如何解决400状态码和意外字符（字符1处）错误？

Question

我有一个机器学习模型，可以拍摄图片并返回预测结果，我将其上传到flask服务器，并在我使用HTML创建的网站上进行测试，结果成功，但是，当我尝试将它与flifter应用程序一起使用时，当我尝试从应用程序上载图片时，会出现以下错误：未处理的异常：FormatException：意外字符（在字符1处）。另外，我得到的状态码是400。。这是我的密码。任何帮助都将不胜感激。

这是我在其中创建flask实例的python代码（当我尝试使用Flatter应用程序建立连接时，我在cmd上运行它）

from flask import Flask, render_template, request, jsonify
import cv2
import numpy as np
import pandas as pd
import tensorflow
from tensorflow import keras
from tensorflow.python.keras import backend as k

app = Flask(__name__)

model = tensorflow.keras.models.load_model('model.h5')


@app.route('/')
def index():
    return render_template("index.html", data="hey")


@app.route("/prediction", methods=["POST"])
def prediction():

    
    img = request.files['img']  
    img.save("img.jpg")
    image = cv2.imread("img.jpg")
    image = cv2.resize(image, (224,224))
    image = np.reshape(image, (1,224,224,3))
    pred = model.predict(image)

    pred = pred > 0.5
    if(pred):
        predd="Malignant"
    else:
        predd="Benign"

    return jsonify({"prediction": predd,})
    
#this ip is my network's IPv4
#(I connected both my laptop and mobile to this WiFi while establishing the connection)
if __name__ == "__main__":
    app.run(debug=False,host='192.168.1.103',port=5000)

这是我在颤振中用来建立连接的代码

  sendImageToServer(File imageFile) async {
    var stream = new http.ByteStream(imageFile.openRead());
    stream.cast();
    var length = await imageFile.length();
    print(length);
    
    //this ip is my network's IPv4 
    //(I connected both my laptop and mobile 
    //to this WiFi while establishing the connection) 

    var uri = Uri.parse('http://192.168.1.103:5000/prediction');
    var request = new http.MultipartRequest("POST", uri);
    var multipartFile = new http.MultipartFile('file', stream, length,
        filename:
            basename(imageFile.path));

    request.files.add(multipartFile);
    request.headers["Content-Type"] = 'multipart/form-data';
    var streamedResponse = await request.send();
    var response = await http.Response.fromStream(streamedResponse);
    
    print(response);

    final Map<String, dynamic> responseJson =
        json.decode(response.toString()) as Map<String, dynamic>;
    print(responseJson.toString());

    pre = responseJson["prediction"];
    print(pre);

    setState(() {
      prediction = pre;
    });
  }

  Future getImage() async {
    final image = await picker.getImage(source: ImageSource.gallery);
    sendImageToServer(File(image.path));

    setState(() {
      fileImage = File(image.path);
      sendImageToServer(fileImage);
    });
  }

这也是我发送请求时在cmd上得到的信息

192.168.1.108 - - [20/Mar/2021 19:14:39] "←[1m←[31mPOST /prediction HTTP/1.1←[0m" 400 -

Answer 1

所以你得到的例外是告诉你服务器无法理解你发送的请求，我认为现在的问题是你在这里发送的请求的密钥

var multipartFile = new http.MultipartFile('file', stream, length,
        filename:
            basename(imageFile.path));

在您的后端，您正在使用密钥"img"

img = request.files['img']  

所以这些应该是一样的，试试下面的

var multipartFile = new http.MultipartFile('img', stream, length,
        filename:
            basename(imageFile.path));

我真的希望这将工作，也检查下面的链接，它可能会帮助你理解更多的问题，

https://dev.to/carminezacc/advanced-flutter-networking-part-1-uploading-a-file-to-a-rest-api-from-flutter-using-a-multi-part-form-data-post-request-2ekm

还有这个问题

错误消息没有明确错误的确切位置，所以这样做，将您的代码添加到一个尝试捕获块中，这样您就可以捕获抛出的任何异常并知道错误是什么。

try {
sendImageToServer(File imageFile) async {
    var stream = new http.ByteStream(imageFile.openRead());
    stream.cast();
    var length = await imageFile.length();
    print(length);
    
    //this ip is my network's IPv4 
    //(I connected both my laptop and mobile 
    //to this WiFi while establishing the connection) 

    var uri = Uri.parse('http://192.168.1.103:5000/prediction');
    var request = new http.MultipartRequest("POST", uri);
    var multipartFile = new http.MultipartFile('file', stream, length,
        filename:
            basename(imageFile.path));

    request.files.add(multipartFile);
    request.headers["Content-Type"] = 'multipart/form-data';
    var streamedResponse = await request.send();
    var response = await http.Response.fromStream(streamedResponse);
    
     //print(response);//this will print instance of response not the response itself
     // i cleared this a bit so we can trace where the error started from
     print(response.statusCode); // the status code of your response
     print(response.reasonPhrase); // the reason it failed "in case it did"
     print(response.body); // the body of the response -> this what you really interested in

  }
} catch (e,s) {
  // this is your life saver, it will tell you everything 
  // you need to know to trace the error (well mostly :D)
  // note : this will catch any exceptions that is thrown in the code, this not related to the response
  // the (e) is error it self so your should print that to know what is the error
  print(e.toString());
  // the (s) is the stackTrace, it will tell you when and where the error 
  // happened, so you should also print it
  print(e.toString());
}