I am trying to use awk to cut with multi-character delimiter
echo 'f1##f2' |awk -F '##' '{print $2}'
It prints f2. Whereas
echo 'f1||f2' |awk -F '||' '{print $1}'
prints f1||f2.
Can anyone explain what is going on?
1 Answer
The vertical bar char | is treated as special character in your case and should be escaped:
echo "f1||f2" |awk -F'\\|\\|' '{print $1}'
f1
As an alternative you may put | into the character class:
echo "f1||f2||f3" |awk -F'[|]{2}' '{print $3}'
f3
answered Sep 12, 2017 at 15:41
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1@pfnuesel, because with this
-F'\\||'you'll get empty{print $2}(second field) Sep 12, 2017 at 16:02 -
2@ctrl-alt-delor, FYI: linguistically
|IS "vertical bar" char (not OR character), semantically - in some context it's logicalORoperator, in some other context - regex alternation operator Sep 12, 2017 at 16:34 -
1@ctrl-alt-delor Adding to the comment of RomanPerekhrest: Calling
|an OR character is confusing because there is an actual OR character which is different: "∨" (Logical Or, U+2228, in Unicode block Mathematical Operators ). (In some fonts, it may look almost like the alphabet letter V in uppercase. It should be larger and have a larger angle.) Sep 13, 2017 at 1:27 -
@VolkerSiegel I was talking about its meaning in the language not the character name. In a regexp it is apparently called alternation, though I call it OR (not logical OR). Sep 13, 2017 at 8:01
awkallows that at all. BSDawkthrows an error,awk: illegal primary in regular expression || at |, as doesmawk,mawk: line 0: regular expression compile failed (missing operand).(foo|bar|)be like(foo|bar)?rather than returning an error. That's also what perl REs do.?in zsh globs is done that (?meaning something different in globs)