mjhjmtu

I have a file something like this:

dn: danan

cn: danian

cn: danian1

dn: danian2

cn: danian2

dn: danian3

cn: danian3

cn: danian4

all I want to do is grep the pattern "dn:" and print just the next 2 lines, but when I run grep -A 2 "dn:" the result is everything in the file, I just want to print the next 2 lines not the next 1 line

This might work:

$ awk -v RS="dn: " -v FS="n" -v ORS="" 'NF>3 print "dn:",$0' input.txt

We define dn: as a record separator. The fields in one record are delimited by a new line character n.

In case you have two lines following a line starting with dn:, before a new record starts with dn:, you will have 3 times a "n" leading to 4 fields in the record. This is why we check if there are more than 3 fields in the record (NF>3). If that is the case, we print out the whole record, but need to prepend the dn:.

Note that NF>3 will find all records with more than two following lines in one record. If you want only those with exact two change it to NF==3.

If the blocks are delimited by a blank line (as it seems in one of your later comments) use this instead:

$ awk -v RS="nn" -v FS="n" -v ORS="nn" 'NF>2 print $0' input.txt

I run the command above and works perfectly,

awk -v RS="nn" -v FS="n" -v ORS="nn" 'NF>2 print $0' 

the blocks delimited are blanck, it works perfectly.