运行 Redis 3.2.1 和最新的 Hedis 库,我有以下发布程序:
{-# LANGUAGE OverloadedStrings #-}
module Main where
import Database.Redis
import Control.Monad
import Control.Concurrent
import Control.Monad.Trans
import Data.ByteString as BS
import System.Posix.Process
import Data.String.Conv
main :: IO ()
main = do
conn <- connect defaultConnectInfo
runRedis conn run
run = do
liftIO $ threadDelay $ 1000 * 1000
pid <- liftIO getProcessID
publish "chan1" (toS $ show pid)
publish "chan2" (toS $ show pid)
liftIO $ Prelude.putStrLn "\n\n%%%%%%%\n\nnext\n\n%%%%%%%%\n\n"
run
订阅者看起来像这样:
{-# LANGUAGE OverloadedStrings #-}
module Main where
import Database.Redis
main :: IO ()
main = do
conn <- connect defaultConnectInfo
runRedis conn $ do
pubSub (subscribe ["chan1"]) $ \msg -> do
putStrLn $ "chan1 " ++ show (msgChannel msg) ++ ": " ++ show (msgMessage msg)
return mempty
pubSub (subscribe ["chan2"]) $ \msg -> do
putStrLn $ "chan2" ++ show (msgChannel msg) ++ ": " ++ show (msgMessage msg)
return mempty
输出是:
%%%%%%%
next
%%%%%%%%
chan1 "chan1": "21542"
%%%%%%%
next
%%%%%%%%
chan1 "chan1": "21542"
%%%%%%%
next
%%%%%%%%
chan1 "chan1": "21542"
%%%%%%%
next
%%%%%%%%
现在,看起来只要订阅者阅读了第一个 channel ,就不会阅读发送到第二个 channel 的消息。换句话说,似乎只是忽略了订阅 chan2
的命令。
为了完整起见,这是我的 Cabal
文件:
name: pub-sub-exp
version: 0.1.0.0
synopsis: Simple project template from stack
description: Please see README.md
homepage: https://github.com/githubuser/pub-sub-exp#readme
license: BSD3
license-file: LICENSE
author: Author name here
maintainer: example@example.com
copyright: 2016 Author name here
category: Web
build-type: Simple
cabal-version: >=1.10
executable pub
hs-source-dirs: src
main-is: Pub.hs
default-language: Haskell2010
build-depends: base >= 4.7 && < 5,
hedis,
mtl,
bytestring,
unix,
string-conv
executable sub
hs-source-dirs: src
main-is: Sub.hs
default-language: Haskell2010
build-depends: base >= 4.7 && < 5,
hedis,
mtl,
bytestring
我正在使用 stack-lts-6.6
。
为了澄清,我希望订阅者指明消息已发送到 channel 1 和 2。
这是 Redis 的一个众所周知的属性吗?我是否遗漏了一些 Haskell 陷阱?
最佳答案
您需要一次订阅两个 channel 。
pubSub (subscribe ["chan1", "chan2"]) $ \msg -> do
Hedis 没有接通您对 pubSub
的第二次调用。从pubSub
's definition可以看出除非订阅计数和未决消息都被耗尽,否则该函数不会返回。另请注意,没有 fork 或其他启用并发的方法。
关于haskell - 如何在同一个函数中订阅多个 Redis channel ?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38228263/