如何使用邮递员将文件发送到 fastapi 端点

本文介绍了如何使用邮递员将文件发送到 fastapi 端点的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我遇到了使用 postman 测试 api 的困难.通过 swagger 文件上传功能正常工作，我在硬盘上保存了一个文件.我想了解如何与邮递员一起执行此操作.我使用标准方式来处理我在使用 Django、flask 时使用的文件.

I faced the difficulty of testing api using postman. Through swagger file upload functionality works correctly, I get a saved file on my hard disk. I would like to understand how to do this with the postman. I use the standard way to work with files which I use when working with Django, flask.

Body -> form-data: key=file, value=image.jpeg

但是使用 fast API，我得到一个错误

But with fast API, I get an error

127.0.0.1:54294 - "POST /uploadfile/ HTTP/1.1" 422 Unprocessable Entity

ma​​in.py

@app.post("/uploadfile/") async def create_upload_file(file: UploadFile = File(...)): img = await file.read() if file.content_type not in ['image/jpeg', 'image/png']: raise HTTPException(status_code=406, detail="Please upload only .jpeg files") async with aiofiles.open(f"{file.filename}", "wb") as f: await f.write(img) return {"filename": file.filename}

我也试过 body ->二进制:image.jpeg .但得到了相同的结果

I also tried body -> binary: image.jpeg . But got the same result

推荐答案

我的代码:

from fastapi import FastAPI, UploadFile, File app = FastAPI() @app.post("/file/") async def create_upload_file(file: UploadFile = File(...)): return {"filename": file.filename}

在 Postman 中设置

Setup in Postman

如 github/tiangolo/fastapi/issues/1653，文件的参数名称是您必须使用的键值.在您使用 key=file 和 value=image.png (或其他)之前.相反，FastAPI 接受 file=image.png.因此错误，因为该文件是必需的，但它不存在(至少，具有该名称的键不存在).

As stated in github/tiangolo/fastapi/issues/1653, the parameter name for the file is the key value that you have to use. Before you were using key=file and value=image.png (or whatever). Instead, FastAPI accepts file=image.png. Thus the error, since the file is necessary, but it is not present (at least, the key with that name is not present).

附:我用 Postman v7.16.1 测试过

P.S. I tested it with Postman v7.16.1