Regex validation for custom command format

Question

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

Isn't it easier to parse the command line (http://stackoverflow.com/questions/491595/best-way-to-parse-command-line-arguments-in-c), 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. — CompuChip, Nov 15 '13 at 09:27
Good suggestion, but I'm in a sort of time crysis and I need this fixed kind of asap... — Alex Barac, Nov 15 '13 at 09:32

Alex Filipovici · Accepted Answer · 2013-11-15T10:05:08.567

2

Try this one:

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

Regular expression visualization

Debuggex Demo

edited Nov 15 '13 at 10:05

answered Nov 15 '13 at 09:51

Alex Filipovici

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Thanks, exactly what I wanted! I wish I could up your answer twice for the link to the visual regex tester. – Alex Barac Nov 15 '13 at 09:54
1

Nice regular expression representation. +1 – Saeed Neamati Nov 15 '13 at 10:05
`a*? Zero or more a's (lazy)`, actually, it's not needed, I removed it. – Alex Filipovici Nov 15 '13 at 10:05

Andrew Cheong · Answer 2 · 2013-11-15T09:52:39.280

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:

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).
That's why I escaped the . that you actually mean to mean a dot, near the end.
Parentheses are also special characters, which is why I've escaped them, near the end.
\d is shorthand for [0-9].
The outer parentheses are so you can apply the {0,1} quantifier on a group.
? 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.

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? — Alex Barac, Nov 15 '13 at 09:50

Regex validation for custom command format

2 Answers2