Use grep for a long line to get the part of the line

Question

I want to search in a long sentence (more than 1024 letters).

I have one text file (test.txt) which has one long sentence, like this:

afdafglwqgkjrldjl;ewqje;'k;g;je;;;fdsgalsdkf;akslg;safdas.....dasfsd

Now I want to check which line contains the word saf. This command just shows the whole sentence:

less test.txt | grep saf

Is it possible to get a part of the sentence or should I use a command other than grep?

Answer 1

Not exactly what you were looking for: show the matching lines and highlight the occurences in those lines:

grep --color 'saf' test.txt

Options for searching saf and displaying up to 15 characters before and after the occurences found using:

• the standard regex syntax, first mentioned by @kamil-maciorowski in his comment on the question:

  grep -o '.\{0,15\}saf.\{0,15\}' test.txt | grep saf --color
  

• Perl-compatible regex syntax with the -P option, if available:

  grep -o -P '.{0,15}saf.{0,15}' test.txt | grep --color saf
  

• extended regex syntax with the -E option, if your grep has no -P option (like e.g. on macOS):

  grep -o -E '.{0,15}saf.{0,15}' test.txt | grep --color saf