3

I need to validate some commands with the format similar to this

"/a foo.bar /b bar.foo /c 01(.01)"

where the final (.01) is optional (brackets are there to mark the contents as optional). Any digits can be set instead of the 0's and 1's. Also the switches /a, /b, /c are fixed For the moment, I've developed this regular expression:

@"/a\s*([\w\W]*)\s*/b\s*([\w\W]*)\s*/c\s*[0-9,0-9,(.,0-9,0-9){0,1}]

but for some reason, if the command is for example 

"/a foo.bar /b bar.foo /c 01.",

it still validates against the regex. Valid commands should end either with 2digits.2digits or simple 2digits.

Can someone help me to get this fixed?

Cheers,

Alex Barac

Improve this question
2
  • 1
    Isn't it easier to parse the command line (stackoverflow.com/questions/491595/…), letting the library take care of parsing and checking the types? Then you only have to check the result for options a, b and c being set. Commented Nov 15, 2013 at 9:27
  • Good suggestion, but I'm in a sort of time crysis and I need this fixed kind of asap... Commented Nov 15, 2013 at 9:32

2 Answers 2

Reset to default
2

Try this one:

^/a\s*(.*)\s*/b\s*(.*)\s*/c\s*((\d{2}\.\d{2})|(\d{2}))$

Regular expression visualization

Debuggex Demo

Improve this answer
Sign up to request clarification or add additional context in comments.

3 Comments

Thanks, exactly what I wanted! I wish I could up your answer twice for the link to the visual regex tester.
Nice regular expression representation. +1
a*? Zero or more a's (lazy), actually, it's not needed, I removed it.
0

You're misunderstanding character classes. For example, this:

[0-9,0-9,(.,0-9,0-9){0,1}]

and this:

[0-9,.(){}]

mean the same thing. In other words, no matter what you put within the square brackets, that expression matches a single character, unless you quantify the character class. What you want, instead, is this:

@"/a\s*([\w\W]*)\s*/b\s*([\w\W]*)\s*/c\s*[0-9][0-9](\(\.[0-9][0-9])){0,1}"

But let me help you simplify and fix this further:

  1. You can probably use . (a wildcard) rather than [\w\W] (unless you're doing that specifically to match newline characters as well—by default, . doesn't match newlines).
  2. That's why I escaped the . that you actually mean to mean a dot, near the end.
  3. Parentheses are also special characters, which is why I've escaped them, near the end.
  4. \d is shorthand for [0-9].
  5. The outer parentheses are so you can apply the {0,1} quantifier on a group.
  6. ? is shorthand for {0,1}.

Taking all that into account:

@"/a\s*(.*)\s*/b\s*(.*)\s*/c\s*\d\d(\(\.\d\d))?"

Last two things. First, if there should be at least one space between arguments, you should use \s+ instead of \s*. Second, if you want to allow more than two digits, then you can do \d+ instead of \d\d.

Improve this answer

2 Comments

Well, that's the part i'm having problems with. How should I correctly quantify an expression of type .digitdigit to appear once or not at all?
This should help in my future regex development. Thanks!

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.