I cannot find a way to represent the inverse of a character class in sed. If this were a perl-like environment I would use [^\s]. However in sed this appears to match non-s, not non-whitespace.
On a line of text (from gdrive) I need to capture the first non-whitespace token and ignore everything after (and including) the first whitespace on the line.
Here's a fake but representative example of the input I am trying to parse:
19845fake-FaKeE-xbk534sWsbBQ mydir dir 2019-01-01 19:10:44
My original attempt at doing this was the line:
sed -rn 's/^([^\s]*).*$/\1/p'
At first it seemed to work until I noticed that this is cutting off at the first 's', rather than the first whitespace.
I have since attempted various permutations like:
#matches up to the first 's'
sed -rn 's/([^\\s]*).*$/\1/p'
#matches only the first character
sed -rn 's/^([^[[:blank:]]]*).*$/\1/p'
#matches nothing at all
sed -rn 's/^([[^:blank:]]*).*$/\1/p'
sed -rn 's/^\s*([^\s]*).*$/\1/p'
Expected: 19845fake-FaKeE-xbk534sWsbBQ
Actual: 19845fake-FaKeE-xbk534
The character class is [:blank:] so to match the opposite of it, you just need [^[:blank:]]. This should work:
sed -rn 's/^([^[:blank:]]*).*$/\1/p'