题描述】：

我已经为 Flutter 中的 API POST 请求编写了一个代码，该代码在应用程序中不起作用，但是，与 https://jsonplaceholder.typicode.com/ 相同的代码正在运行。

这是我的项目的代码。（不工作）

  Future<void> _doSignIn() async 
    String apiUrl = "http://prelive.sewer-viewer.com/api_daily_logs/user/signIn";
    Map<String, String> headers = "Content-type": "multipart/form-data";
    final json =  convert.jsonEncode("email": "[e-mail]", "password": "PASSWORD_HERE");
    http.Response response = await http.post(apiUrl,headers: headers, body: json);
    var jsonResponse = convert.jsonDecode(response.body);
    print(jsonResponse);
  

总是返回以下响应。

  "status": 0,
  "msg": "userToken: , status: 0, msg: Invalid username and/or password. Please try again"

和 jsonplaceholder.typeicode.com 的代码是这样的。 （工作正常）

  Future<void> _doSignIn() async 
    Map<String, String> header = "Content-type": "multipart/form-data";
    String testUrl = "https://jsonplaceholder.typicode.com/posts";
    final test = convert.jsonEncode(
      "userId": 11,
      "title": "Test Title",
      "body": "Test Body Test Body Test Body Test Body Test Body Test Body Test Body Test Body Test Body Test Body "
    );
    http.Response resp = await http.post(testUrl,headers: header, body: test);
    var jsonResp = convert.jsonDecode(resp.body);
    print(jsonResp);

总是返回

id: 101

但在邮递员应用程序中，它工作正常。这是原始输出。

POST /api_daily_logs/user/signIn HTTP/1.1
User-Agent: PostmanRuntime/7.26.2
Accept: */*
Postman-Token: b60ecfe7-7ccb-44bd-93f8-d95995bc5eb8
Host: prelive.sewer-viewer.com
Accept-Encoding: gzip, deflate, br
Connection: keep-alive
Content-Type: multipart/form-data; boundary=--------------------------071795598422914636012534
Content-Length: 304
----------------------------071795598422914636012534
Content-Disposition: form-data; name="email"

[e-mail]
----------------------------071795598422914636012534
Content-Disposition: form-data; name="password"

PASSWORD_HERE
----------------------------071795598422914636012534--
HTTP/1.1 200 OK
Date: Fri, 07 Aug 2020 20:38:13 GMT
Server: Apache
X-Powered-By: php/7.1.33
Status: 200
Content-Length: 405
Keep-Alive: timeout=5, max=100
Connection: Keep-Alive
Content-Type: application/json
"userToken":"eyJ0eXAiOiJKV1QiLCiOjE1OTY4MjkxNjUsImp0aSI6InNUWmphR3VaN3p1RTlRPT0iLCJpc3MiOiJwcmVsaXZlLnNld2VyLXZpZXdlci5jb20iLCJuYmYiOjE1OTY4cCI6MTSFDGSDF9070SDFLK8LK78L7KGDFL5LLK54SI6eyJ1c2VySWQiOiIxMDE0IiwidXNlck5hbWUiOiJheml6YV9hcGlAY2FsbS1zb2xASDFSFA89ASDF._mLNOkKJHKkjkj34tcluD1w8w","status":1,"msg":"User authentication has been successfully processed."

我的包括这些。

import 'package:flutter/material.dart';
import 'dart:async';
import 'package:http/http.dart' as http;
import 'dart:convert' as convert;

我无法弄清楚是什么问题。我已经尝试将 contentType 更改为几乎所有可能的值。

如果需要任何进一步的细节，请告诉我。

Flutter Doctor 输出。

[√] Flutter (Channel stable, v1.17.0, on Microsoft Windows [Version 10.0.18362.959], locale en-US)
    • Flutter version 1.17.0 at F:\flutter-sdk
    • Framework revision e6b34c2b5c (3 months ago), 2020-05-02 11:39:18 -0700
    • Engine revision 540786dd51
    • Dart version 2.8.1


[√] android toolchain - develop for Android devices (Android SDK version 30.0.1)
    • Android SDK at C:\Users\MRCOM\AppData\Local\Android\Sdk
    • Platform android-30, build-tools 30.0.1
    • ANDROID_HOME = C:\Users\MRCOM\AppData\Local\Android\Sdk
    • Java binary at: C:\Program Files\Android\Android Studio\jre\bin\java
    • Java version OpenJDK Runtime Environment (build 1.8.0_242-release-1644-b01)
    • All Android licenses accepted.

