有服务器2的CPU负载 [英] Load of server having 2 CPUs

问题描述

我有2 * CPU的服务器。当我在服务器上低于code运行它给出了两个CPU的LoadPercentage值：

I have a server with 2*CPU. when i run below code on server it gives two CPU LoadPercentage value as:

code：

@echo off
setlocal enabledelayedexpansion

(For /F "tokens=1,* delims==" %%A in ('"wmic cpu get LoadPercentage  /value  |find "P" "') do (
set "line=%%A %%B"
set "line=!line:~0,-1!"
echo !line!
))>output.txt

output.txt的：

LoadPercentage 2
LoadPercentage 4

虽然在任务管理器 - 服务器>性能其仅示出一个值（显然，如服务器为1）。
这怎么一个值显示，而CPU的两个......？它是平均二的的..？我如何使用批处理获得CPU负载百分比的单一价值？

While on server in task manager->performance its showing only one value (obviously, as server is 1). How this one value is showing while CPU's are two...? Is it average of two's..? How i can get that single value of CPU Load percentage using batch ?

推荐答案

是的，这是由负载处理器数量除以总和。

Yes, it's sum of loads divided by the number of processors.

在为循环，当你的负荷，总结他们并计算CPU的数量。在为退出，设置/一平均= ... 将给出答案。

In for loop, as you get the loads, sum them and count the number of cpus. On for exit, set /a average=... will give the answer.

编辑（在OP自己的答案）

EDIT (over the OP own answer)

@echo off
    setlocal enableextensions enabledelayedexpansion

    rem Initialize variables
    set sum=0
    set num=0

    rem Iterate over wmic output 
    for /f "tokens=2 delims==" %%1 in ('wmic cpu get LoadPercentage /value ^|find "P" ') do (
        set /a sum=!sum!+%%1
        set /a num=!num!+1
    )

    rem test if we get any data
    if %num% gtr 0 (
        set /a avg=%sum%/%num%
    ) else (
        set avg=0
    )

    echo CPULoadPercentage %avg% > output.txt

    exit /b

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