Removing all strings before first occurrence of a pattern in bash-script

Question

so i've got a problem. I want to remove the whole content of a file before a certain pattern has been found, but only at the first occurrence of it. the pattern : ([0-9]{2}:[0-9]{2}:[0-9]{2}).([0-9]{6}) (it is fits the date part of the string).

For example, this content:

-- 10:17:40.614568 00:00:00:00:00:00 > ff:ff:ff:ff:ff:ff, ethertype IPv4 (0x0800), length 303: (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto UDP (17), length 289) 0.0.0.0.68 > 255.255.255.255.67: BOOTP/DHCP, Request from 00:00:00:00:00:00, length 261, xid 0x1466f606, Flags [Broadcast]

should be parsed to :

10:17:40.614568 00:00:00:00:00:00 > ff:ff:ff:ff:ff:ff, ethertype IPv4 (0x0800), length 303: (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto UDP (17), length 289) 0.0.0.0.68 > 255.255.255.255.67: BOOTP/DHCP, Request from 00:00:00:00:00:00, length 261, xid 0x1466f606, Flags [Broadcast]

Answer 1

EDIT: Since OP mentioned condition to be met only for very first regex match in entire Input_file so adding this solution now.

awk --re-interval '
match($0,/[0-9]{2}:[0-9]{2}:[0-9]{2}\.[0-9]{6}/)  && !index{
  print substr($0,RSTART)
  index=1
  next
}
index ' Input_file

---

---

Could you please try following.

awk --re-interval '
match($0,/[0-9]{2}:[0-9]{2}:[0-9]{2}\.[0-9]{6}/){
  print substr($0,RSTART)
}' Input_file

It will only print those lines which have a regex match found. In case you want to print those lines too which NOT have a match then do following.

awk --re-interval '
match($0,/[0-9]{2}:[0-9]{2}:[0-9]{2}\.[0-9]{6}/){
  print substr($0,RSTART)
  next
}
1' Input_file

Since I am using an OLD awk version so I have out --re-interval remove it in case for you above code works(new versions of awk it works)

Explanation of 1st code:

awk --re-interval '                                ##Starting awk program from here and --re-interval enables ERE for OLD versions of awk.
match($0,/[0-9]{2}:[0-9]{2}:[0-9]{2}\.[0-9]{6}/){  ##Using match utility of awk here which will check REGEX provided to it either it presents on line or not.
  print substr($0,RSTART)                          ##match utility/function has variable named RSTART which denotes the starting of REGEX point, so I am using substring function to print from starting of that point to till end of line, since OP want to remove everything before REGEX match.
}
' Input_file                                       ##Mentioning Input_file name here.

For 2nd code's explanation will be same as 1st code only difference is 2nd code has next will skip lines whose regex is matched and 1 will print non-matched lines in Input_file.