Find a string enclosed in spaces or coming from the beginning or end with grep

Question

How can I match some string enclosed in spaces or coming from the beginning or end?

I need match for -someword in the following sentences: word1 -someword word2, -someword word1, word1 -someword, -someword. And don't need match in the sentences: s-someword, -somewordd

I tried grep the above with regex grep -r [^ ]-someword[$ ] (i.e before -someword must be a space or -someword must start the sentence, after -someword must be a space or -someword must end the sentence), but it finds nothing.

Answer 1

Try:

grep -w -e -someword

From man grep:

-w, --word-regexp

          Select only those lines containing matches that form whole
          words.  The test is that the matching substring must
          either be at the beginning of the line, or preceded by a
          non-word constituent character.  Similarly, it must be
          either at the end of the line or followed by a non-word
          constituent character.  Word-constituent characters are
          letters, digits, and the underscore.  This option has no
          effect if -x is also

Please note that it would also match -someword if it's surrounded by some other non alphanumeric character except for white spaces, such as # or ,. If you want to ensure it's surrounded only by spaces or beginning/end of line, you can use:

egrep '(^|[[:space:]])-someword([[:space:]]|$)'

# Which is equivalent to:

grep -E '(^|[[:space:]])-someword([[:space:]]|$)'

# Or without extended regex:

grep '\(^\|[[:space:]]\)-someword\([[:space:]]\|$\)'

Answer 2

You may use the following to avoid complicated regular expressions. It matches any line that has -someword at the start with a space after it, -someword somewhere in the middle with spaces on either side, or -someword at the end with a space before it:

grep -e '^-someword ' -e ' -someword ' -e ' -someword$'

This all assumes that -someword is a string that does not contain any characters that are special in regular expressions. If it does, the string must be rewritten to match these literal characters, e.g., by escaping them in the pattern.

If you also want to match lines containing nothing but the string -someword, add -e '^-someword$'.

If you by "space" mean "a blank character", which would include both tabs and spaces, then change the literal spaces in the pattern with [[:blank:]]. Would you need to match a wider variety of space-like characters, such as vertical tabs and carriage returns, use [[:space:]] instead.

Answer 3

The Regex pattern for this is (https://regexr.com/7b8g0):

(\s|^)-someword(\s|$)

A lot of shells freak out about these ()\| so you'll have to fiddle with quotes and escapes a bit. For example fish seems to like \\s but zsh likes \s.

With ripgrep and fish, it's pretty easy to implement:

$ bat word.txt --style=numbers
   1 word1 -someword word2
   2 -someword word1
   3 word1 -someword
   4 -someword
   5 s-someword
   6 -somewordd
   7 \s-someword

$ bat word.txt | rg '(\\s|^)-someword(\\s|$)' --only-matching --line-number
1: -someword
2:-someword
3: -someword
4:-someword

$ bat word.txt | rg '(\\s|^)-someword(\\s|$)' --line-number -v
5:s-someword
6:-somewordd
7:\s-someword

(I added \s-someword and -v to address some concerns in comments.)

Note that the spaces become part of the match. To fix that in the regex, you'd have to add a capturing group around -someword (easy) and then then tell rg to return the first group (pffft...).

Because grep is ancient it won't process this "advanced" regex syntax by default. You need to add -E.

$ bat word.txt | grep -E '(\\s|^)(-someword)(\\s|$)' -n
1:word1 -someword word2
2:-someword word1
3:word1 -someword
4:-someword

$ bat word.txt | grep -v -E '(\\s|^)(-someword)(\\s|$)' -n
5:s-someword
6:-somewordd
7:\s-someword

Frankly, if you live in a year starting with 2, you should probably alias grep to grep -E anyway. Or just use rg.