[SOLVED] Search/Grep only Public IP Address with Regex in Linux

Issue

I've been reading about Regex to ONLY match PUBLIC IPv4 address and tried all solution given, but none of them actually can match the Public IP Address accurately.

Sample IP

[user@linux ~]$ cat ip.txt
1.1.1.1
8.8.8.8
10.1.1.1
127.0.0.1
[user@linux ~]$

Solution 1 - https://stackoverflow.com/a/39195704/11392987

[user@linux ~]$ egrep '^([0-9]|[1-9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5])(?<!172\.(16|17|18|19|20|21|22|23|24|25|26|27|28|29|30|31))(?<!127)(?<!^
10)(?<!^0)\.([0-9]|[1-9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5])(?<!192\.168)(?<!172\.(16|17|18|19|20|21|22|23|24|25|26|27|28|29|30|31))\.([0-9]|[1-
9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5])\.([0-9]|[1-9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5])(?<!\.255$)$' ip.txt
[user@linux ~]$

Sol 2 - https://stackoverflow.com/a/33453740/11392987

[user@linux ~]$ egrep '(\d+)(?<!10)\.(\d+)(?<!192\.168)(?<!172\.(1[6-9]|2\d|3[0-1]))\.(\d+)\.(\d+)' ip.txt
[user@linux ~]$

Sol 3 - https://stackoverflow.com/a/46399203/11392987

[user@linux ~]$ egrep '(^0\.)|(^10\.)|(^100\.6[4-9]\.)|(^100\.[7-9]\d\.)|(^100\.1[0-1]\d\.)|(^100\.12[0-7]\.)|(^127\.)|(^169\.254\.)|(^172\.1[6-
9]\.)|(^172\.2[0-9]\.)|(^172\.3[0-1]\.)|(^192\.0\.0\.)|(^192\.0\.2\.)|(^192\.88\.99\.)|(^192\.168\.)|(^198\.1[8-9]\.)|(^198\.51\.100\.)|(^203.0\
.113\.)|(^22[4-9]\.)|(^23[0-9]\.)|(^24[0-9]\.)|(^25[0-5]\.)' ip.txt
10.1.1.1
127.0.0.1
[user@linux ~]$

10.1.1.1 & 127.0.0.1 are Private IP Address, not Public IP

Sol 4 - https://stackoverflow.com/a/57077560/11392987

[user@linux ~]$ egrep '^(?!^0\.)(?!^10\.)(?!^100\.6[4-9]\.)(?!^100\.[7-9]\d\.)(?!^100\.1[0-1]\d\.)(?!^100\.12[0-7]\.)(?!^127\.)(?!^169\.254\.)(?
!^172\.1[6-9]\.)(?!^172\.2[0-9]\.)(?!^172\.3[0-1]\.)(?!^192\.0\.0\.)(?!^192\.0\.2\.)(?!^192\.88\.99\.)(?!^192\.168\.)(?!^198\.1[8-9]\.)(?!^198\.
51\.100\.)(?!^203.0\.113\.)(?!^22[4-9]\.)(?!^23[0-9]\.)(?!^24[0-9]\.)(?!^25[0-5]\.)(([0-9]|[1-9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5])\.([0-9]|[1-
9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5])\.([0-9]|[1-9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5])\.([0-9]|[1-9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5]))$' ip
.txt
[user@linux ~]$

Sol 5 - https://www.bigdatamark.com/regexp-for-extracting-public-ip-address/

[user@linux ~]$ egrep '\b(?!(10)|192\.168|172\.(2[0-9]|1[6-9]|3[0-2]))[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}' ip.txt
[user@linux ~]$

Solution

Try this regex. It matches all public IPs and doesn't match reserved IPs.

Edit - Try using grep -P instead of egrep because egrep doesn't support lookahead.

grep -P '^(?!^0\.)(?!^10\.)(?!^100\.6[4-9]\.)(?!^100\.[7-9]\d\.)(?!^100\.1[0-1]\d\.)(?!^100\.12[0-7]\.)(?!^127\.)(?!^169\.254\.)(?!^172\.1[6-9]\.)(?!^172\.2[0-9]\.)(?!^172\.3[0-1]\.)(?!^192\.0\.0\.)(?!^192\.0\.2\.)(?!^192\.88\.99\.)(?!^192\.168\.)(?!^198\.1[8-9]\.)(?!^198\.51\.100\.)(?!^203.0\.113\.)(?!^22[4-9]\.)(?!^23[0-9]\.)(?!^24[0-9]\.)(?!^25[0-5]\.)(([0-9]|[1-9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5])\.([0-9]|[1-9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5])\.([0-9]|[1-9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5])\.([0-9]|[1-9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5]))$' ip.txt

Output

1.1.1.1
8.8.8.8

Answered By - Karmveer Singh

Answer Checked By - Senaida (WPSolving Volunteer)