[SOLVED] How to use grep/sed/awk, to remove a pattern from beginning of a text file

Issue

I have a text file with the following pattern written to it:

TIME[32.468ms]  -(3)-............."TEXT I WANT TO KEEP"

I would like to discard the first part of each line containing

TIME[32.468ms]  -(3)-.............

To test the regular expression I've tried the following:

cat myfile.txt | egrep "^TIME\[.*\]\s\s\-\(3\)\-\.+"

This identifies correctly the lines I want. Now, to delete the pattern I've tried:

cat myfile.txt | sed s/"^TIME\[.*\]\s\s\-\(3\)\-\.+"//

but it just seems to be doing the cat, since it shows the content of the complete file and no substitution happens.

What am I doing wrong?

OS: CentOS 7

Solution

With your shown samples, please try following grep command. Written and tested with GNU grep.

grep -oP '^TIME\[\d+\.\d+ms\]\s+-\(\d+\)-\.+\K.*' Input_file

Explanation: Adding detailed explanation for above code.

^TIME\[          ##Matching string TIME from starting of value here.
\d+\.\d+ms\]     ##Matching digits(1 or more occurrences) followed by dot digits(1 or more occurrences) followed by ms ] here.
\s+-\(\d+\)-\.+  ##Matching spaces91 or more occurrences) followed by - digits(1 or more occurrences) - and 1 or more dots.
\K               ##Using \K option of GNU grep to make sure previous match is found in line but don't consider it in printing, print next matched regex part only.
.*               ##to match till end of the value.

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2nd solution: Adding awk program here.

awk 'match($0,/^TIME\[[0-9]+\.[0-9]+ms\][[:space:]]+-\([0-9]+\)-\.+/){print substr($0,RSTART+RLENGTH)}' Input_file

Explanation: using match function of awk, to match regex ^TIME\[[0-9]+\.[0-9]+ms\][[:space:]]+-\([0-9]+\)-\.+ which will catch text which we actually want to remove from lines. Then printing rest of the text apart from matched one which is actually required by OP.

Answered By - RavinderSingh13