我从this question获取了上传视频文件的代码;但是会发生错误。
下面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。
第一次
以下是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
1条答案
按热度按时间huus2vyu1#
您正在尝试发送
multipart/form-data（请参阅FastAPI documentation for uploading Files），但您已将Content-Type标头设置为application/json。因此，您应该删除/更改它。另外，在客户端，您应该使用与服务器端相同的
form密钥来上传文件，即files（因为您将其定义为files: List[bytes] = File(...)），而不是video。This answer演示了如何将单个或多个文件上传到FastAPI服务器。请查看它。