Here are my potential inputs:
brian#muck.co, brian#gmail.com
brian#gmail.com, brian#muck.co
What I want to do is extract the #muck.co email address.
What I have tried is:
\s.*#muck.co
The problem is that this only grabs an email address if it is preceded by a space (so it would only match the second example input above). . . How would I write a Regex expression to match either inputs?
\s matches for a space, so you should wanted to use something like [^\s]*#muck.co - this means any number of not space caracters. [] - for a set of symbols, ^ - for negate effect.
It does not work for me, because \s in my regex flavour seems to not contain regular space, but this works [^[:space:]]\+#muck\.co. Also \+ instead of * for one or more non-space characters instead of any number and escape dot \. which unescaped stands for any single character.
You can use a negated character class to not cross the # and use either a word boundary at the end to prevent a partial word match:
[^\s#]+#muck\.co\b
Regex demo
I want to design an expression for not allowing whitespace at the beginning and at the end of a string, but allowing in the middle of the string.
The regex I've tried is this:
\^[^\s][a-z\sA-Z\s0-9\s-()][^\s$]\
This should work:
^[^\s]+(\s+[^\s]+)*$
If you want to include character restrictions:
^[-a-zA-Z0-9-()]+(\s+[-a-zA-Z0-9-()]+)*$
Explanation:
the starting ^ and ending $ denotes the string.
considering the first regex I gave, [^\s]+ means at least one not whitespace and \s+ means at least one white space. Note also that parentheses () groups together the second and third fragments and * at the end means zero or more of this group.
So, if you take a look, the expression is: begins with at least one non whitespace and ends with any number of groups of at least one whitespace followed by at least one non whitespace.
For example if the input is 'A' then it matches, because it matches with the begins with at least one non whitespace condition. The input 'AA' matches for the same reason. The input 'A A' matches also because the first A matches for the at least one not whitespace condition, then the ' A' matches for the any number of groups of at least one whitespace followed by at least one non whitespace.
' A' does not match because the begins with at least one non whitespace condition is not satisfied. 'A ' does not matches because the ends with any number of groups of at least one whitespace followed by at least one non whitespace condition is not satisfied.
If you want to restrict which characters to accept at the beginning and end, see the second regex. I have allowed a-z, A-Z, 0-9 and () at beginning and end. Only these are allowed.
Regex playground: http://www.regexr.com/
This RegEx will allow neither white-space at the beginning nor at the end of your string/word.
^[^\s].+[^\s]$
Any string that doesn't begin or end with a white-space will be matched.
Explanation:
^ denotes the beginning of the string.
\s denotes white-spaces and so [^\s] denotes NOT white-space. You could alternatively use \S to denote the same.
. denotes any character expect line break.
+ is a quantifier which denote - one or more times. That means, the character which + follows can be repeated on or more times.
You can use this as RegEx cheat sheet.
In cases when you have a specific pattern, say, ^[a-zA-Z0-9\s()-]+$, that you want to adjust so that spaces at the start and end were not allowed, you may use lookaheads anchored at the pattern start:
^(?!\s)(?![\s\S]*\s$)[a-zA-Z0-9\s()-]+$
^^^^^^^^^^^^^^^^^^^^
Here,
(?!\s) - a negative lookahead that fails the match if (since it is after ^) immediately at the start of string there is a whitespace char
(?![\s\S]*\s$) - a negative lookahead that fails the match if, (since it is also executed after ^, the previous pattern is a lookaround that is not a consuming pattern) immediately at the start of string, there are any 0+ chars as many as possible ([\s\S]*, equal to [^]*) followed with a whitespace char at the end of string ($).
In JS, you may use the following equivalent regex declarations:
var regex = /^(?!\s)(?![\s\S]*\s$)[a-zA-Z0-9\s()-]+$/
var regex = /^(?!\s)(?![^]*\s$)[a-zA-Z0-9\s()-]+$/
var regex = new RegExp("^(?!\\s)(?![^]*\\s$)[a-zA-Z0-9\\s()-]+$")
var regex = new RegExp(String.raw`^(?!\s)(?![^]*\s$)[a-zA-Z0-9\s()-]+$`)
If you know there are no linebreaks, [\s\S] and [^] may be replaced with .:
var regex = /^(?!\s)(?!.*\s$)[a-zA-Z0-9\s()-]+$/
See the regex demo.
