If I have a string
This is a #really nice#day.
On first pass I should get as an output/result words really and day (results should not contain dots or any other punctuation signs, also you should not just match A-Z,a-z and everything else ignore because string could contain international characters so keep that in mind).
On second pass I should get out everything except those two words and punctuation for ex.
This is a nice
RegEx is done via PHP.
EDIT: #hochl
The problem with ([##]\w+) is that it doesn't catch international characters like šđžćč so #dayš is recognized only as #day.
To catch the international characters, you could use the following:
[##]\p{L}+
You would need to use the unicode modifier /u for this to work in php.
Note:
The \p{L} is telling it to match unicode "letters"
You don't need to wrap the whole thing in parentheses () as the whole match is always the first group
Related
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 trying to get a strip a text from all punctuation but since the text is in Spanish I can't use [A-Za-z0-9].
I have found this regex:
trim(preg_replace('#[^\p{L}\p{N}]+#u', ' ', $str)
which seems to do the job, but I would like to keep two special characters # and #, how can I achieve that?
Extra question: How can I delete all strings that are just numbers? e.g. 123 would be deleted but not as5623.
Thanks in advance!
You can simply add those characters to your negated class to retain them. And be sure to change your pattern delimiters to something other than # as well.
~[^\p{L}\p{N}##]+~u
To remove all strings that are numbers, you can place word boundaries \b around your pattern.
\b\d+\b
Note: A word boundary does not consume any characters. It asserts that on one side there is a word character, and on the other side there is not.
You can use posix character classes too.
/[^[:alnum:]##]+/
But for the two special character, you just have to add it inside character class.
To delete all the only number containing words following regex would work.
/\b[[:digit:]]+\b/
I have the following regex meant to test against valid name formats:
^[a-zA-Z]+(([\'\,\.\- ][a-zA-Z ])?[a-zA-Z]*)*$
it seems to work fine with all the expected odd name possibilities, including the following:
o'Bannon
Smith, Jr.
Double-barreled
I'm having problem when I plug this into my PHP code. If the first character is a number it passes through as valid.
If the last character is a space, comma, full-stop or other special allowed character, it's failing as invalid.
My PHP code is :
$v = 'Tested Value';
$value = (filter_var($v, FILTER_VALIDATE_REGEXP,array("options"=>array("regexp"=>"^[a-zA-Z]+(([\'\,\.\-,\ ][a-zA-Z ])?[a-zA-Z]*)*$^"))));
if (strlen($value) <2 && strlen($v) !=0) {
return "not valid";
}
What am I doing wrong here?
^[a-zA-Z]+(([\'\,\.\-,\ ][a-zA-Z ])?[a-zA-Z]*)*$^
The carets (^) at the beginning and end of the regex are being interpreted as regex deliminators, not as anchors. The regex isn't really matching the digits at the beginning of the string, it's skipping over them so it can start matching at the first letter it finds. You can use almost any ASCII punctuation character as the regex deliminator, but most people use # or ~, which are relatively uncommon and have no special meaning in regexes.
As for not allowing punctuation at the end, that's how the regex is written. Specifically, [\'\,\.\- ][a-zA-Z ] requires that each apostrophe, comma, period or hyphen be followed by a letter or a space. If you really want to allow any of those characters at the end, it's pretty simple:
~^(?:[a-z]+[',. -]*)+$~i
Of course, that's not a particularly good regex for validating names, but I have nothing better to offer; it's a job for which regexes are particularly ill-suited. And do you really want to be the one to tell your users their own names are invalid?
Your regex is way to complex
/^[a-z]+[',. a-z-]*$/i
should do the same thing
when I try preg_match with the following expression: /.{0,5}/, it still matches string longer than 5 characters.
It does, however, work properly when trying in online regexp matcher
The site you reference, myregexp.com, is focussed on Java.
Java has a specific function for matching an exact pattern, without needing to use anchor characters. This is the function which myregexp.com uses.
In most other languages, in order to match an exact pattern, you would need to add the anchoring characters ^ and $ at the start and end of the pattern respectively, otherwise the regex assumes it only needs to find the matched pattern somewhere within the string, rather than the whole string being the match.
This means that without the anchors, your pattern will match any string, of any length, because whatever the string, it will contain within it somewhere a match for "zero to five of any character".
So in PHP, and Perl, and virtually any other language, you need your pattern to look like this:
/^.{0,5}$/
Having explained all that, I would make one final observation though: this specific pattern really doesn't need to be a regular expression -- you could achieve the same thing with strlen(). In addition, the dot character in regex may not work exactly as you expect: it typically matches almost any character; some characters, including new line characters, are excluded by default, so if your string contains five characters, but one of them is a new line, it will fail your regex when you might have expected it to pass. With this in mind, strlen() would be a safer option (or mb_strlen() if you expect to have unicode characters).
If you need to match any character in regex, and the default behaviour of the dot isn't good enough, there are two options: One is to add the s modifier at the end of the expression (ie it becomes /^.{0,5}$/s). The s modifier tells regex to include new line characters in the dot "any character" match.
The other option (which is useful for languages that don't support the s modifier) is to use an expression and its negative together in a character class - eg [\s\S] - instead of the dot. \s matches any white space character, and \S is a negative of \s, so any character not matched by \s. So together in a character class they match any character. It's more long winded and less readable than a dot, but in some languages it's the only way to be sure.
