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I am complete novice at regex and Javascript. I have the following problem: need to check into a textfield the existence of one (1) or many (n) consecutive * (asterisk) character/characters eg. * or ** or *** or infinite (n) *. Strings allowed eg. *tomato or tomato* or **tomato or tomato** or as many(n)*tomato many(n)*. So, far I had tried the following:

var str = 'a string'
var value = encodeURIComponent(str);
var reg = /([^\s]\*)|(\*[^\s])/;

if (reg.test(value) == true ) {
    alert ('Watch out your asterisks!!!')
}
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Your question isn't really clear... Are you after prefix/suffix asterisks so that the number at beginning and ending match? So the number can be any given number as long as there are the same number of asterisks at the beginning and at the end? –  Robert Koritnik Jul 30 '12 at 11:27
 
Could you define a little more what's the expected behavior? If I have a * anywhere in the word, should it fire the alert? Or only if it's at start and/or end of the string? –  Fabrício Matté Jul 30 '12 at 11:27
 
What are you trying to do with the matches once you have them? And are they compulsory? Is 'the string' the whole of the textarea, or text-input, value? Or should this be checked word-by-word? –  David Thomas Jul 30 '12 at 11:28
 
From your conditions, it looks like you are matching any string which has an asterisk in it. You could simplify the regex in that case. –  Foxish Jul 30 '12 at 11:33
 
@DarkXphenomenon: His regular expression is definitely not what he needs, hence the question on SO. –  Robert Koritnik Jul 30 '12 at 11:45
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2 Answers

up vote 2 down vote accepted

By your question it's hard to decipher what you're after... But let me try:

Only allow asterisks at beginning or at end

If you only allow an arbitrary number (at least one) of asterisks either at the beginning or at the end (but not on both sides) like:

  • *****tomato
  • tomato******
  • but not **tomato*****

Then use this regular expression:

reg = /^(?:\*+[^*]+|[^*]+\*+)$/;

Match front and back number of asterisks

If you require that the number of asterisks at the biginning matches number of asterisks at the end like

  • *****tomato*****
  • *tomato*
  • but not **tomato*****

then use this regular expression:

reg = /^(\*+)[^*]+\1$/;

Results?

It's unclear from your question what the results should be when each of these regular expressions match? Are strings that test positive to above regular expressions fine or wrong is on you and your requirements. As long as you have correct regular expressions you're good to go and provide the functionality you require.

I've also written my regular expressions to just exclude asterisks within the string. If you also need to reject spaces or anything else simply adjust the [^...] parts of above expressions.

Note: both regular expressions are untested but should get you started to build the one you actually need and require in your code.

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up vote 1 down vote

If I understand correctly you're looking for a pattern like this:

var pattern = /\**[^\s*]+\**/;

this won't match strings like ***** or ** ***, but will match ***d*** *d or all of your examples that you say are valid (***tomatos etc).
If I misunderstood, let me know and I'll see what I can do to help. PS: we all started out as newbies at some point, nothing to be ashamed of, let alone apologize for :)

After the edit to your question I gather the use of an asterisk is required, either at the beginning or end of the input, but the string must also contain at least 1 other character, so I propose the following solution:

var pattern = /^\*+[^\s*]+|[^\s*]+\*+$/;
'****'.match(pattern);//false
' ***tomato**'.match(pattern);//true

If, however *tomato* is not allowed, you'll have to change the regex to:

var pattern = /^\*+[^\s*]+$|^[^\s*]+\*+$/;

Here's a handy site to help you find your way in the magical world of regular expressions.

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Many many thanks!!! –  anarchos78 Jul 30 '12 at 12:02
 
Isn't it that you don't have to escape characters within a set? so your set could be replaced by [^ *]? Just a suggestion. Not saying that anything's wrong with your regex whatsoever. –  Robert Koritnik Jul 30 '12 at 13:38
 
@RobertKoritnik you're absolutely right, I'll edit my regex's. Thanks –  Elias Van Ootegem Jul 30 '12 at 14:00
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