i need to write a regex for make a double check: if a string contains empty spaces at the beginning, at the end, and if all string it's composed by empty spaces, and if string contains only number.
I've write this regex
$regex = '/^(\s+ )| ^(\d+)$/';
but it doesn't' work. What's wrong ?
First things first: get your spaces right!
For example (\s+ ) will match a minimum of one space (\s+) followed by another space ()! Same applies for the space between | and ^. This way you will match the space literally every time and this leads to wrong results.
If I get you right and you want to match on strings which
start with one or more spaces OR
end with one or more spaces OR
consist only of spaces OR
consist only of numbers
I'd use
/^(?:\s+.*|.*\s+$|\d+$)/
Demo # regex101
This way you match spaces at the start of the string (\s+.*) or (|) spaces at the end of the string (.*\s+$) or a completely numeric string (\d+$).
Insert capturing groups as needed.
This will match in case the whole string consists of spaces, too, because technically the string then starts with spaces.
The space before ^(\d+) make your regex can't catch the numeric string.
It should be like below:
$regex = '/^\s*\d*\s*$/';
First if all, remove the space between | and ^. You are trying to match a space before the beginning of the line (^), so that can not work.
I do not exactly understand what you want. Either a string that only consists of white spaces, or a number that may have white spaces at the beginning or end? Try this:
$regex = '/^\s*\d*\s*$/';
Related
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)+$)
This expression will match the first and last spaces of the article..
I have the following requirements for validating an input field:
It should only contain alphabets and spaces between the alphabets.
It cannot contain spaces at the beginning or end of the string.
It cannot contain any other special character.
I am using following regex for this:
^(?!\s*$)[-a-zA-Z ]*$
But this is allowing spaces at the beginning. Any help is appreciated.
For me the only logical way to do this is:
^\p{L}+(?: \p{L}+)*$
At the start of the string there must be at least one letter. (I replaced your [a-zA-Z] by the Unicode code property for letters \p{L}). Then there can be a space followed by at least one letter, this part can be repeated.
\p{L}: any kind of letter from any language. See regular-expressions.info
The problem in your expression ^(?!\s*$) is, that lookahead will fail, if there is only whitespace till the end of the string. If you want to disallow leading whitespace, just remove the end of string anchor inside the lookahead ==> ^(?!\s)[-a-zA-Z ]*$. But this still allows the string to end with whitespace. To avoid this look back at the end of the string ^(?!\s)[-a-zA-Z ]*(?<!\s)$. But I think for this task a look around is not needed.
This should work if you use it with String.matches method. I assume you want English alphabet.
"[a-zA-Z]+(\\s+[a-zA-Z]+)*"
Note that \s will allow all kinds of whitespace characters. In Java, it would be equivalent to
[ \t\n\x0B\f\r]
Which includes horizontal tab (09), line feed (10), carriage return (13), form feed (12), backspace (08), space (32).
If you want to specifically allow only space (32):
"[a-zA-Z]+( +[a-zA-Z]+)*"
You can further optimize the regex above by making the capturing group ( +[a-zA-Z]+) non-capturing (with String.matches you are not going to be able to get the words individually anyway). It is also possible to change the quantifiers to make them possessive, since there is no point in backtracking here.
"[a-zA-Z]++(?: ++[a-zA-Z]++)*+"
Try this:
^(((?<!^)\s(?!$)|[-a-zA-Z])*)$
This expression uses negative lookahead and negative lookbehind to disallow spaces at the beginning or at the end of the string, and requiring the match of the entire string.
I think the problem is there's a ? before the negation of white spaces, which means it is optional
This should work:
[a-zA-Z]{1}([a-zA-Z\s]*[a-zA-Z]{1})?
at least one sequence of letters, then optional string with spaces but always ends with letters
I don't know if words in your accepted string can be seperated by more then one space. If they can:
^[a-zA-Z]+(( )+[a-zA-z]+)*$
If can't:
^[a-zA-Z]+( [a-zA-z]+)*$
String must start with letter (or few letters), not space.
String can contain few words, but every word beside first must have space before it.
Hope I helped.
I am looking to find comma after Oliver and before Roy in the following string
"Username,Hello Person Oli,"Oliver,Who" roy,Roy"
I want to replace this one Oliver,Who
I am using the following regex
'/^"(?:[a-zA-Z]+)(?:,+)(?:[a-zA-Z]+)"$/'
However its not working with preg_replace
This is my code
$pregData = preg_replace('/^"(?:[a-zA-Z]+)(?:,+)(?:[a-zA-Z]+)"$/',';',$csv);
Any ideas why ?
Sorry for the poor first message.
