Question

我有一个针对大学的BASH项目，我有两个我不理解的错误

这是我的脚本BASH：

#!/bin/bash

proc_name=`cat /proc/cpuinfo | grep 'model name' | cut -d':' -f2 |cut -d'@' -f1 | uniq`;
proc_freq=`cat /proc/cpuinfo | grep 'model name' | cut -d':' -f2 |cut -d'@' -f2 | uniq`;
proc_core=`cat /proc/cpuinfo | grep 'cpu cores' | cut -d':' -f2 | uniq`;
proc_hyperthreading=`cat /proc/cpuinfo | grep 'siblings' | cut -d':' -f2 | uniq`;
proc_architecture=`lscpu | grep '64-bit' | cut -d',' -f2 | cut -d'-' -f1`;
proc_cache_L1=`lscpu | grep 'Cache L1i' | cut -d':' -f2 | sed "s/\ \ */\ /g"`;
proc_cache_L2=`lscpu | grep 'Cache L2' | cut -d':' -f2 | sed "s/\ \ */\ /g"`;
proc_cache_L3=`lscpu | grep 'Cache L3' | cut -d':' -f2 | sed "s/\ \ */\ /g"`;
proc_virtualisation=`lscpu | grep 'Virtualisation' | cut -d':' -f2 |sed "s/\ \ */\ /g"`;
proc_load_average=`w | head -1 | cut -d" " -f12 | cut -d"," -f1-2` | tr ',' '.'`

ip_infos_addr_ipv4=`/sbin/ifconfig eth0 | awk '/inet adr:/{print $2}' | awk -F ':' '{print $2}'`;
ip_infos_addr_ipv6=`/sbin/ifconfig eth0 | awk '/adr inet6:/{print $3}'`;
ip_publique_addr=`dig +short myip.opendns.com @resolver1.opendns.com`;
carte_reseau=`lspci |grep Ethernet | cut -d":" -f3`;


echo -e "$proc_name\n$proc_freq\n$proc_core\n$proc_hyperthreading\n$proc_architecture\n$proc_cache_L1\n$proc_cache_L2\n$proc_cache_L3\n$proc_virtualisation\n$proc_load_average\n$ip_infos_addr_ipv4\n$ip_infos_addr_ipv6\n$ip_publique_addr\n$carte_reseau" > Collecteur/collecteur_cpu_reseau.txt;

以下是我遇到的两个错误：

./collecteur_cpu_reseau: line 17: unexpected EOF while looking for matching ``'
./collecteur_cpu_reseau: line 21: syntax error: unexpected end of file

提前感谢您的帮助

Answer 1

proc_load_average=`w | head -1 | cut -d" " -f12 | cut -d"," -f1-2` | tr ',' '.'`

......需要......

# remove the extra backtick
proc_load_average=`w | head -1 | cut -d" " -f12 | cut -d"," -f1-2 | tr ',' '.'`

......或更好......

# use parens, not backticks
proc_load_average=$(w | head -1 | cut -d" " -f12 | cut -d"," -f1-2 | tr ',' '.')

......或更好......

# read the content straight from procfs without the big silly pipeline
read -r loadavg_1min loadavg_5min loadavg_10min _ </proc/loadavg
echo "1-minute load average: $loadavg_1min"

也就是说，这个剧本作为一个整体是无可挽回的。

例如，考虑以下替代/proc/cpuinfo中的所有混乱：

declare -A cpuinfo=( ) # create an associative array

# read line-by-line; see http://mywiki.wooledge.org/BashFAQ/001
while IFS=$'\t:' read -r k v || [[ $k ]]; do
  [[ $v ]] || continue
  cpuinfo[$k]=${v# } # trim leading space; see http://wiki.bash-hackers.org/syntax/pe
done </proc/cpuinfo

echo "Model name: ${cpuinfo['model name']%@*}"
echo "Model freq: ${cpuinfo['model name']#*@}"
echo "Actual frequency: ${cpuinfo['cpu MHz']}"
echo "Siblings: ${cpuinfo['siblings']}"

......不是那么容易吗？您可以通过将lscpu替换为< <(lscpu)来部署读取</proc/cpuinfo之类命令输出的相同策略。

BASH意外的EOF，同时寻找匹配的``'

1 个答案: