CODEWITHSUNDEEP
javajavascriptregex
aditya86c
aditya86c

Reputation: 57

Regex match for alphanumeric chars and one * which should be a total length of 3

How can I write a regex for alphanumeric chars allowing one or two stars and restricting the total string length to 3.

Ex : the below strings length is 3

*12  or *2* 0r *a* or *B* or **2

So, the * symbol can occur at last or middle or at the first of *12. Similarly, if you take the last example **2 you see more than one * symbol and that can occur in any order of that string.

Upvotes: 0

Views: 495

Answers (5)

user557597
user557597

Reputation:

You could always use a lookahead assertion in javascript. Its a little tricky but it's better suited to fine-tune any specific permutations.

/^(?=(?:[^*]*\*){1,2}[^*]*$)[a-zA-Z0-9*]{3}$/

Expanded: 

^                        # beginning of line
  (?=                         # start lookahead
      (?:                       # non-capture group
          [^*]*                    # optional not '*' characters
          \*                       # '*' character
      ){1,2}                    # end group, do 1 or 2 times
      [^*]*                     # optional not '*' characters
      $                         # end of line
  )                          # end lookahead
  [a-zA-Z0-9*]{3}            # back at begining of line. at this point there will
                             # be only 1 or 2 '*' characters in the line.
                             # match exactly 3 of alphanumeric characters or '*'
$                        # end of line

Substitute any requirements you need.
Below is a perl test case, javascript is not my strong point. 

@samps = qw(
 *12  1*2  12*  **1  *1*  1**  ***
 a*12 a1*2 a12* **a1 *a1* a1** ****
 *2   *2   2*   *1   1*   **   
);

for $teststr (@samps) {
   if ($teststr =~ /^(?=(?:[^*]*\*){1,2}[^*]*$)[a-zA-Z0-9*]{3}$/) {
      print "$teststr passed\n";
   }
   else {
      print "$teststr failed\n";
   }
}

Output:

*12 passed
1*2 passed
12* passed
**1 passed
*1* passed
1** passed
*** failed
a*12 failed
a1*2 failed
a12* failed
**a1 failed
*a1* failed
a1** failed
**** failed
*2 failed
*2 failed
2* failed
*1 failed
1* failed
** failed

Edit For @bozdoz

I didn't realize a string might be scraped for multiple instances of this. If so, the regex can be generalized to be used with/without delimeters.
The important thing is that this scales up very well if the requirements change to, for example 8 total characters and only 2-4 asterisks.

Examples:

no delimeters other than begin/end of string: 

  /
    ^
     (?= [a-z0-9*]{3} $ )
     (?:[a-z0-9]*\*){1,2} [a-z0-9]*
    $
  /xi

delimeter is \s, the context is single-line and global. Data is captured in group 1

 /
   (?:^|\s)
    (?= [a-z0-9*]{3} (?:$|\s) )
    ( (?:[a-z0-9]*\*){1,2} [a-z0-9]* )
   (?=$|\s)
 /xig

delimeter is [^a-z0-9*], the context is single-line and global. Data is captured in group 1

 /
   (?:^|[^a-z0-9*])
    (?= [a-z0-9*]{3} (?:$|[^a-z0-9*]) )
    ( (?:[a-z0-9]*\*){1,2} [a-z0-9]* )
   (?=$|[^a-z0-9*])
 /xig

Upvotes: 2

Simon
Simon

Reputation: 32893

What about this:

[a-zA-Z0-9*]{2}[^*]|[a-zA-Z0-9*][^*][a-zA-Z0-9*]|[^*][a-zA-Z0-9*]{2}

Upvotes: 0

MetaEd
MetaEd

Reputation: 3881

There are three cases:

\w[*\w]{2}  # case 1, string begins with word character, last 2 can be stars
\*\w[*\w]   # case 2, string begins with 1 star, last can be a star
\*{2}\w     # case 3, string begins with 2 stars, last cannot be a star

Taken together, and adding the necessary start and end of string assertions, we get:

^(\w[*\w]{2}|\*\w[*\w]|\*{2}\w)$

But this solution is not quite correct because the \w character class allows not only alphanumerics but also the _ character. Therefore, we substitute a bracketed character class [a-zA-Z0-9] for \w and get:

^([a-zA-Z0-9][*a-zA-Z0-9]{2}|\*[a-zA-Z0-9][*a-zA-Z0-9]|\*{2}[a-zA-Z0-9])$

Upvotes: 0

bozdoz
bozdoz

Reputation: 12860

This regex works with a lookbehind. I have tested it with PHP in codepad here.

(?<![\w*])(\w(?!\w\w)|\*(?!\*\*)){3}(?![\w*])

It basically looks for a three character word that doesn't have three word characters or three star characters. (?<![\w*]) removes words that follow a word character or a * and (?![\w*]) removes words that precede them (therefore returning ONLY three character word-segments).

Javascript doesn't exactly have lookbehinds, so I tried to adapt on a technique used here. I then came up with the following regex, tested in jsfiddle here.

/(?![\w*])(.?)(\w(?!\w\w)|\*(?!\*\*)){3}(?![\w*])/g

Hope this helps!!!!!!!! <- regex's drive me a little crazy

Upvotes: 1

nnnnnn
nnnnnn

Reputation: 150040

EDIT: for your updated question without the commas and optional spaces:

/^(\*[A-Z0-9]{2}|\*[A-Z0-9]\*|\*\*[A-Z0-9])$/i

Your examples don't include the alphanumeric character first, e.g., A**, but if you want that I'm sure you can figure it out from what I've already given you.

(see below for comment on mixed case)

My original answer:

/^(\*[A-Z0-9]{2}|\*[A-Z0-9]\*|\*\*[A-Z0-9])(, *(\*[A-Z0-9]{2}|\*[A-Z0-9]\*|\*\*[A-Z0-9]))*$/i

That is the JavaScript syntax with the "i" option to make it case-insensitive. I can't be bothered looking up the Java equivalent for case-insensitive matching, but if necessary you could always change each [A-Z0-9] part to [A-Za-z0-9].

Also you can use \w instead of [A-Za-z0-9] if you extend your definition of "alphanumeric" to include underscores.

Upvotes: 0

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