我有这个非常简单的代码，它获取一个字典作为输入并返回一个字典:
app = FastAPI()
class User(BaseModel):
user_name: dict
@app.post("/")
def main(user: User):
## some processing
return user
import requests
import json
import pandas as pd
df = pd.read_csv("myfile.csv")
data = {}
data["user_name"] = df.to_dict()
headers = {'content-type': 'application/json'}
url = 'http://localhost:8000/'
resp = requests.post(url,data=json.dumps(data), headers=headers )
resp
我想上传一个本地文件，将它发送到 API 并获取一个处理过的 .csv 文件。
现在我有以下代码来上传文件:
from typing import Optional
from fastapi import FastAPI
from fastapi import FastAPI, File, UploadFile
from pydantic import BaseModel
from typing import List
from fastapi.responses import HTMLResponse
app = FastAPI()
class User(BaseModel):
user_name: dict
@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(...)):
return {"filenames": [file.filename for file in files]}
@app.get("/")
async def main():
content = """
<body>
<form action="/files/" enctype="multipart/form-data" method="post">
<input name="files" type="file" multiple>
<input type="submit">
</form>
</body>
"""
return HTMLResponse(content=content)
这允许我选择和加载文件:
但是当我点击发送时，我被重定向到:http://localhost:8000/uploadfiles/并收到以下错误:
{"detail":"Method Not Allowed"}
最佳答案
我复制了您的代码，但遇到了同样的错误。经过FastAPI documentation for requesting files , 提到
To receive uploaded files, first install python-multipart.
E.g. pip install
python-multipart.This is because uploaded files are sent as "form data".
安装库后
python-multipart ，代码完美运行。引用文献:
关于python - 将 csv 文件发送到 fastAPI 并取回新文件，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/62749365/