python - 快速 API - 如何在 GET 中显示来自 POST 的图像？

Question

我正在使用 FastAPI 创建一个应用程序，该应用程序应该生成已上传图像的调整大小版本。上传应该通过 POST/images 完成，在调用路径/images/800x400 后，它应该显示来自数据库的随机图像，大小为 800x400。我在尝试显示图像时遇到错误。

 from fastapi.responses import FileResponse
 import uuid

 app = FastAPI()

 db = []

@app.post("/images/")
async def create_upload_file(file: UploadFile = File(...)):

    contents = await file.read() 

    db.append(file)

    with open(file.filename, "wb") as f:
        f.write(contents)

    return {"filename": file.filename}

@app.get("/images/")
async def show_image():  
     return db[0]

作为回应，我得到:

{
  "filename": "70188bdc-923c-4bd3-be15-8e71966cab31.jpg",
  "content_type": "image/jpeg",
  "file": {}
}

我想使用:return FileResponse(some_file_path)并在文件路径中放置上面的文件名。这是正确的思维方式吗？

Answer 1

首先，您将 File 对象添加到您的数据库列表中，这解释了您得到的响应。

您想将文件的内容写入您的数据库。

如果您将其用作“持久性”，您也不需要将其写入文件系统(当然，如果您关闭或重新加载应用程序，所有文件都会消失)。

from fastapi import FastAPI, File, UploadFile
from fastapi.responses import Response
import os
from random import randint
import uuid

app = FastAPI()

db = []


@app.post("/images/")
async def create_upload_file(file: UploadFile = File(...)):

    file.filename = f"{uuid.uuid4()}.jpg"
    contents = await file.read()  # <-- Important!

    db.append(contents)

    return {"filename": file.filename}


@app.get("/images/")
async def read_random_file():

    # get a random file from the image db
    random_index = randint(0, len(db) - 1)

    # return a response object directly as FileResponse expects a file-like object
    # and StreamingResponse expects an iterator/generator
    response = Response(content=db[random_index])

    return response

如果您想将文件实际保存到磁盘，这是我会使用的方法(对于完整的应用程序，仍然首选真正的数据库)

from fastapi import FastAPI, File, UploadFile
from fastapi.responses import FileResponse
import os
from random import randint
import uuid

IMAGEDIR = "fastapi-images/"

app = FastAPI()


@app.post("/images/")
async def create_upload_file(file: UploadFile = File(...)):

    file.filename = f"{uuid.uuid4()}.jpg"
    contents = await file.read()  # <-- Important!

    # example of how you can save the file
    with open(f"{IMAGEDIR}{file.filename}", "wb") as f:
        f.write(contents)

    return {"filename": file.filename}


@app.get("/images/")
async def read_random_file():

    # get a random file from the image directory
    files = os.listdir(IMAGEDIR)
    random_index = randint(0, len(files) - 1)

    path = f"{IMAGEDIR}{files[random_index]}"
    
    # notice you can use FileResponse now because it expects a path
    return FileResponse(path)

引用:

(FastAPI 继承了 Starlette 的响应)

• Starlette Response
• Starlette StreamingResponse
• Starlette FileResponse

(Tiangolo的文档还是很不错的)

• FastAPI Response
• FastAPI StreamingResponse
• FastAPI FileResponse