Redis如何根据两个不同的排序集进行排序? [英] How can Redis sort according to two different sorted sets?

问题描述

我有两个不同的排序集.

I have two different sorted sets.

一个用于编辑者ID:

article_id  editor_id
101         10
102         11
103         10
104         10

另一个排序集用于日期排序:

The other sorted set is for date sorting:

article_id  day
101         29
102         27
103         25
104         27

我想合并显示第一编辑器第二天排序状态的这些集合. 我应该使用哪些命令?

I want to merge these sets which shows first editor second day sorted state. Which commands should I use?

推荐答案

假定article_id是您成员的值，并且editor_id/day是相应的有序集中的得分，并假设每个出现在两个排序集中，您可以执行以下操作:

Assuming that article_id is your members' value and that editor_id/day are the scores in the respective Sorted Set, and assuming each article_id is present in both Sorted Sets, you can do the following:

ZINTERSTORE t 2 k1 k2 WEIGHTS 100 1 AGGREGATE SUM

说明:

• t是一个临时键，将保存结果
• k1是存储editor_id
的排序集
• k2是存储day
的排序集
• 权重100乘editor_id乘以100(即，将其右移两个位置)
• AGGREGATE SUM的得分如下:editor_id * 100 + day

• t is a temporary key that will hold the result
• k1 is the Sorted Set that stores the editor_id
• k2 is the Sorted Set that stores the day
• the weight 100 multiplies editor_id by 100 (i.e. "shifts" it two places to the right)
• the AGGREGATE SUM results in the following score: editor_id * 100 + day

注意:

• 您可以使用ZUNIONSTORE代替以获得相同的结果
• 权重100的使用假定day是2位数的值

• you can use ZUNIONSTORE instead for the same result
• the use of weight 100 assumes that day is a 2-digit value

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