JS demo:
var strs = ['a b c', ' a b b', 'a b c '];
var regex = /^(?!\s)(?![\s\S]*\s$)[a-zA-Z0-9\s()-]+$/;
for (var i=0; i<strs.length; i++){
console.log('"',strs[i], '"=>', regex.test(strs[i]))
}
if the string must be at least 1 character long, if newlines are allowed in the middle together with any other characters and the first+last character can really be anyhing except whitespace (including ##$!...), then you are looking for:
^\S$|^\S[\s\S]*\S$
explanation and unit tests: https://regex101.com/r/uT8zU0
This worked for me:
^[^\s].+[a-zA-Z]+[a-zA-Z]+$
Hope it helps.
How about:
^\S.+\S$
This will match any string that doesn't begin or end with any kind of space.
^[^\s].+[^\s]$
That's it!!!! it allows any string that contains any caracter (a part from \n) without whitespace at the beginning or end; in case you want \n in the middle there is an option s that you have to replace .+ by [.\n]+
pattern="^[^\s]+[-a-zA-Z\s]+([-a-zA-Z]+)*$"
This will help you accept only characters and wont allow spaces at the start nor whitespaces.
This is the regex for no white space at the begining nor at the end but only one between. Also works without a 3 character limit :
\^([^\s]*[A-Za-z0-9]\s{0,1})[^\s]*$\ - just remove {0,1} and add * in order to have limitless space between.
As a modification of #Aprillion's answer, I prefer:
^\S$|^\S[ \S]*\S$
It will not match a space at the beginning, end, or both.
It matches any number of spaces between a non-whitespace character at the beginning and end of a string.
It also matches only a single non-whitespace character (unlike many of the answers here).
It will not match any newline (\n), \r, \t, \f, nor \v in the string (unlike Aprillion's answer). I realize this isn't explicit to the question, but it's a useful distinction.
Letters and numbers divided only by one space. Also, no spaces allowed at beginning and end.
/^[a-z0-9]+( [a-z0-9]+)*$/gi
I found a reliable way to do this is just to specify what you do want to allow for the first character and check the other characters as normal e.g. in JavaScript:
RegExp("^[a-zA-Z][a-zA-Z- ]*$")
So that expression accepts only a single letter at the start, and then any number of letters, hyphens or spaces thereafter.
use /^[^\s].([A-Za-z]+\s)*[A-Za-z]+$/. this one. it only accept one space between words and no more space at beginning and end
If we do not have to make a specific class of valid character set (Going to accept any language character), and we just going to prevent spaces from Start & End, The must simple can be this pattern:
/^(?! ).*[^ ]$/
Try on HTML Input:
input:invalid {box-shadow:0 0 0 4px red}
/* Note: ^ and $ removed from pattern. Because HTML Input already use the pattern from First to End by itself. */
<input pattern="(?! ).*[^ ]">
Explaination
^ Start of
(?!...) (Negative lookahead) Not equal to ... > for next set
Just Space / \s (Space & Tabs & Next line chars)
(?! ) Do not accept any space in first of next set (.*)
. Any character (Execpt \n\r linebreaks)
* Zero or more (Length of the set)
[^ ] Set/Class of Any character expect space
$ End of
Try it live: https://regexr.com/6e1o4
^[^0-9 ]{1}([a-zA-Z]+\s{1})+[a-zA-Z]+$
-for No more than one whitespaces in between , No spaces in first and last.
^[^0-9 ]{1}([a-zA-Z ])+[a-zA-Z]+$
-for more than one whitespaces in between , No spaces in first and last.
Other answers introduce a limit on the length of the match. This can be avoided using Negative lookaheads and lookbehinds:
^(?!\s)([a-zA-Z0-9\s])*?(?<!\s)$
This starts by checking that the first character is not whitespace ^(?!\s). It then captures the characters you want a-zA-Z0-9\s non greedily (*?), and ends by checking that the character before $ (end of string/line) is not \s.
Check that lookaheads/lookbehinds are supported in your platform/browser.
Here you go,
\b^[^\s][a-zA-Z0-9]*\s+[a-zA-Z0-9]*\b
\b refers to word boundary
\s+ means allowing white-space one or more at the middle.
(^(\s)+|(\s)+$)
This expression will match the first and last spaces of the article..
I'm trying to create a simple Grammar correction tool.
I want to create a regular expression that finds fullstops (" . ") that are not followed by a space so I can replace that with a fullstop and space.
For e.g. This is a sentence.This is another sentence.
