【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": "EMAILADDRESS@GMAIL.COM", "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"
EMAILADDRESS@GMAIL.COM
----------------------------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