How to grep for tabs without using literal tabs and why does \t not work?

When I search for tabs in a file with (e)grep I use the litteral tab (^v + <tab>). I can not utilize \t as a replacement for tabs in regular expressions. With e.g. sed this expression works very well.

So is there any possibility to use a non-litteral replacement for <tab> and what are the backgrounds for a non working / not interpreted \t ?

1

4 Answers

grep is using regular expressions as defined by POSIX. For whatever reasons POSIX have not defined \t as tab.

You have several alternatives:

• tell grep to use the regular expressions as defined by perl (perl has \t as tab):

  grep -P "\t" foo.txt

  the man page warns that this is an "experimental" feature. at least \t seems to work fine. but more advanced perl regex features may not.

• use printf to print a tab character for you:

  grep "$(printf '\t')" foo.txt

• use the literal tab character:

  grep "^V<tab>" foo.txt

  that is: type grep ", then press ctrl+v, then press tab, then type " foo.txt. pressing ctrl+v in the terminal causes the next key to be taken verbatim. that means the terminal will insert a tab character instead of triggering some function bound to the tab key.

• use the ansi c quoting feature of bash:

  grep $'\t' foo.txt

  this does not work in all shells.

• use awk:

  awk '/\t/'

• use sed:

  sed -n '/\t/p'

See the wikipedia article about regular expressions for an overview of the defined character classes in POSIX and other systems.

3

It is not exactly the answer you would want to hear, but a possible use of escape sequences is provided by bash

command | grep $'\t'

(do not put it into double quotes!).

6

awk '/\t/' is my favorite workaround:

printf 'a\t\nb' | awk '/\t/'

Output: a\t.

One can always resort to using ascii hex-code for tab:

$ echo "one"$'\t'"two" > input.txt
$ grep -P "\x9" input.txt
one two
$ grep $'\x9' input.txt
one two

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