python - 为 API 构建异步 Python 函数

Question

我正在创建一个与 API 通信的模块，并且我正在考虑使用 asyncio 创建我的函数。现在我的代码看起来像这样：

def getMethod(self, stuff:str=None): #yes it in a class
    header = {'Content-Type': 'application/json'}
    endpoint = f"https://{self.endpoint}/"
    params = {"myParam": stuff}
    res = requests.get(endpoint, params=params, headers=header)
    return res

如您所见，我正在使用请求模块来发送数据。我认为我需要这样做：

import asyncio
async def getMethod(self, stuff:str=None): #yes it in a class
    header = {'Content-Type': 'application/json'}
    endpoint = f"https://{self.endpoint}/"
    params = {"myParam": stuff}
    res = requests.get(endpoint, params=params, headers=header)
    yield res

这会使我的函数异步吗？还是我也应该使用 httpx（或 aiohttp）之类的东西来等待结果？像下面这样

async def getMethod(self, stuff:str=None): #yes it in a class
    header = {'Content-Type': 'application/json'}
    endpoint = f"https://{self.endpoint}/"
    params = {"myParam": stuff}
    await res = httpx.get(endpoint, params=params, headers=header)
    return res

谢谢你的帮助。试图在那里学习异步python。

Answer 1

建议使用 httpx，异步代码如下：

async def getMethod(self, stuff: str = None):  # yes it in a class
    header = {'Content-Type': 'application/json'}
    endpoint = f"https://{self.endpoint}/"
    params = {"myParam": stuff}
    async with httpx.AsyncClient(headers=header, params=params) as client:
        res = await client.get(endpoint, params=params, headers=header)
    return res