To take into account other characters than just a-z, you could use:
/"([^,]+?),(.+?)"/
" = quote
[^,]+?, = anything that's not a comma until the first comma
.+?" = anything else until the next quote
Note that the string on the left and on the right of the comma are captured by (...) constructs. That means that, if the expression matches, then the string on the left will be assigned to \1, while the string on the right will be assigned to \2. Therefore, if you want to replace this with something like left; right, you could use:
preg_replace('/"([^,]+?),(.+?)"/', '\1; \2', $csv)
If you just want to keep the left and right parts without the comma in the middle, you can simply replace the expression with the left part, followed by a space, followed by the right part.
preg_replace('/"([^,]+?),(.+?)"/', '\1 \2', $csv)
Your regex is looking for a text with only a-z at the beginning, a comma, and a-z again. That doesn't match with your string, you should remove the beginning (^) and end ($) of string characters.
/"(?:[a-zA-Z]+)(?:,+)(?:[a-zA-Z]+)"/
If you want to replace only the comma, then you need, additionally to the two wrong anchors, to make the outer groups capturing.
'/"([a-zA-Z]+)(?:,+)([a-zA-Z]+)"/'
So that you can reuse the captured words around the comma using $1 and $2
See it here on Regexr
I just want to find unknown text around my string between two spaces.
For example:
$mystring = "blabalbla <b>sometext</b> <b>ssssss</b>"
What I want to do with this:
I know the "sometext" but I want to put in a string the
<b>sometext</b>.
But my string is always changing, forexample it can be:
<s><b>sometext</b></s>
Now I need to put the whole into a string
<s><b>sometext</b></s>.
So I can't use simply attaching my variable to
<b>.mystring.</b>
beacuse in cases I can have unknown strings around it.
How can I do this? Or is there another way to find and delete those
<b><s><i></b></s></i> etc.... around my string?
Thnaks, Creep.
You could use a regex:
$mystring = preg_replace('/(^|\s)(?:<[^>]*>)*sometext(?:<[^>]*>)*(\s|$)/i', '$1'.$some_new_text.'$2', $mystring);
I tested this against what you provided, is should work pretty well. It handles the text being on it's own, at the start or end of the string, and surrounded by an unlimited number of html entities.
Description
Match either the start of the string, or a space (^|\s)
Followed by zero or more html nodes (?:<[^>]*>)*
These are in a non-capturing group, so they don't get assigned a group number
Followed by the known string
If you plan on having this string be dynamic, you will need to use the preg_quote method to escape any special characters
Followed by zero or more html nodes (?:<[^>]*>)*
Followed by either a space or the end of the string (\s|$)
Match case-insensitvely /i (optional)
Notice that the leading and trailing spaces (if any) are added back in on the replacement string $1 and $2.
How would I go about removing numbers and a space from the start of a string?
For example, from '13 Adam Court, Cannock' remove '13 '
Because everyone else is going the \d+\s route I'll give you the brain-dead answer
$str = preg_replace("#([0-9]+ )#","",$str);
Word to the wise, don't use / as your delimiter in regex, you will experience the dreaded leaning-toothpick-problem when trying to do file paths or something like http://
:)
Use the same regex I gave in my JavaScript answer, but apply it using preg_replace():
preg_replace('/^\d+\s+/', '', $str);
Try this one :
^\d+ (.*)$
Like this :
preg_replace ("^\d+ (.*)$", "$1" , $string);
Resources :
preg_replace
regular-expressions.info
On the same topic :
Regular expression to remove number, then a space?
regular expression for matching number and spaces.
I'd use
/^\d+\s+/
It looks for a number of any size in the beginning of a string ^\d+
Then looks for a patch of whitespace after it \s+
When you use a backslash before certain letters it represents something...
\d represents a digit 0,1,2,3,4,5,6,7,8,9.
\s represents a space .
Add a plus sign (+) to the end and you can have...
\d+ a series of digits (number)
\s+ multiple spaces (typos etc.)
The same regex I gave you on your other question still applies. You just have to use preg_replace() instead.
Search for /^[\s\d]+/ and replace with the empty string. Eg:
$str = preg_replace(/^[\s\d]+/, '', $str);
This will remove digits and spaces in any order from the beginning of the string. For something that removes only a number followed by spaces, see BoltClock's answer.
If the input strings all have the same ecpected format and you will receive the same result from left trimming all numbers and spaces (no matter the order of their occurrence at the front of the string), then you don't actually need to fire up the regex engine.
I love regex, but know not to use it unless it provides a valuable advantage over a non-regex technique. Regex is often slower than non-regex techniques.
Use ltrim() with a character mask that includes spaces and digits.
Code: (Demo)
var_export(
ltrim('420 911 90210 666 keep this part', ' 0..9')
);
Output:
'keep this part'
It wouldn't matter if the string started with a space either. ltrim() will greedily remove all instances of spaces or numbers from the start of the string intil it can't anymore.