UploadFile
FastAPI 自动在 OpenAPI 规范中为参数生成模式。
例如,这段代码:
from fastapi import FastAPI, File, UploadFile
app = FastAPI()
@app.post("/uploadfile/")
async def create_upload_file(file: UploadFile = File(..., description="The file")):
return {"filename": file.filename}
components:schemas
将在 OpenAPI 规范中生成此模式:
{
"Body_create_upload_file_uploadfile__post": {
"title": "Body_create_upload_file_uploadfile__post",
"required":["file"],
"type":"object",
"properties":{
"file": {"title": "File", "type": "string", "description": "The file","format":"binary"}
}
}
}
如何明确指定 UploadFiles 的架构(或至少其名称)?
我已阅读 FastAPIs 文档并搜索了问题跟踪器,但一无所获。