I can very easily write a regular expression to match a string that contains 2 consecutive repeated characters:
/(\w)\1/
How do I do the complement of that? I want to match strings that don't have 2 consecutive repeated characters. I've tried variations of the following without success:
/(\w)[^\1]/ ;doesn't work as hoped
/(?!(\w)\1)/ ;looks ahead, but some portion of the string will match
/(\w)(?!\1)/ ;again, some portion of the string will match
I don't want any language/platform specific way to take the negation of a regular expression. I want the straightforward way to do this.
The below regex would match the strings which don't have any repeated characters.
^(?!.*(\w)\1).*
(?!.*(\w)\1) negative lookahead which asserts that the string going to be matched won't contain any repeated characters. .*(\w)\1 will match the string which has repeated characters at the middle or at the start or at the end. ^(?!.*(\w)\1) matches all the starting boundaries except the one which has repeated characters. And the following .* matches all the characters exists on that particular line. Note this this matches empty strings also. If you don't want to match empty lines then change .* at the last to .+
Note that ^(?!(\w)\1) checks for the repeated characters only at the start of a string or line.
Lookahead and lookbehind, collectively called "lookaround", are zero-length assertions just like the start and end of line. They do not consume characters in the string, but only assert whether a match is possible or not. Lookaround allows you to create regular expressions that are impossible to create without them, or that would get very longwinded without them.
I have the string
7,456.23%
where I would like to use a regular expression to match BOTH the comma(,) and percent(%) characters and remove them so the result is
7456.23
I can figure out how to match one character or the other, but not both.
Simply use Character Classes or Character Sets
With a "character class", also called "character set", you can tell the regex engine to match only one out of several characters.
Simply place the characters you want to match between square brackets. If you want to match an a or an e, use [ae].
System.out.println("7,456.23%".replaceAll("[,%]",""));
OR try with ORing (Alternation Operator)
System.out.println("7,456.23%".replaceAll(",|%",""));
I want a regular expression that prevents symbols and only allows letters and numbers. The regex below works great, but it doesn't allow for spaces between words.
^[a-zA-Z0-9_]*$
For example, when using this regular expression "HelloWorld" is fine, but "Hello World" does not match.
How can I tweak it to allow spaces?
tl;dr
Just add a space in your character class.
^[a-zA-Z0-9_ ]*$
Now, if you want to be strict...
The above isn't exactly correct. Due to the fact that * means zero or more, it would match all of the following cases that one would not usually mean to match:
An empty string, "".
A string comprised entirely of spaces, " ".
A string that leads and / or trails with spaces, " Hello World ".
A string that contains multiple spaces in between words, "Hello World".
Originally I didn't think such details were worth going into, as OP was asking such a basic question that it seemed strictness wasn't a concern. Now that the question's gained some popularity however, I want to say...
...use #stema's answer.
Which, in my flavor (without using \w) translates to:
^[a-zA-Z0-9_]+( [a-zA-Z0-9_]+)*$
(Please upvote #stema regardless.)
Some things to note about this (and #stema's) answer:
If you want to allow multiple spaces between words (say, if you'd like to allow accidental double-spaces, or if you're working with copy-pasted text from a PDF), then add a + after the space:
^\w+( +\w+)*$
If you want to allow tabs and newlines (whitespace characters), then replace the space with a \s+:
^\w+(\s+\w+)*$
Here I suggest the + by default because, for example, Windows linebreaks consist of two whitespace characters in sequence, \r\n, so you'll need the + to catch both.
Still not working?
Check what dialect of regular expressions you're using.* In languages like Java you'll have to escape your backslashes, i.e. \\w and \\s. In older or more basic languages and utilities, like sed, \w and \s aren't defined, so write them out with character classes, e.g. [a-zA-Z0-9_] and [\f\n\p\r\t], respectively.
* I know this question is tagged vb.net, but based on 25,000+ views, I'm guessing it's not only those folks who are coming across this question. Currently it's the first hit on google for the search phrase, regular expression space word.
One possibility would be to just add the space into you character class, like acheong87 suggested, this depends on how strict you are on your pattern, because this would also allow a string starting with 5 spaces, or strings consisting only of spaces.
