Is there any cut command to cut based on a word eg :
171212 16082784 6264 XXX xxxxxxxx Transaction XXXXX abend ABCD. The task has terminated abnormally because of a program check. 16:08:27I want the output as :
171212 16082784 6264 XXX xxxxxxxx Transaction XXXXX abend ABCD.How to approach?
35 Answers
I suggest:
awk -F 'ABCD' '{print $1 FS "."}' fileOutput:
171212 16082784 6264 XXX xxxxxxxx Transaction XXXXX abend ABCD.
FS contains your delimiter "ABCD". "ABCD" is a regex.
sed version:
sed 's/ABCD.*/ABCD./' input.txt
171212 16082784 6264 XXX xxxxxxxx Transaction XXXXX abend ABCD.Source:
2awk can do the job if you provide -F flag with the word which you want to use:
$ awk -F 'test' '{print $1;print $2}' <<< "onetesttwotest"
one
twoIn your particular case, that would be:
$ awk -F 'ABCD' '{print $1,FS}' input.txt
171212 16082784 6264 XXX xxxxxxxx Transaction XXXXX abend ABCDJudging from your example, you're only trying to print stuff up to ABCD so deleting everything after that is also an option:
$ awk '{print substr($0,0,match($0,/ABCD/)+4);}' input.txt
171212 16082784 6264 XXX xxxxxxxx Transaction XXXXX abend ABCD.grep solution:
grep -o '^.*ABCD\.' input.txtThe regular expression will match to each line that begin ^ with any character . repeated number of times * and ends with the string ABCD. The backslash \ escapes the special meaning of the dot . at the end.
The option -o will tell the grep command to print only the matched (non-empty) parts of a matching line.
cut can itself perform this job. Not based on a word, but based on . delimiter.
Use :
cut -f 1 -d '.' input.txt | xargs -I "%" echo %.Output
rooney@bond-pc:~$ cat input.txt
171212 16082784 6264 XXX xxxxxxxx Transaction XXXXX abend ABCD. The task has terminated abnormally because of a program check. 16:08:27
rooney@bond-pc:~$
rooney@bond-pc:~$ cut -f 1 -d '.' input.txt | xargs -I "%" echo %.
171212 16082784 6264 XXX xxxxxxxx Transaction XXXXX abend ABCD.xargs is used here only to append . at the end of string ABCD.
