Go+redis实现简单的抽奖系统

redis的集合是 无序、不重复的集合，如果一个抽奖系统只需要简单的从一堆数据里抽出一部分，那么redis就可以简单的实现

主要用到了SPOP(随机移除并返回集合中一个或多个元素) 和 SRANDMEMBER(随机返回集合中一个或多个元素)

package main

import (
	"fmt"
	"github.com/gomodule/redigo/redis"
)

// 简单抽奖

func main() {
	var allPeople = []string{
		"Tom",
		"Jerry",
		"Linda",
		"John",
		"Peter",
		"Cooper",
		"Amy",
		"Bob",
		"Luck",
	}
	var keyName = "lucky:1"
	args := []interface{}{keyName}
	for _, person := range allPeople {
		args = append(args, person)
	}
	// sadd 存入 lucky:1
	_, err := Redis().Do("sadd", args...)
	defer Redis().Close()
	if err != nil {
		panic(err)
	}
	// 随机抽三个幸运用户
	//SRANDMEMBER(随机返回集合中一个或多个元素)
	lotteryResult := getLottery(keyName, 3)
	fmt.Printf("三个幸运用户是:%v \n", lotteryResult)
	// 分层抽，抽三次，每次抽，1,2,3 个人分别为1等奖，2等奖，3等奖,且一个人只能获得一种奖，不重复抽取
	// SPOP(随机移除并返回集合中一个或多个元素)
	noRepeatLottery := getNoRepeatLottery(keyName)
	for i, strings := range noRepeatLottery {
		fmt.Printf("%v等奖:%v \n", i+1, strings)
	}
}
func getNoRepeatLottery(keyName string) [][]string {
	var lotteryPeopleList [][]string
	//先抽一个一等奖
	reply, err := redis.Strings(Redis().Do("SPOP", keyName, 1))
	if err != nil {
		panic(err)
	}
	lotteryPeopleList = append(lotteryPeopleList, reply)
	//抽两个二等奖
	reply2, err := redis.Strings(Redis().Do("SPOP", keyName, 2))
	if err != nil {
		panic(err)
	}
	lotteryPeopleList = append(lotteryPeopleList, reply2)
	//抽三个三等奖
	reply3, err := redis.Strings(Redis().Do("SPOP", keyName, 3))
	if err != nil {
		panic(err)
	}
	lotteryPeopleList = append(lotteryPeopleList, reply3)
	return lotteryPeopleList
}

func getLottery(keyName string, luckyPeopleCount int) []string {
	//SRANDMEMBER(随机返回集合中一个或多个元素)
	reply, err := redis.Strings(Redis().Do("SRANDMEMBER", keyName, luckyPeopleCount))
	defer Redis().Close()
	if err != nil {
		panic(err)
	}
	return reply
}

func Redis() redis.Conn {
	dial, err := redis.Dial("tcp", "127.0.0.1:6379")
	if err != nil {
		panic(err)
	}
	return dial

}