The other possibility is to define a pattern:
I will use \w this is in most regex flavours the same than [a-zA-Z0-9_] (in some it is Unicode based)
^\w+( \w+)*$
This will allow a series of at least one word and the words are divided by spaces.
^ Match the start of the string
\w+ Match a series of at least one word character
( \w+)* is a group that is repeated 0 or more times. In the group it expects a space followed by a series of at least one word character
$ matches the end of the string
This one worked for me
([\w ]+)
Try with:
^(\w+ ?)*$
Explanation:
\w - alias for [a-zA-Z_0-9]
"whitespace"? - allow whitespace after word, set is as optional
I assume you don't want leading/trailing space. This means you have to split the regex into "first character", "stuff in the middle" and "last character":
^[a-zA-Z0-9_][a-zA-Z0-9_ ]*[a-zA-Z0-9_]$
or if you use a perl-like syntax:
^\w[\w ]*\w$
Also: If you intentionally worded your regex that it also allows empty Strings, you have to make the entire thing optional:
^(\w[\w ]*\w)?$
If you want to only allow single space chars, it looks a bit different:
^((\w+ )*\w+)?$
This matches 0..n words followed by a single space, plus one word without space. And makes the entire thing optional to allow empty strings.
This regular expression
^\w+(\s\w+)*$
will only allow a single space between words and no leading or trailing spaces.
Below is the explanation of the regular expression:
^ Assert position at start of the string
\w+ Match any word character [a-zA-Z0-9_]
Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
1st Capturing group (\s\w+)*
Quantifier: * Between zero and unlimited times, as many times as possible, giving back as needed [greedy]
\s Match any white space character [\r\n\t\f ]
\w+ Match any word character [a-zA-Z0-9_]
Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
$ Assert position at end of the string
Just add a space to end of your regex pattern as follows:
[a-zA-Z0-9_ ]
This does not allow space in the beginning. But allowes spaces in between words. Also allows for special characters between words. A good regex for FirstName and LastName fields.
\w+.*$
For alphabets only:
^([a-zA-Z])+(\s)+[a-zA-Z]+$
For alphanumeric value and _:
^(\w)+(\s)+\w+$
If you are using JavaScript then you can use this regex:
/^[a-z0-9_.-\s]+$/i
For example:
/^[a-z0-9_.-\s]+$/i.test("") //false
/^[a-z0-9_.-\s]+$/i.test("helloworld") //true
/^[a-z0-9_.-\s]+$/i.test("hello world") //true
/^[a-z0-9_.-\s]+$/i.test("none alpha: ɹqɯ") //false
The only drawback with this regex is a string comprised entirely of spaces. " " will also show as true.
It was my regex: #"^(?=.{3,15}$)(?:(?:\p{L}|\p{N})[._()\[\]-]?)*$"
I just added ([\w ]+) at the end of my regex before *
#"^(?=.{3,15}$)(?:(?:\p{L}|\p{N})[._()\[\]-]?)([\w ]+)*$"
Now string is allowed to have spaces.
This regex allow only alphabet and spaces:
^[a-zA-Z ]*$
Try with this one:
result = re.search(r"\w+( )\w+", text)
How can i match a string that does not finish with / . I know I can do that /\/$/ and it will match if string does finish with /, but how can I test to see if it doesn't?
You can use a negative character class:
/[^\/]$/
This however requires that the string contains at least one character. If you also want to allow the empty string you can use an alternation:
/[^\/]$|^$/
A different approach is to use a negative lookbehind but note that many popular regular expression engines do not support lookbehinds:
/(?<!\/)$/
You can say "not character" by doing [^...]. In this case, you can say "not backslash by doing": /[^\/]$/
[^\/]$
^ will negate any character class expression.
I have a situation where using VBScript I need to check for the presence of multiple spaces.
I want to check for the presence of 2 or more consecutive spaces, so \s+ doesnt work for my needs.
Does anyone know how I can accomplish this using VBScript regular expressions.
Use brackets to specify how many repetitions to match. This matches two or more whitespace characters:
\s{2,}
If you want to match only space characters, just use a space instead of \s, or the character code:
\x20{2,}
This ought to do the trick:
\s{2,}