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)+$)
This expression will match the first and last spaces of the article..
I am trying to write a regex and matches 3 conditions, and would return true if all three conditions are met.
Condition 1) string starts with a "{"
Condition 2) string DOES NOT contain a space somewhere between brackets
Condition 3 string ends with a "}"
So far I have come up with ^{|[ ]|}$, which checks for a space. But I need to make it match if there are no spaces between the brackets. Also this will return true if the string starts with a { but doesn't end with a } and vise versa. I've been messing with regex101.com, but can't figure the no spaces portion out.
Could someone explain how to match if something does not exist in the string?
It seems you need
/^{\S*}$/
See the regex demo.
Details:
^ - start of string
{ - a literal {
\S* - 0+ chars other than whitespace
} - a literal }
$ - end of string.
For a general case, when you want to match if something DOES NOT exist in the string, you can bear in mind 2 approaches:
1) The string that should be absent is a single char: use either a negated character class (e.g., ^[^;]*$ will match a string that has no ; inside), or a "negative" shorthand character class (like here, \S is the reverse of \s and using ^\S*$ you match a string that has no whitespace)
2) The text that the string must not contain is a multicharacter string: then use lookarounds (e.g. the string should not contain like: ^(?!.*like).*$, or even ^(?!.*like)).
I want users input their username with only alphanumeric and dot character.
So I wrote a regex pattern as following:
'/([a-zA-Z0-9\.]+)/'
But I want to know is it the same with:
'/([a-zA-Z0-9.]+)/'
2 below patterns is the same? Thank you for help! :-)
You don't need to escape the dot which was present inside a character class. Inside a character class, dot . and escaped dot \. matches the literal dot. So both regexes are same.
And also for validation purposes, i would suggest you to add anchors like '/^[a-zA-Z0-9.]+$/' . Anchors would be used to do a exact string match. That is , /[a-zA-Z0-9.]+/ regex would match the substring foo in this ()foo input string but if you add start and end anchors to your regex like /^[a-zA-Z0-9.]+$/, it won't match even a single character in the above mentioned string. It's allowed to match only one or more alphanumeric or dot characters , if it finds a character other than dot or alphanumeric, then the regex engine won't match the corresponding string.
I’m trying to create a regex for form validation but it always returns true. The user must be able to add something like {user|2|S} as input but also use brackets if they are escaped with \.
This code checks for the left bracket { for now.
$regex = '/({(?=([a-zA-Z0-9]+\|[0-9]*\|(S|D[0-9]*)}))|[^{]|(?<=\\\){)*/';
if (preg_match($regex, $value)) {
return TRUE;
} else {
return FALSE;
}
A possible correct input would be:
Hello {user|1|S}, you have {amount|2|D2}
or
Hello {user|1|S}, you have {amount|2|D2} in \{the_bracket_bank\}
However, this should return false:
Hello {user|1|S}, you have {amount|2}
and this also:
Hello {user|1|S}, you have {amount|2|D2} in {the_bracket_bank}
A live example can be found here: http://regexr.com?37tpu Note that there is a \ in the lookbehind at the end, PHP was giving me error messages because I had to escape it an extra time in my code.
The main error is that you do not specify that the regex should match from the beginning to the of the checked string. Use the ^ and $ assertions.
I think you have to escape { and } in your regex as they have special meaning. Together they form a quantifier.
The (?<=\\\) is better written (?<=\\\\). The backslash has to be double escaped as it has special meaning in both single-quoted string and PCRE regex. Using \\\ works too, because if single-quoted string contains any escape sequence except \\ and \', it handles it as literal backslash and letter, therefore \) is taken literally. But explicitly escaping the backslash twice seems easier to read to me.
The regex should be
$regex = '/^(\{(?=([a-zA-Z0-9]+\|[0-9]*\|(S|D[0-9]*)\}))|[^{]|(?<=\\\\)\{)*$/';
But notice that the look-around assertions are not necessary. This regex should do the job too:
$regex = '/^([^{]|\\\{|\{[a-zA-Z0-9]+\|[0-9]*\|(S|D[0-9]*)\})*$/';
Any non-{ characters are matched by the first alternative. When a { is read, one of the remaining two alternatives is used. Either the pattern for the brace thing matches, or the regex engine backtracks one character and tries to match \{ character sequence. If it fails, both ways, it backtracks further till it reaches string start and fails completely.
Matching without lookbehind
You can make a regex for this without using lookbehind/lookaheads (which is usually recommended).
For example, if your requirement is that you can match any character but a { and a } unless it's preceded by a \. You can also say:
Match any character but a { and a } OR match a \{ or a \}. To match any character but a { and a } use:
[^{}]
To match a \{ use:
\\\{
One backslash is for escaping the { (which might not be necessary, depending on your regex compiler) and one backslash is for escaping the other backslash.
You would end up with this:
(?:
[^{}]
|
\\\{
|
\\\}
)+
I nicely formatted this regex so that it's readable. If you want to use it in your code like this make sure to use the [PCRE_EXTENDED][1] modifier.
Looks more of a job for a lookbehind to me:
/((?<!\\\\)\{[a-zA-Z0-9]+\|[0-9]+\|[SD][0-9]*\})/
However, the obfuscation factor is so high that I would rather recognize all bracketed strings and parse them later.
I'm creating a password validator which takes any character but whitespaces and with at least 6 characters.
After searching the best I came up is this is this example:
What is the regular expression for matching that contains no white space in between text?
It disallows any spaces inbetween but does allow starting and ending with a space. I want to disallow any space in the string passed.
I tried this but it doesn't work:
if (preg_match("/^[^\s]+[\S+][^\s]{6}$/", $string)) {
return true;
} else {
return false;
}
Thanks.
Something like this:
/^\S{6,}\z/
Can be quoted like:
preg_match('/^\S{6,}\z/', $string)
All answers using $ are wrong (at least without any special flags). You should use \z instead of $ if you do not want to allow a line break at the end of the string.
$ matches end of string or before a line break at end of string (if no modifiers are used)
\z matches end of string (independent of multiline mode)
From http://www.pcre.org/pcre.txt:
^ start of subject
also after internal newline in multiline mode
\A start of subject
$ end of subject
also before newline at end of subject
also before internal newline in multiline mode
\Z end of subject
also before newline at end of subject
\z end of subject
The simplest expression:
^\S{6,}$
^ means the start of the string
\S matches any non-whitespace character
{6,} means 6 or more
$ means the end of the string
In PHP, that would look like
preg_match('/^\S{6,}$/', $string)
Edit:
>> preg_match('/^\S{6,}$/', "abcdef\n")
1
>> preg_match('/^\S{6,}\z/', "abcdef\n")
0
>> preg_match('/^\S{6,}$/D', "abcdef\n")
0
Qtax is right. Good call! Although if you're taking input from an HTML <input type="text"> you probably won't have any newlines in it.
I think you should be fine using the following, which would match any string longer than 1 character with no whitespace:
^[^\s]+$
You can see the test here: http://regexr.com?2ua2e.
Try this. This will match at least 6 non whitespace characters followed by any number of additional non whitespace characters.
^[^\s]{6}[^\s]*$
\S - Matches any non-white-space character. Equivalent to the Unicode character categories [^\f\n\r\t\v\x85\p{Z}]. If ECMAScript-compliant behavior is specified with the ECMAScript option, \S is equivalent to [^ \f\n\r\t\v].
The start of string you can do : ^[ \t]+, and for end : [ \t]+$ (tab and spaces)
ETA:
By the way, you regex [\S+], i think you're looking for : [\S]+