Please can someone help me understand the first line arguements perpose and use ?
ports=$(nmap -p- --min-rate=1000 -T4 10.10.10.27 | grep [1] | cut -d ‘/’ -f 1 | tr ‘\n’ ‘,’ | sed s/,$//)
-p- = ?
–min-rate=1000 = ?
-T4 = ?
| grep [2] = ?
cut -d ‘/’ -f 1 = ?
tr ‘\n’ ‘,’ = ?
| sed s/,$//) = ?
Help explanation would be apprecaited.
@khalmxj said:
Please can someone help me understand the first line arguements perpose and use ?
This is (IMHO) not a great way to use nmap but YMMV.
ports=$(nmap -p- --min-rate=1000 -T4 10.10.10.27 | grep [1] | cut -d ‘/’ -f 1 | tr ‘\n’ ‘,’ | sed s/,$//)
-p- = ?
This is telling nmap to scan ports 0 - 65535.
–min-rate=1000 = ?
Send at least 1000 packets per second.
-T4 = ?
Use the aggressive speed template in nmap.
| grep [2] = ?
This pipes (|) the output to grep and uses a regular expression to filter on lines starting with a number.
cut -d ‘/’ -f 1 = ?
This pipes the output to cut which sets a delimiter of / and selects the first field.
tr ‘\n’ ‘,’ = ?
This pipes the output to tr to remove line breaks
| sed s/,$//) = ?
This pipes the output to sed which changes commas the at the end of a line to /
Help explanation would be apprecaited.
For more information, you can try:
man nmap
man grep
man tr
man sed
or https://linux.die.net/man/1/nmap
Thanks for a very clear direction and help.
@khalmxj said:
Thanks for a very clear direction and help.
Welcome.
Really, I’d suggest using nmap as a single line rather than sending it to the $ports variable and calling it a second time, but it depends on what automation you have in place.
Personally, I’d use:
nmap -Pn -sC -sV -oA all_tcp -T4 --min-rate=1000 -vvvvvvv --reason -p- IPADDRESSGOESHERE on HTB
Thanks for that, i’m kinda new to linux and was hoping to find just that.
1 more question though: why the “s” in “sed s/,$//” ? i know the comma and the slash are escaped with the “/”, $ is end of line, but still can’t figure out what the “s” is for. is it just to prevent it from reading the first “/” as an argument? like it could in fact be any letter?
@Popi said:
Thanks for that, i’m kinda new to linux and was hoping to find just that.
1 more question though: why the “s” in “sed s/,$//” ?
Because you are using the search and replace functionality.
You can find out more about sed with examples here: https://linuxconfig.org/learning-linux-commands-sed
i know the comma and the slash are escaped with the “/”, $ is end of line,
Well, not really here. An escape in bash is normally \.
Doing a replace with sed is basically the format:
sed 's/search/replace/g'
So they aren’t escapes in this context. It is search for , at the end of a line and replace with /.
but still can’t figure out what the “s” is for. is it just to prevent it from reading the first “/” as an argument? like it could in fact be any letter?
No.