这是我的代码尝试上载图像列表:
import requests
import glob
import cv2
path = glob.glob("test_folder/*", recursive=True) # a list of image's path
lst_img = []
for p in path[:3]:
# img = cv2.imread(p)
lst_img.append((p, open(p, 'rb'), "image/jpeg"))
data = {"files": lst_img}
url = "http://localhost:6789/" # url api of app
res = requests.post(url=url, data=data)
print(res.status_code)
print(res.text)
我正在尝试通过url api上传图像列表。在这里,我使用请求(python程序包),但也许我的请求格式不正确,那么我无法发布到url api。我收到错误代码422:
"detail":[{"loc":["body","files",0],"msg":"Expected UploadFile, received: <class 'str'>","type":"value_error"}
这是我的请求格式:
{'files': [('test_folder/image77.jpeg', <_io.BufferedReader name='test_folder/image77.jpeg'>, 'image/jpeg'), ('test_folder/image84.jpeg', <_io.BufferedReader name='test_folder/image84.jpeg'>, 'image/jpeg'), ('test_folder/image82.jpeg', <_io.BufferedReader name='test_folder/image82.jpeg'>, 'image/jpeg')]}
我尝试了很多方法,但总是失败。非常感谢你们帮助解决这个问题。
我尝试过但仍然无法正常工作
lst_img.append(("file", (p, open(p, 'rb'), "image/jpeg")))
我的FastAPI main.py
from typing import List
from fastapi import FastAPI, File, UploadFile
from fastapi.responses import StreamingResponse, FileResponse
app = FastAPI()
@app.post("/")
async def main(files: List[UploadFile] = File(...)):
# file_like = open(video_path, mode="rb")
# return StreamingResponse(file_like, media_type="video/mp4")
return {"filenames": [file.filename for file in files]}
答案 0 :(得分:0)
您应该使用files
模块的request
参数发送文件
import requests
import glob
path = glob.glob("test_folder/*", recursive=True) # a list of image's path
lst_img = []
for p in path[:3]:
lst_img.append({"files": open(p, 'rb')})
url = "http://localhost:6789/" # url api of app
for data in lst_img:
res = requests.post(url=url, files=data)
print(res.status_code)
print(res.text)