iOS - 无法将视频文件从 Swift 上传到 FastAPI 后端

Question

我从这个问题获取了上传视频文件的代码；但是，出现错误。

这是从相册中选择视频的代码：

var videoURL: URL?

extension ViewController: UIImagePickerControllerDelegate, UINavigationControllerDelegate {
    func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey : Any]) {

        picker.dismiss(animated: true, completion: nil)
        if let pickedVideo = info[UIImagePickerController.InfoKey.mediaURL] as? URL {
            videoURL = pickedVideo
}

这是将视频上传到服务器的代码：

func uploadImage(imageURL: URL?) {
        
        let URL = "http://127.0.0.1:8000/uploadfiles"
        let header : HTTPHeaders = [
            "Content-Type" : "application/json"]
    
    let timestamp = NSDate().timeIntervalSince1970
    
    AF.upload(multipartFormData: { multipartFormData in
        multipartFormData.append(imageURL!, withName: "video", fileName: "\(timestamp).mp4", mimeType: "\(timestamp).mp4")
        
    }, to: URL, usingThreshold: UInt64.init(), method: .post, headers: header).response { response in
        guard let statusCode = response.response?.statusCode,
              statusCode == 200
        else { return }

@IBAction func ButtonClicked(_ sender: UIButton) {
        do {
            uploadImage(imageURL: videoURL!)
 
        } catch {
        }
    }

这是FastAPI服务器代码：

from fastapi import FastAPI, File, UploadFile
from typing import List
import os

app = FastAPI()

@app.get("/")
def read_root():
  return { "Hello": "World" }

@app.post("/files/")
async def create_files(files: List[bytes] = File(...)):
    return {"file_sizes": [len(file) for file in files]}

@app.post("/uploadfiles")
async def create_upload_files(files: List[UploadFile] = File(...)):
    print('here')
    UPLOAD_DIRECTORY = "./"
    for file in files:
        contents = await file.read()
        with open(os.path.join(UPLOAD_DIRECTORY, file.filename), "wb") as fp:
            fp.write(contents)
        print(file.filename)
    return {"filenames": [file.filename for file in files]}

这是错误：

INFO:     127.0.0.1:65191 - "POST /uploadfiles HTTP/1.1" 422 Unprocessable Entity

Answer 1

您正在尝试发送

multipart/form-data

（请参阅FastAPI 文档上传文件），但您已将

Content-Type

标头设置为

application/json

。因此，您应该删除/更改它。

此外，在客户端，您应该使用服务器端为上传文件提供的相同

form

键，即

files

（因为您将其定义为

files: List[bytes] = File(...)

），而不是

video

。 这个答案演示了如何将单个或多个文件上传到 FastAPI 服务器。请看一下。