Created
December 24, 2012 03:18
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实现一个API,API内部有两个http接口A,B并发调用,如果都在200ms内返回,则合并结果输出,如果B比A慢,且B耗时超过200ms,则丢弃B调用只返回A结果
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-module(multi_try). | |
-export([dothese/1]). | |
dothese(FuncArray) -> | |
Me = self(), | |
lists:foreach(fun(F) -> spawn(fun() -> F(Me) end) end, FuncArray), | |
spawn(fun() -> timer:sleep(1000), Me ! timeout end), | |
Result = get_result("",length(FuncArray)), | |
io:format("result: ~p~n", [Result]). | |
get_result(Result, 0) -> | |
Result; | |
get_result(Result, RemainTimes) -> | |
receive | |
timeout -> | |
Result; | |
Sth -> get_result(Result ++ Sth, RemainTimes - 1) | |
end. |
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#!/usr/bin/env escript | |
main(_) -> | |
A = fun(Pid) -> timer:sleep(200), Pid ! "A" end, | |
B = fun(Pid) -> timer:sleep(1200), Pid ! "B" end, | |
C = fun(Pid) -> timer:sleep(500), Pid ! "C" end, | |
multi_try:dothese([A,B,C]). |
200毫秒吗?我写成1000毫秒了,逻辑在这里
spawn(fun() -> timer:sleep(1000), Me ! timeout end),
修改成 200ms 了,更新到 https://gist.github.com/4367307
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没有看到如果B比A慢,且超过200ms时候,丢弃B调用的相关逻辑