How do I test to see the length of a string using regex?
For example, how do i match a string if it contains only 1 character?
4 Answers
^.$
But most frameworks include methods that will return string length, which you should use instead of regex for this.
answered May 20, 2011 at 22:00
Yuriy Faktorovich
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4 Comments
CanSpice
The
{1} is really extraneous and just adds line-noise.
Seth Robertson
$ also can match newline, so the re would match "x" and "x\n"
Phrogz
@Seth In what regex flavor does
$ match after a newline? The OP didn't mention an application. In JavaScript, for example, /foo$/.test("foo\n") returns false.Seth Robertson
@Phrogz: No, the point is it matches before a newline and end of line. As for what regex flavors, System V, POSIX, Perl, Ruby, PHP, Python are the ones I know of. I'm not actually aware of any that do not work that way. Well, I wasn't until I saw your javascript example.
Anchor to the start and end of string and match one character. In many languages:
^.{1}$
In Ruby's Regex:
\A.{1}\z
2 Comments
Seth Robertson
The {1} doesn't add anything. The latter is correct while the former will match "x" and "x\n". You also need to ensure that . can match \n, which it often will not without the /s modifier (/m in Ruby)
Phrogz
@Seth The
{1} doesn't add anything in the "for example" case mentioned by the OP, but the OP asked for how to test the length in general. I've added the {1} (as have others) to make it clear how to match a different number of characters. Further, whether or not $ matches after a "\n" depends on the regex flavor. As the OP has not yet specified the flavor, your statement is not absolutely correct.Matching a single character would be (using Perl regex):
/\A.\z/s
\A means "start of the string", . means "any character", and \z means "end of the string". Without the \A and \z you'll match any string that's longer than one character.
Edit: But really you should be doing something like:
if( length($string) == 1 ) {
...
}
(using Perl as an example)
Edit2: Previously I had /^.$/ but, as Seth pointed out, this allows matches on strings that are two characters long where the last character is \n. The \A...\z construct fixes that.
2 Comments
Seth Robertson
The listed regex will match "x" and "x\n"
Seth Robertson
Except . typically means "any character but \n" unless you use the /s modifier (/m in Ruby), so "\n" would not match.
/^.\z/s
The above requires perl compatible regexps. The trick is that /^.$/ might match "x" and "x\n". Adding /s modifier doesn't help there.
length()orstrlen()?/^.$/, but I can't believe whatever language you're using doesn't have a better way of doing this.