[英]how to Replace IP Address from in specified file
我有以下腳本為此,它替換了指定文件中的每個ip,但是邏輯不好
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 $OPTARG"
else
echo "fail to update"
fi ;;
h) ;;
esac
done
`
想要將所有搜索到的IP地址存儲在陣列中替換它如何執行此操作
輸入文件是
ome log file entries
some log file entries
some log file entries
some log file entries
This system ip is not found
some log file entries
some log file entries
some log file entries
This system IP is 122.0.0.0
some log file entries
some log file entries
This system IP:122.0.0.0
some log file entries
some log file entries
some log file entries
Hostname:ip-172.31.18.255.ec2.internal
some log file entries
some log file entries
以下解決方案將允許您將所有匹配的IP地址存儲在陣列中,並在腳本的其他位置使用它們。 根據我的觀察(使用您的腳本),源文件中的所有IP地址都被替換了,但EC2專用主機名中包含的IP地址除外。 尚不清楚這是否是您想要的。 下面將替換所有IP地址(包括EC2主機名中包含的IP)。 我正在使用的正則表達式摘自本文中的可接受答案。
#!/usr/bin/env bash
#
# Directory where this script is located
#
DIR="$( 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
如果您不想替換EC2主機名中包含的IP地址,則如下所示的正則表達式(我是根據本文的答案構建的 )
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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