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 have this pattern.. The matches must not cross multiple lines (there must not be any newline char in the matches) so I added the m modifier..
But sometimes there is a \n in the matches.. How to prevent this?
preg_match_all('/(?<!\d|\d\D)(?:dk)?([\d\PL]{8,})/m', $input, $matches, PREG_PATTERN_ORDER);
The \PL pattern matches any char but a Unicode letter and also matches digits and whitespace chars. So, [\d\PL] can be shortened to \PL and since you need to subtract line breaks from it, replace it with the reverse shorthand character class (\pL) and use it inside a negated bracket expression, [^\pL], and add \r and \n there:
'/(?<!\d|\d\D)(?:dk)?([^\pL\r\n]{8,})/u'
The m modifier is redundant since it only redefines the behavior of ^ and $ anchors. You might need the u modifier though, for the Unicode property class to work safely with Unicode strings in PHP/PCRE. Change \d to [0-9] and \D to [^0-9] if you only want to match ASCII digits.
How to write a regex with matches whitespace but no tabs and new line?
thanks everything
[[:blank:]]{2,} <-- Even though this isn't good for me because its whitespace or tab but not newlines.
As per my original comment, you can use this.
Code
See regex in use here
Note: The link contains whitespace characters: tab, newline, and space. Only space is matched.
[^\S\t\n\r]
So your regex would be [^\S\t\n\r]{2,}
Explanation
[^\S\t\n\r] Match any character not present in the set.
\S Matches any non-whitespace character. Since it's a double negative it will actually match any whitespace character. Adding \t, \n, and \r to the negated set ensures we exclude those specific characters as well. Basically, this regex is saying:
Match any whitespace character except \t\n\r
This principle in regex is often used with word characters \w to negate the underscore _ character: [^\W_]
[ ]{2,} works normally (not sure about php)
or even / {2,}/
I have a string stored in variable $text:
$text = '
I should not be removed.
I should not be removed.
I should not be removed?
I should not be removed!
I should be removed
I should be removed-
I should not be removed?
';
I want to remove all lines in the string that do not end with ., ? or !. How do I do this effectively? Maybe a preg_replace() approach?
If there is no whitespace at the end of the lines, you can use
'~^.*(?<![.?!])$\R?~m'
See regex demo
Explanation:
^ - start of line (as /m modifier indicates the multiline mode when ^ and $ match start and end of line, not string)
.* - any characters but a newline up to...
(?<![.?!])$ - the end of the string that is not preceded with a . or ! or ?
\R? - optional line break
To ignore the trailing whitespace, use a lookahead based regex:
'~^(?!.*[.?!]\h*$).*$\R?~m'
See regex demo
Explanation:
^ - start of a line
(?!.*[.?!]\h*$) - a negative lookahead that fails a match if there is a ., ? or ! at the end of the string followed by optional horizontal whitespace (\h*)
.*$ - any characters but a newline, 0 or more occurrences, up to the end of the line
\R? - optional newline sequence (optional, as the last line may not be followed with a newline character).
PHP code demo:
$re = '~^(?!.*[.?!]\h*$).*$\R?~m';
$str = "I should not be removed. \nI should not be removed.\nI should not be removed?\nI should not be removed! \nI should be removed\nI should be removed-\nI should not be removed? ";
$result = preg_replace($re, "", $str);
echo $result;
If you need to ignore the whitespace and punctuation, just add a [\p{P}\h] character class to the lookahead:
^(?!.*[.?!][\p{P}\h]*$).*$\R?
See demo. Now, the lookahead looks like (?!.*[.?!][\p{P}\h]*$). It fails a match if there is a ., ?, or ! followed by punctuation (\p{P}) or horizontal whitespace (\h), zero or more occurrences (*).
AND FINAL UPDATE: If you need to also ignore all non-word symbols (including Unicode letters) and all HTML entities, you can use
'~^(?!.*[.?!](&\w+;|\W)*$).*$\R?~m'
See another regex demo and an IDEONE demo. The lines ending with .  and .  do not get removed.
The difference here is (&\w+;|\W)* that matches 0 or more substrings starting with & and followed by 1 or more word characters (letters [A-Za-z], digits ([0-9]) or an underscore) and then a semi-colon, or non-word characters (\W). You can unroll the pattern as [^\w&]*(?:&\w+;\W*)* so that the regex performance might improve.
Note that you can use \W to match all Unicode letters and symbols other than ASCII since the /u modifier is not used here.
Consider the following strings
breaking out a of a simple prison
this is b moving up
following me is x times better
All strings are lowercased already. I would like to remove any "loose" a-z characters, resulting in:
breaking out of simple prison
this is moving up
following me is times better
Is this possible with a single regex in php?
$str = "breaking out a of a simple prison
this is b moving up
following me is x times better";
$res = preg_replace("#\\b[a-z]\\b ?#i", "", $str);
echo $res;
How about:
preg_replace('/(^|\s)[a-z](\s|$)/', '$1', $string);
Note this also catches single characters that are at the beginning or end of the string, but not single characters that are adjacent to punctuation (they must be surrounded by whitespace).
If you also want to remove characters immediately before punctuation (e.g. 'the x.'), then this should work properly in most (English) cases:
preg_replace('/(^|\s)[a-z]\b/', '$1', $string);
As a one-liner:
$result = preg_replace('/\s\p{Ll}\b|\b\p{Ll}\s/u', '', $subject);
This matches a single lowercase letter (\p{Ll}) which is preceded or followed by whitespace (\s), removing both. The word boundaries (\b) ensure that only single letters are indeed matched. The /u modifier makes the regex Unicode-aware.
The result: A single letter surrounded by spaces on both sides is reduced to a single space. A single letter preceded by whitespace but not followed by whitespace is removed completely, as is a single letter only followed but not preceded by whitespace.
So
This a is my test sentence a. o How funny (what a coincidence a) this is!
is changed to
This is my test sentence. How funny (what coincidence) this is!
You could try something like this:
preg_replace('/\b\S\s\b/', "", $subject);
This is what it means:
\b # Assert position at a word boundary
\S # Match a single character that is a “non-whitespace character”
\s # Match a single character that is a “whitespace character” (spaces, tabs, and line breaks)
\b # Assert position at a word boundary
Update
As raised by Radu, because I've used the \S this will match more than just a-zA-Z. It will also match 0-9_. Normally, it would match a lot more than that, but because it's preceded by \b, it can only match word characters.
As mentioned in the comments by Tim Pietzcker, be aware that this won't work if your subject string needs to remove single characters that are followed by non word characters like test a (hello). It will also fall over if there are extra spaces after the single character like this
test a hello
but you could fix that by changing the expression to \b\S\s*\b
Try this one:
$sString = preg_replace("#\b[a-z]{1}\b#m", ' ', $sString);