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up vote 0 down vote favorite

This looks so simple but it isn't:

[[ "1234+5678" =~ [0-9]+(\s*(\-|\*)\s*[0-9]+)* ]] && echo $?

returns a 0. However, it actually should not do that, as only minus (-) and multiplication (*) operator are allowed. Also, I grabbed some regex tool in the net and tried to match this pattern: result was the null string. (as expected)

In prose, this extended regex reads:

  • look for number (mandatory)
  • check if there is some white space
  • operator must be either a - or a *
  • check if there is some white space (again)
  • look for another number (must be there if preceded by an operator)

Also, the asterisk following the expression in parentheses says that 2nd to nth operator-number pair is optional.

Where is my mistake in thinking here?

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3 Answers 3

up vote 3 down vote

Without markers the regexp (right part) can match any part of string. So your variant match 1234. To satisfy requirements you have to use markers:

[[ "1234+5678" =~ ^[0-9]+(\s*(\-|\*)\s*[0-9]+)*$ ]] ; echo $?

And shorter (if you'd like):

[[ "1234+5678" =~ ^[0-9\ *-]+*$ ]] ; echo $?
share|improve this answer
    
SO simple? Just putting the ^ at the beginning did the trick. Thank you very much! (also for not inverting the logic for no reason ;)) Again: single numbers must work as well as numbers with addends. So it is not that simple as replacing * by +. –  syntaxerror Jun 23 at 21:09
    
the shorter expression also matches 123-*5* –  ikrabbe Jun 23 at 21:12
    
Note that \s* stands for various whitespace like e. g. tabulators as well. Your shorter solution blindly assumes that whitespace = space, which is a little naughty ;) Plus, whatever you write into [ ] is arbitrary in order, whilst my expression dictates a definite order - for a reason! Your expression would even allow a * or - as first characters, which is plainly wrong, as they are arithmetic operators and must be preceded by a number. Shorter is not always better, as it may unwantedly alter the logic. –  syntaxerror Jun 23 at 21:15
    
@ikrabbe Yes, but the choice to use (or do not use) some variant or its modification depends of operated data and users forecast. –  Costas Jun 23 at 21:21
    
@syntaxerror I have stated: if you'd like. Certainly, if you expect data with a * or - as first characters and so on you are free to choice 1 variant but in most cases shorter is enough –  Costas Jun 23 at 21:25
up vote 1 down vote

Your expression returns 0 because it matches the first number. Anchor the regexp to make it do what you want:

[[ "1234+5678" =~ ^[0-9]+( *(-|\*) *[0-9]+)*$ ]] && echo $?

On a side note: (1) you don't need to quote -, and (2) \s is not recognized as an ERE.

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Thanks, also for the info about \s. Well, I wanted to support both tabulators (et al.) and "plain" spaces, hence I used \s on purpose. However, I wasn't aware that this is not part of ERE. Most peculiar... –  syntaxerror Jun 23 at 21:24
1  
It isn't peculiar: \s and friends come from Perl's lineage of regexps, not the much older BRE / ERE. –  lcd047 Jun 24 at 3:25
up vote 0 down vote 
[[ "1234+5678" =~ [0-9]+(\s*(\-|\*)\s*[0-9]+) ]] ; echo $?

The * regex operator at the end of your regex allows any number of repeats of the expression inside the () also none. So you match [0-9]+ from the start and the rest is optional.

If you remove the * or replace it by a + the expression returns as expected.

Also, the asterisk following the expression in parentheses says that 2nd to nth operator-number pair is optional.

That is the mistake: The * means the any operator is option. Use the + operator!

Ok, that was not the answer, but you can consume the expressions and see if the result is empty:

echo "1234*5678"|sed 's/[0-9]\+\([ \*\-][0-9]\+\)*//'
share|improve this answer
    
Thanks, but it's meant to be that way. :) It should match on "1234" as well as "2345*6789*4321", or in prose, on single numbers as well as on "formulae" (i. e. additions in this case). However, the problem is that it matches also on additions + and divisions /, even though I'm only allowing subtractions and multiplications in this sample one-liner. Your answer will also exclude standalone numbers, as now you do no longer allow that the addend be omitted. –  syntaxerror Jun 23 at 20:52
    
Just add an end anchor ($) and you are set: [[ "1234+5678" =~ [0-9]+(\s*(\-|\*)\s*[0-9]+)*$ ]] || echo $? –  user1794469 Jun 23 at 20:52
    
@ user1794469 No. Doesn't help me at all. By replacing && by ||, you've just inverted the logic. Coolish. But that's totally pointless, as I need the result of the expression itself, not the result if the expression is false. –  syntaxerror Jun 23 at 20:54
    
I thought so, but it doesn't work, I inverted the logic just to get some output from a expression that is true. –  ikrabbe Jun 23 at 20:56
    
@ikrabbe It definitely doesn't. :) So I now get it, you were the initiator of inverting the logic, and user179... just jumped on the bandwagon. LOL. –  syntaxerror Jun 23 at 20:57

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