Only the first fullstop in the above example should be matched in the expression.
I've tried /\.[^\s]/g but it returns an additional character after the matched fullstop. I would like to match only the fullstop.
How can I do this?
The negated character class [^\s] in the pattern expects a match (any character except a whitespace character), that is why you have the additional character.
If you want to match the dot only, you could use a negative lookahead to assert what is on the right is not a whitspace char or the end of the string:
\.(?!\s|$)
Regex demo
To not match a dot that is not followed by a whitespace char excluding a newline:
\.(?![^\S\r\n])
Regex demo
You can look for all dots using:
(\.)
This will match all dots on below examples:
This is a sentence.This is another sentence.
i am looking. for dots. . ...
You can add a |$ to seek for end of line, and with a little tweak, you get a regex that match all dots not followed by whitespace nor being on the end of a line:
(\.(?!\ |$))
Note that there's a whitespace as literal here. The "must-work-everywhere" example will be like:
(\.(?![[:space:]]|$))
If not, search on the regex reference on the language you use.
Given the following string
span.a.b this.is.really.confusing
I need to return the matches a and b. I've been able to get close with the following regex:
(?<=\.)[\w]+
But it's also matching is, really, and confusing. When I include a negative lookahead I get even closer, but I'm still not there.
(?<=\.)[\w]+(?=\s) # matches b, confusing
How can I match words after a dot until a whitespace occurs?
How can I match words after a dot until a whitespace occurs?
NB: this is language agnostic pseudo-code, but should work.
regex = "^[^\s.]+.(\S+).*"
targets = <extracted_group>.split(".")
Regex explanation:
"^": beings with
"[^\s.]+." 1 or more non-whitespace, non-period characters, followed by a period.
"(\S+)": group and capture all of the following non-whitespace characters
".*": matches 0 or more of any non-newline character
If the split function takes a regex instead of a string, you'll need to escape the '.' or use a character class.
NB: You can do it without the split, but I think that the split is more transparent.
I am not sure if this is good enough for all your possible cases, but it should work with the provided example:
\.([\w]+)\.([\w]+)\s
$1 = a, $2 = b
I have this regex:
(?:\S)\++(?:\S)
Which is supposed to catch all the pluses in a query string like this:
?busca=tenis+nike+categoria:"Tenis+e+Squash"&pagina=4&operador=or
It should have been 4 matches, but there are only 3:
s+n
e+c
s+e
It is missing the last one:
e+S
And it seems to happen because the "e" character has participated in a previous match (s+e), because the "e" character is right in the middle of two pluses (Teni s+e+S quash).
If you test the regex with the following input, it matches the last "+":
?busca=tenis+nike+categoria:"Tenis_e+Squash"&pagina=4&operador=or
(changed "s+e" for "s_e" in order not to cause the "e" character to participate in the match).
Would someone please shed a light on that?
Thanks in advance!
In a consecutive match the search for the next match starts at the position of the end of the previous match. And since the the non-whitespace character after the + is matched too, the search for the next match will start after that non-whitespace character. So a sequence like s+e+S you will only find one match:
s+e+S
\_/
You can fix that by using look-around assertions that don’t match the characters of the assumption like:
\S\++(?=\S)
This will match any non-whitespace character followed by one or more + only if it is followed by another non-whitespace character.
But tince whitespace is not allowed in a URI query, you don’t need the surrounding \S at all as every character is non-whitespace. So the following will already match every sequence of one or more + characters:
\++
You are correct: The fourth match doesn't happen because the surrounding character has already participated in the previous match. The solution is to use lookaround (if your regex implementation supports it - JavaScript doesn't support lookbehind, for example).
Try
(?<!\s)\++(?!\s)
This matches one or more + unless they are surrounded by whitespace. This also works if the plus is at the start or the end of the string.
Explanation:
(?<!\s) # assert that there is no space before the current position
# (but don't make that character a part of the match itself)
\++ # match one or more pluses
(?!\s) # assert that there is no space after the current position
If your regex implementation doesn't support lookbehind, you could also use
\S\++(?!\s)
That way, your match would contain the character before the plus, but not after it, and therefore there will be no overlapping matches (Thanks Gumbo!). This will fail to match a plus at the start of the string, though (because the \S does need to match a character). But this is probably not a problem.
You can use the regex:
(?<=\S)\++(?=\S)
To match only the +'s that are surrounded by non-whitespace.