redis lua 脚本与单次调用 [英] redis lua script vs. single calls

问题描述

我有以下设置:

• 2 种不同的数据结构:集合、字符串
• 它们在不同的命名空间 *:collections:*, *:resources:*
• 客户不知道这一点，我每次都尝试获取两个命名空间.
• 基于exists，我决定最终要获得哪种数据结构.
• 对 redis 的所有调用都是异步完成的 (vert.x redis-mod)

• 2 different datastructures: Sets, Strings
• They are in different namespaces *:collections:*, *:resources:*
• The client doesn't know about this and I try to get both namespaces every time.
• Based on exists I decide which datastructure to get finally.
• all calls to redis are done asynchronous (vert.x redis-mod)

现在我必须决定是将它作为 lua 脚本还是作为单个命令来执行.

Now I have to decide if I execute this as lua script or as single commands.

我想出的lua脚本:

local path = KEYS[1]
local resourcesPrefix = ARGV[1]
local collectionsPrefix = ARGV[2]

if redis.call('exists',resourcesPrefix..path) == 1 then
    return redis.call('get',resourcesPrefix..path)
elseif redis.call('exists',collectionsPrefix..path) == 1 then
    return redis.call('smembers',collectionsPrefix..path)
else
    return "notFound"
end 

单次调用或lua脚本有什么优缺点吗?

Are there any pros and cons for single calls or lua script?

推荐答案

是的，LUA 脚本是 EVALSHA 调用的最佳解决方案:

Yes, LUA script is a best solution in case of EVALSHA call:

• 您正在使用 redis 异步工作.所以 LUA 可以帮助您减少代码数量和代码可读性.
• LUA 情况更快，因为减少了网络通信.

• You are working woth redis asynchronous. So LUA helps you to reduce number of code and code readability.
• LUA case is faster becouse of reduce network communication.

我认为您只需使用 2 个命令即可编写代码.您的代码中不需要 exists.

I think you can write your code with just 2 commands. You do not need exists in your code.

local path = KEYS[1]
local resourcesPrefix = ARGV[1]
local collectionsPrefix = ARGV[2]
local ret

set ret = redis.call('get',resourcesPrefix..path)
if ret then
   return ret
end  
set ret = redis.call('smembers',collectionsPrefix..path)
if ret then
   return ret
end  

return "notFound" 

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