You can find out more about this here: http://www.regular-expressions.info/dot.html
Hope that helps.
You need to anchor it with ^$. These symbols match the beginning and end of the string respectively, so it must be 0-5 characters between the beginning and end. Leaving out the anchors will match anywhere in the string so it could be longer.
/^.{0,5}$/
For better readability, I would probably also enclose the . in (), but that's kind of subjective.
/^(.){0,5}$/
This question already has answers here:
How to validate an email address in PHP
(15 answers)
Closed 2 years ago.
Regex is blowing my mind. How can I change this to validate emails with a plus sign? so I can sign up with test+spam#gmail.com
if(!preg_match("/^[_a-z0-9-]+(\.[_a-z0-9-]+)*#[a-z0-9-]+(\.[a-z0-9-]+)*$/i", $_GET['em'])) {
It seems like you aren't really familiar with what your regex is doing currently, which would be a good first step before modifying it. Let's walk through your regex using the email address john.robert.smith#mail.com (in each section below, the bolded part is what is matched by that section):
^ is the start of string
anchor.
It specifies that any match must
begin at the beginning of the
string. If the pattern is not
anchored, the regex engine can match
a substring, which is often
undesired.
Anchors are zero-width, meaning that
they do not capture any characters.
[_a-z0-9-]+ is made up of two
elements, a character
class
and a repetition
modifer:
[...] defines a character class, which tells the regex engine,
any of these characters are valid matches. In this case the class
contains the characters a-z, numbers
0-9 and the dash and underscore (in
general, a dash in a character class
defines a range, so you can use
a-z instead of
abcdefghijklmnopqrstuvwxyz; when
given as the last character in the
class, it acts as a literal dash).
+ is a repetition modifier that specifies that the preceding token
(in this case, the character class)
can be repeated one or more times.
There are two other repetition
operators: * matches zero or more
times; ? matches exactly zero or
one times (ie. makes something
optional).
(captures
john.robert.smith#mail.com)
(\.[_a-z0-9-]+)* again contains a
repeated character class. It also
contains a
group,
and an escaped character:
(...) defines a group, which allows you to group multiple tokens
together (in this case, the group
will be repeated as a
whole).Let's say we wanted to
match 'abc', zero or more times (ie.
abcabcabc matches, abcccc doesn't).
If we tried to use the pattern
abc*, the repetition modifier
would only apply to the c, because
c is the last token before the
modifier. In order to get around
this, we can group abc ((abc)*),
in which case the modifier would
apply to the entire group, as if it
was a single token.
\. specifies a literal dot character. The reason this is needed
is because . is a special
character in regex, meaning any
character.
Since we want to match an actual dot
character, we need to escape it.
(captures
john.robert.smith#mail.com)
# is not a special character in
regex, so, like all other
non-special characters, it matches
literally.
(captures john.robert.smith#mail.com)
[a-z0-9-]+ again defines a repeated character class, like item #2 above.
(captures john.robert.smith#mail.com)
(\.[a-z0-9-]+)* is almost exactly the same pattern as #3 above.
(captures john.robert.smith#mail.com)
$ is the end of string anchor. It works the same as ^ above, except matches the end of the string.
With that in mind, it should be a bit clearer how to add a section with captures a plus segment. As we saw above, + is a special character so it has to be escaped. Then, since the + has to be followed by some characters, we can define a character class with the characters we want to match and define its repetition. Finally, we should make the whole group optional because email addresses don't need to have a + segment:
(\+[a-z0-9-]+)?
When inserted into your regex, it'd look like this:
/^[_a-z0-9-]+(\.[_a-z0-9-]+)*(\+[a-z0-9-]+)?#[a-z0-9-]+(\.[a-z0-9-]+)*$/i
Save your sanity. Get a pre-made PHP RFC 822 Email address parser
I've used this regex to validate emails, and it works just fine with emails that contain a+:
/^(([^<>()[\]\\.,;:\s#\"]+(\.[^<>()[\]\\.,;:\s#\"]+)*)|(\".+\"))#((\[[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\])|(([a-zA-Z\-0-9]+\.)+[a-zA-Z]{2,}))$/
\+ will match a literal + sign, but be aware: You still won't be close to matching all possible email addresses according to the RFC spec, because the actual regex for that is madness. It's almost certainly not worth it; you should use a real email parser for this.
This is another solution (is similar to the solution found by David):
//Escaped for .Net
^[_a-zA-Z0-9-]+((\\.[_a-zA-Z0-9-]+)*|(\\+[_a-zA-Z0-9-]+)*)*#[a-zA-Z0-9-]+(\\.[a-zA-Z0-9-]+)*(\\.[a-zA-Z]{2,4})$
//Native
^[_a-zA-Z0-9-]+((\.[_a-zA-Z0-9-]+)*|(\+[_a-zA-Z0-9-]+)*)*#[a-zA-Z0-9-]+(\.[a-zA-Z0-9-]+)*(\.[a-zA-Z]{2,4})$
This is the another solution
/^[_a-z0-9-+]+(\.[_a-z0-9-+]+)*(\+[a-z0-9-]+)?#[a-z0-9-.]+(\.[a-z0-9]+)$/
or For razor page(#=\u0040)
/^[_a-z0-9-+]+(\.[_a-z0-9-+]+)*(\+[a-z0-9-]+)?\u0040[a-z0-9-.]+(\.[a-z0-9]+)$/