nmap command help required

Please can someone help me understand the first line arguements perpose and use ?

ports=$(nmap -p- --min-rate=1000 -T4 10.10.10.27 | grep ^[0-9] | cut -d ‘/’ -f 1 | tr ‘\n’ ‘,’ | sed s/,$//)

-p- = ?
–min-rate=1000 = ?
-T4 = ?
| grep ^[0-9] = ?
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=$(nmap -p- --min-rate=1000 -T4 10.10.10.27 | grep ^[0-9] | cut -d ‘/’ -f 1 | tr ‘\n’ ‘,’ | sed s/,$//)

-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 ^[0-9] = ?

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 nmap(1) - Linux man page

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.

@TazWake Thanks a lot for the clarification. It helps.
Much appreciated!