创建视图函数而不在Flask中返回响应 [英] Creating a view function without returning a response in Flask

问题描述

我是网络编程和Flask的新手，我最近遇到了一个我想要创建的网站的问题。我目前有一个jquery过程，它向Flask中的视图函数发送一个post请求。这个函数只是增加了我数据库中的一个值，并且在增加这个值之后我没有必要返回一个响应。但是，据我所见，Flask中的视图函数需要返回一个Response对象。我当然可以简单地返回某种json已更新的响应，但它对我的应用程序来说真的不重要。有谁知道解决这个问题的正确方法？谢谢。

I'm rather new to web programming and Flask, and I've recently run into a problem with a website I am trying to create. I currently have a jquery procedure which sends a post request to a view function in Flask. This function simply increments a value in my database, and it's not really necessary for me to return a response after incrementing this value. However, as far as I've seen, view functions in Flask are required to return a Response object. I could of course trivially return some sort of json "was-updated" response, but it's really not important to my application. Does anyone know if the proper way to resolve this issue? Thanks.

推荐答案

这很简单，只需返回一个包含204状态代码的空字符串：

It's quite easy, just return an empty string with 204 status code:

@app.route('/')
def hello():
    ...
    return '', 204

这篇关于创建视图函数而不在Flask中返回响应的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！