MySQL评分系统-查找评分 [英] MySQL Rating System - Find Rating

问题描述

我正在尝试通过用户给出的最佳评分来对水果类型"进行评分.

I'm trying to rate Types of Fruit by the best possible rating given by users.

我有一个名为Ratings的Fruit表，用于通过以下信息提交评分

I have a Fruit table called Ratings just for submitting ratings with the following Information

Fruit_id int
From_ID int
Rating int

现在我正在尝试运行如下所示的SQL命令

now I'm trying to run a SQL command like the following

select From_ID, AVG(Rating) AS Rating FROM Ratings Group BY `Fruit_ID` ORDER by Rating DESC

但是，如果我得到1个将水果的苹果评级为5的用户，然后又有1000个用户将橙色的评级为4的用户，则苹果突然被认为更好.

This works however, if i get 1 user who rates a fruit an apple a 5, and then 1000 users who rate an orange a 4 the apple is suddenly considered better.

我该如何考虑投票数，以确保投票得最多和得票最多?

How can i take number of votes into account, to ensure that the most voted along with best is first.

推荐答案

为什么不只是选择票数并将其显示给用户.因此，是的，一个苹果以1票对5票将是只有1票的5票.这将为用户提供一种简单的方法来查看给定评级的受欢迎程度".

Why not just select the number of votes and display that to the user. So yes, 1 vote of 5 for an apple would be an overall 5 BUT with only 1 vote. This would give the users a simple way to see how "popular" a given rating is.

如果我没有记错的话，这就是当今大多数五星级评级系统在网上工作的方式.

This is how most of the fivestar rating systems work on the net today if I am not mistaken.

祝你好运！

尼克

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