0

The sed command can use a full stop to find a wild card symbol, and so .* retrieves everything on the line.

Is it possible to say that I wish any symbol at all except (in this case) spaces? So:

FredSmith - would be valid

Fred&Ginger - would be valid

6times7 - would be valid

One too many - would not be valid

4

Yes, it's possible.

echo "One too many" | sed -n '/^[^[:space:]]*$/p'

The above command returns nothing


echo "FredSmith" | sed -n '/^[^[:space:]]*$/p'
FredSmith

While the latter prints FredSmith


  • [:space:] - POSIX character class for space characters

  • /^[^[:space:]]*$/p - means "print match if it contains only non-space characters [^[:space:]]" (used negated character class)

  • Ideal - thank you! Out if curiosity can you check against other characters, so a string that has no 'Z's in it for example? – HugMyster Jun 8 '17 at 11:11
  • @HugMyster, welcome. Yes, to match the string containing any character(including space) except Z character - use pattern /^[^Z]*$/ – RomanPerekhrest Jun 8 '17 at 11:14
1

Use negated character classes ([^ ] or [^[:space:]]) as Roman showed, or invert the match as below. Which one is better depends on the surrounding structure.

$ cat foo.txt
with space
spaceless
$ sed '/ /d'  < foo.txt     # (d)elete the line on match
spaceless
$ sed -n '/ /!p' < foo.txt  # (p)rint if the pattern does not (!) match
spaceless

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