我只想分享一个我用来增强docker stats命令的小脚本。
我不确定这种方法的正确性。
我可以假设整个Docker部署消耗的内存总量是每个容器消耗的内存之和吗?
请分享您的修改和/或更正。此命令在此处记录:https://docs.docker.com/engine/reference/commandline/stats/
运行docker stats时,输出如下所示:
$ docker stats --all --format "table {{.MemPerc}}\t{{.CPUPerc}}\t{{.MemUsage}}\t{{.Name}}"
MEM % CPU % MEM USAGE / LIMIT NAME
0.50% 1.00% 77.85MiB / 15.57GiB ecstatic_noether
1.50% 3.50% 233.55MiB / 15.57GiB stoic_goodall
0.25% 0.50% 38.92MiB / 15.57GiB drunk_visvesvaraya
我的脚本将在末尾添加以下行:
2.25% 5.00% 350.32MiB / 15.57GiB TOTAL
docker_stats.sh
#!/bin/bash
# This script is used to complete the output of the docker stats command.
# The docker stats command does not compute the total amount of resources (RAM or CPU)
# Get the total amount of RAM, assumes there are at least 1024*1024 KiB, therefore > 1 GiB
HOST_MEM_TOTAL=$(grep MemTotal /proc/meminfo | awk '{print $2/1024/1024}')
# Get the output of the docker stat command. Will be displayed at the end
# Without modifying the special variable IFS the ouput of the docker stats command won't have
# the new lines thus resulting in a failure when using awk to process each line
IFS=;
DOCKER_STATS_CMD=`docker stats --no-stream --format "table {{.MemPerc}}\t{{.CPUPerc}}\t{{.MemUsage}}\t{{.Name}}"`
SUM_RAM=`echo $DOCKER_STATS_CMD | tail -n +2 | sed "s/%//g" | awk '{s+=$1} END {print s}'`
SUM_CPU=`echo $DOCKER_STATS_CMD | tail -n +2 | sed "s/%//g" | awk '{s+=$2} END {print s}'`
SUM_RAM_QUANTITY=`LC_NUMERIC=C printf %.2f $(echo "$SUM_RAM*$HOST_MEM_TOTAL*0.01" | bc)`
# Output the result
echo $DOCKER_STATS_CMD
echo -e "${SUM_RAM}%\t\t\t${SUM_CPU}%\t\t${SUM_RAM_QUANTITY}GiB / ${HOST_MEM_TOTAL}GiB\tTOTAL"
最佳答案
从上面链接的文档中,
然后,
根据您的问题,看起来您可以这样假设,但也不要忘记,它也是存在但未运行的容器中的因素。
关于bash - 增强的docker stats命令,具有RAM和CPU总量,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/47331106/