Cut command with word as delimiter

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. 16:08:27

I want the output as :

171212 16082784       6264 XXX     xxxxxxxx Transaction XXXXX abend ABCD.

How to approach?

Solution 1:

I suggest:

awk -F 'ABCD' '{print $1 FS "."}' file

Output:

171212 16082784       6264 XXX     xxxxxxxx Transaction XXXXX abend ABCD.

FS contains your delimiter "ABCD". "ABCD" is a regex.

Solution 2:

sed version:

sed 's/ABCD.*/ABCD./' input.txt
171212 16082784       6264 XXX     xxxxxxxx Transaction XXXXX abend ABCD.

Source: https://unix.stackexchange.com/questions/257514/how-to-delete-everything-after-a-certain-pattern-or-a-string-in-a-file

Solution 3:

awk 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
two

In your particular case, that would be:

$ awk -F 'ABCD' '{print $1,FS}' input.txt                                                                                
171212 16082784       6264 XXX     xxxxxxxx Transaction XXXXX abend  ABCD

Judging 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.