I want to validate using regex the user input to see if match the following rules.
Valid input:
2-000000000000
1-234342324342
...
Rules:
It has to be 13 digit numbers, no string.
Allow hyphen - come after the first character.
Allow space before and after the hyphen - character.
Here's what I tried in PHP, but still not correct:
if(preg_match("/[0-9?\-[0-9]/i])) {
echo "matched";
}
Note your double-quoted string literal is malformed, the closing " is missing.
Also, mind that [0-9?\-[0-9] is a malformed regex that matches a digit, ?, - or [ char. The i flag is irrelevant here, since there are no letters in the pattern.
I suggest using
preg_match("/^\d *- *\d{12}\z/", $string)
If the spaces can be any whitespace, replace the literal spaces with \s.
Note the use of \z anchor, I prefer it to $ in validation scenarios, since $ can match before the final line feed char in the string.
See the regex demo (\z replaced with $ since the input is a single multiline string there).
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)+$)
This expression will match the first and last spaces of the article..
I use a regex pattern i preg_match php function. The pattern is let's say '/abc$/'. It matches both strings:
'abc'
and
'abc
'
The second one has the line break at its end. What would be the pattern that matches only this first string?
'abc'
The reason why /abc$/ matches both "abc\n" and "abc" is that $ matches the location at the end of the string, or (even without /m modifier) the position before the newline that is at the end of the string.
You need the following regex:
/abc\z/
where \z is the unambiguous very end of the string, or
/abc$/D
where the /D modifier will make $ behave the same way as \z. See PHP.NET:
The meaning of dollar can be changed so that it matches only at the very end of the string, by setting the PCRE_DOLLAR_ENDONLY option at compile or matching time.
See the regex demo
I want to check if the input is Persian and it's at least 3 characters.
When I use below regex it works. It checks if the beginning of the word is Persian, but I want it to end with Persian, too, and to contain no English alphabet and no numbers.
/^[ئابپتثجچحخدذرزژسشصضطظعغفقکگلمنوهآی\s]{3,}/
For ensuring it ends with Persian I add $ after ] but I get this error:
Warning: preg_match(): Compilation failed: nothing to repeat at offset 77
/^[ئابپتثجچحخدذرزژسشصضطظعغفقکگلمنوهآی\s]${3,}/
Also, what is \s before the closing bracket ]? Is it for new line?
Use
/^[ئابپتثجچحخدذرزژسشصضطظعغفقکگلمنوهآی\s]{3,}$/u
$ is the end of line/string anchor and is not quantifiable (i.e. you cannot use a + or * or {1,} after it).
See demo
As for \s, it just matches whitespace from this set: [\r\n\t\f ].
EDIT: Using Rishida Unicode Converter, I re-wrote your expression using \x blocks:
/^[\x{626}-\x{628}\x{67E}\x{62A}-\x{62C}\x{686}\x{62D}\x{62E}\x{62F}-\x{632}\x{698}\x{633}-\x{63A}\x{641}\x{642}\x{6A9}\x{6AF}\x{644}-\x{648}\x{622}\x{6CC}]{3,}$/u
It does not allow spaces and looks nicer than ^[ئابپتثجچحخدذرزژسشصضطظعغفقکگلمنوهآی]{3,}$ that should also work (BTW, in MS Word it will look prettier :)).
To match arabic letters you can use:
^[\x{600}-\x{6FF}]{3,}$
if (preg_match('/^[\x{600}-\x{6FF}]{3,}$/u', $value)) {
# match
} else {
# no match
}
I am trying to build a regex to match these strings:
jfldfldf ldjfdlf ldfl
ldfldf 8998 dfjldjf 89dfdf dfdf899
ljdljf [dff]dfdf (fdfdf) 898
Requirements:
String should starts only with any small or capital character (A-Z)
It may contain spaces or brackets (( ) [ ])
Any other special characters are not allowed
I tried /^[a-zA-Z]+[\sa-zA-Z0-9\[\]\(\)].+/m, but it is still accepting other special characters.
So close.
/^[a-zA-Z]+[\sa-zA-Z0-9\[\]\(\)].+/m
^ ^ ^-- missing $
^ ^-- delete this dot
^-- you could also delete this plus, but that's not as important
/^[a-zA-Z]{1}[a-zA-Z0-9\ \[\]\(\)]+$/m
\s = allows whitespaces like spaces tabs and new lines, so this should probably be "\ "
Because the rule is only the first letter needs to be a capital or lowercase letter, strictly it's {1} as + means one or more.
Needed a $ at the end to show this is the end of the line, and nothing else can follow it
The biggest thing that is failing in that regex is the single '.'. That serves as a wildcard matching any value aside from a new line. The plus symbols are not needed and the end of string character '$' is missing.
/^[a-zA-Z][\sa-zA-Z0-9\[\]\(\)]$/m