How to grep without leading whitespaces

command linegrepsedtext processing

I'm greping through a large codebase, and leading whitespaces and tabulation seem to be quite annoying. Is there any way to get rid of it?

grep -R "something" ./

E.g, instead of:

foo/bar.cpp:                       qwertyuiosomethingoi
foo/bar/baz.h:                          43rfsgsomethingdrfg
bar/bar.cpp:            1234edwssomethingczd

I want to get something like:

foo/bar.cpp: qwertyuiosomethingoi
foo/bar/baz.h: 43rfsgdsomethingrfg
bar/bar.cpp: 1234edwssomethingczd

Or better:

foo/bar.cpp:   qwertyuisomethingooi
foo/bar/baz.h: 43rfsgdrsomethingfg
bar/bar.cpp:   1234edwssomethingczd

Best Answer

Create test files

echo -e "\t   foo-somethingfoo" >something.foo
echo "    bar-bar-somethingbar" >something.bar_bar
echo "baz-baz-baz-somethingbaz" >something.baz_baz_baz
echo "  spaces    something  s" >something.spaces

produce full glorious colour :)

grep --colour=always "something" something.* | 
 sed -re  's/^([^:]+):(\x1b\[m\x1b\[K)[[:space:]]*(.*)/\1\x01\2\3/' |
   column -s $'\x01' -t

output (run it to get the colour).

something.bar_bar      bar-bar-somethingbar
something.baz_baz_baz  baz-baz-baz-somethingbaz
something.foo          foo-somethingfoo
something.spaces       spaces    something  s

Tested in gnome-terminal, konsole, terminator, xterm

Related Solutions

Bash – counting multiple patterns in a single pass with grep

IFS=$'\n'
gzip -dc file.gz | grep -v '^>' | grep -Foe "${tri[*]}" | sort | uniq -c

But by the way, AAAC matches both AAA and AAC, but grep -o will output only one of them. Is that what you want? Also, how many occurrences of AAA in AAAAAA? 2 or 4 ([AAA]AAA, A[AAA]AA, AA[AAA]A, AAA[AAA])?

Maybe you want instead:

gzip -dc file.gz | grep -v '^>' | fold -w3 | grep -Fxe "${tri[*]}" | sort | uniq -c

That is split the lines in groups of 3 characters and count the occurrences as full lines (would find 0 occurrence of AAA in ACAAATTCG (as that's ACA AAT TCG)).

Or on the other hand:

gzip -dc file.gz | awk '
  BEGIN{n=ARGC;ARGC=0}
  !/^>/ {l = length - 2; for (i = 1; i <= l; i++) a[substr($0,i,3)]++}
  END{for (i=1;i<n;i++) printf "%s: %d\n", ARGV[i], a[ARGV[i]]}' "${tri[@]}"

(would find 4 occurrences of AAA in AAAAAA).

Shell – Empty a file without grep subsequently treating it like a binary file

If the only problem is that grep treats it as binary, tell grep to search it regardless:

$ head /bin/bash > out
$ echo "test" >> out 
$ grep test out 
Binary file out matches
$ grep -a test out 
test

From man grep:

   -a, --text
          Process  a binary file as if it were text; this is equivalent to
          the --binary-files=text option.

Best Answer

Related Solutions

Bash – counting multiple patterns in a single pass with grep

Shell – Empty a file without grep subsequently treating it like a binary file

Related Question