[英]how to Replace IP Address from in specified file
我有以下脚本为此,它替换了指定文件中的每个ip,但是逻辑不好
while getopts i:h: opt
do
case $opt in
i)
echo "Proposed ip will $OPTARG"
if [[ $OPTARG =~ ^[0-9]+\.[0-9]+\.[0-9]+\.[0-9]+$ ]]; then
sed -i 's/[0-9]\{1,3\}\.[0-9]\{1,3\}\.[0-9]\{1,3\}\.[0-9]\{1,3\}$/'$OPTARG'/g' sample.txt
echo "proposed ip is $OPTARG"
else
echo "fail to update"
fi ;;
h) ;;
esac
done
`
想要将所有搜索到的IP地址存储在阵列中替换它如何执行此操作
输入文件是
ome log file entries
some log file entries
some log file entries
some log file entries
This system ip is not found
some log file entries
some log file entries
some log file entries
This system IP is 122.0.0.0
some log file entries
some log file entries
This system IP:122.0.0.0
some log file entries
some log file entries
some log file entries
Hostname:ip-172.31.18.255.ec2.internal
some log file entries
some log file entries
以下解决方案将允许您将所有匹配的IP地址存储在阵列中,并在脚本的其他位置使用它们。 根据我的观察(使用您的脚本),源文件中的所有IP地址都被替换了,但EC2专用主机名中包含的IP地址除外。 尚不清楚这是否是您想要的。 下面将替换所有IP地址(包括EC2主机名中包含的IP)。 我正在使用的正则表达式摘自本文中的可接受答案。
#!/usr/bin/env bash
#
# Directory where this script is located
#
DIR="$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd )"
declare ips=()
declare ip_regex='[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}'
while getopts i:h: opt
do
case $opt in
i)
echo "Proposed ip is $OPTARG"
if [[ $OPTARG =~ ${regex} ]]; then
# Store IPs in array for further processingg
ips=$(echo $(cat "${DIR}/sample.txt") | grep -Eo ${ip_regex})
for ip in ${ips[@]}; do
echo "Replacing '${ip}' with '${OPTARG}'"
sed -i 's/'${ip}'/'$OPTARG'/g' "${DIR}/sample.txt"
done
else
echo "Failed to update"
fi ;;
h) ;;
esac
done
如果您不想替换EC2主机名中包含的IP地址,则如下所示的正则表达式(我是根据本文的答案构建的 )
declare ip_regex='(^|[[:space:]])[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}([[:space:]]|$)'
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