FastAPI - 如何在使用 UploadFile 时读取 json 文件
[英]FastAPI - How to read an json file while using UploadFile
问题描述
from typing import List, Optional
from fastapi import FastAPI, File, UploadFile
from fastapi.responses import HTMLResponse
from pydantic import BaseModel
app = FastAPI(debug=True)
@app.post("/uploadfiles/")
def create_upload_files(upload_file: UploadFile = File(...)):
json_data = ?? upload_file ??
result = model().calculate(json_data)
return { "estimation": result}
@app.get("/")
async def main():
content = """
<body>
<form action="/uploadfiles/" enctype="multipart/form-data" method="post">
<input name="upload_file" type="file" multiple>
<input type="submit">
</form>
</body>
"""
return HTMLResponse(content=content)
I have the above FastAPI app.
我有上面的 FastAPI 应用程序。 I need to upload a json file.我需要上传一个json文件。 Thus the upload_file is a json file.因此，upload_file 是一个 json 文件。 Also the model() instance uses a method calculate that takes as input json data.此外，model() 实例使用了一种将 json 数据作为输入的方法。 I struggle on how to decode the upload_file from Fast_API to dictionairy format.我在如何将上传文件从 Fast_API 解码为字典格式方面苦苦挣扎。
I tried upload_file.read() but this returns a bytes array我试过 upload_file.read() 但这返回一个字节数组
Could you please help?能否请你帮忙？
1 个解决方案
解决方案1
3 已采纳 2020-11-20 17:06:03
You can use the standard json module to parse the content by using json.load() --(Doc) from an uploaded JSON file as您可以使用标准json模块通过使用json.load() --(Doc)从上传的 JSON 文件解析内容作为
from fastapi import FastAPI, File, UploadFile
import json
app = FastAPI(debug=True)
@app.post("/uploadfiles/")
def create_upload_files(upload_file: UploadFile = File(...)):
json_data = json.load(upload_file.file)
return {"data_in_file": json_data} Thus, you will have the JSON contents in your json_data variable.因此，您的json_data变量中将包含 JSON 内容。
Alternatively, you can use json.loads() --(Doc) function as或者，您可以使用json.loads() --(Doc)函数作为
json_data = json.loads(upload_file.file.read())
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