[√] Android Studio (version 4.0)
    • Android Studio at C:\Program Files\Android\Android Studio
    • Flutter plugin version 48.0.2
    • Dart plugin version 193.7361
    • Java version OpenJDK Runtime Environment (build 1.8.0_242-release-1644-b01)

[√] VS Code (version 1.47.3)
    • VS Code at C:\Users\MRCOM\AppData\Local\Programs\Microsoft VS Code
    • Flutter extension version 3.13.1

[√] Connected device (1 available)
    • sdk gphone x86 • emulator-5554 • android-x86 • Android 11 (API 30) (emulator)

• No issues found!

API代码如下。

     /* signIn
     *
     * This function will used to verify user login detail and provide an access token which will contains token
     * expiration and other user detail.
     *
     * This function also update user API key in app_users table which will later use to verify user authentication.
     *
     * @author Usman
     */
public function signIn_post()
        //---- variables
        $response_array = array('userToken'=> '', 'status'=>0, 'msg' => $this->lang->parseLine('invalid_detail'));
        $isValidToken = false;

        //---- posted arguments
        $post = $this->input->post();
        $email = !empty($post['email']) ? trim($post['email']) : '';
        $password = !empty($post['password']) ? trim($post['password']) : '';

        //---- if an email or password is empty then return an error message
        if($email == '' || $password == '')
            header('WWW-Authenticate: Basic realm="REST API Required username:password"');
        
        else 
            //---- validate login detail
            $apiUserData = $this->user_model->loginAuthentication($email, $password);

            //---- if user data found then generate access token and update in DB
            if(!empty($apiUserData))
                //---- verify that user is already logged in and don't have password expired
                if(!empty($apiUserData['dailyLogApiKey']))
                    $isValidToken = $this->authorization_server->isValidToken($apiUserData['dailyLogApiKey']);
                

                if(!$isValidToken)
                    $tokenData = array();
                    $tokenData['id'] = $apiUserData['userID'];
                    $tokenData['username'] = $apiUserData['email'];
                    $token = $this->authorization_server->generateToken($tokenData, CLIENT_SECRET_CODE);

                    //---- update newly access token in DB
                    $this->user_model->updateKeyAuthentication($apiUserData['userID'], $token);

                    //---- update sign in user ID in API Call Log
                    $this->user_model->updateSignInApiCallLog($apiUserData['userID']);
                
                else
                    $token = $apiUserData['dailyLogApiKey'];
                

                //---- update response array
                $response_array['userToken'] = $token;
                $response_array['status'] = 1;
                $response_array['msg'] = $this->lang->parseLine('successful_login');

                $this->response($response_array, 200);
            
        

        //---- Authentication Failed
        $this->response($response_array, 401);
    

【问题讨论】：

这可能是你的 API 的问题，它需要一个令牌，你能分享你的 api 代码吗？

因为它说它需要一个头文件格式来授权

@ParitoshYadav，没有明白你所说的头文件格式的意思。你能详细说明一下吗？

我的意思是在标头中使用令牌进行授权

你可以在邮递员中看到他们提供的令牌

【参考方案1】：

1) 网址错误

在颤振中你使用的是 URL：

.../api_daily_logs/user/login

但在邮递员中：

.../api_daily_logs/user/signIn
                          ^
                          |
                  this part is different

将 URL 更改为 signIn 后，响应 (.json) 不同了：

    "userToken": "",
    "status": 0,
    "msg": "Invalid username and/or password. Please try again"

2) 修复主体

首先 - 不要对这个地图进行编码，所以从：

final json = convert.jsonEncode(
  "email": "[e-mail]",
  "password": "PASSWORD_HERE",
);

到：

// It's just Map<String, String>

final json = 
  "email": "[e-mail]",
  "password": "PASSWORD_HERE",
;

其次 - 删除标头，因为如果 body 是 Map，则请求的内容类型将设置为 "application/x-www-form-urlencoded"

工作代码

String apiUrl = "http://prelive.sewer-viewer.com/api_daily_logs/user/signIn";

final json = 
  "email": "[e-mail]",
  "password": "PASSWORD_HERE"
;

http.Response response = await http.post(apiUrl, body: json);

var jsonResponse = jsonDecode(response.body);
print(jsonResponse);

【讨论】：

是的....我也想过..如果你没有发布答案，我会自己回答。

但问题可能是其他问题......因为它使用相同的用户名/密码通过邮递员登录，所以为什么不没有它。

@GoharSahi 什么/哪里有问题？还是不行？

是的，它不起作用。当我从邮递员应用程序登录时，它可以工作..但是当我从颤振登录时，它不起作用。使用相同的用户名/密码。

it doesn't work 是什么意思？错误的答案？当我使用上面的代码接收Invalid username and/or password. Please try again 时，它就像在邮递员中一样。你的回应是什么？