I am having trouble finding a pattern that would detect the following
909-999-9999
909 999 9999
(909) 999-9999
(909) 999 9999
999 999 9999
9999999999
\A[(]?[0-9]{3}[)]?[ ,-][0-9]{3}[ ,-][0-9]{3}\z
I tried it but it doesn't work for all the instances . I was thinking I can divide the problem by putting each character into an array and then checking it. but then the code would be too long.
You have 4 digits in the last group, and you specify 3 in the regex.
You also need to apply a ? quantifier (1 or 0 occurrence) to the separators since they are optional.
Use
^[(]?[0-9]{3}[)]?[ ,-]?[0-9]{3}[ ,-]?[0-9]{4}$
See the demo here
PHP demo:
$re = "/\A[(]?[0-9]{3}[)]?[ ,-]?[0-9]{3}[ ,-]?[0-9]{4}\z/";
$strs = array("909-999-9999", "909 999 9999", "(909) 999-9999", "(909) 999 9999", "999 999 9999","9999999999");
$vals = preg_grep($re, $strs);
print_r($vals);
And another one:
$re = "/\A[(]?[0-9]{3}[)]?[ ,-]?[0-9]{3}[ ,-]?[0-9]{4}\z/";
$str = "909-999-9999";
if (preg_match($re, $str, $m)) {
echo "MATCHED!";
}
BTW, optional ? subpatterns perform better than alternations.
Try this regex:
^(?:\(\d{3}\)|\d{3})[- ]?\d{3}[- ]?\d{4}$
Explaining:
^ # from start
(?: # one of
\(\d{3}\) # '(999)' sequence
| # OR
\d{3} # '999' sequence
) #
[- ]? # may exist space or hyphen
\d{3} # three digits
[- ]? # may exist space or hyphen
\d{4} # four digits
$ # end of string
Hope it helps.
Related
Site users enter numbers in different ways, example:
from 8 000 packs
432534534
from 344454 packs
45054 packs
04 555
434654
54 564 packs
I am looking for a regular expression with which I could highlight words before digits (if there are any), digits in any format and words after (if there are any). It is advisable to exclude spaces.
Now I have such a design, but it does not work correctly.
(^[0-9|a-zA-Z].*?)\s([0-9].*?)\s([a-zA-Z]*$)
The main purpose of this is to put the strings in order, bring them to the same form, format them in PHP digit format, etc.
As a result, I need to get the text before the digits, the digits themselves and the text after them into the variables separately.
$before = 'from';
$num = '8000';
$after = 'packs';
Thank you for any help in this matter)
I think you may try this:
^(\D+)?([\d \t]+)(\D+)?$
group 1: optional(?) group that will contain anything but digit
group 2: mandatory group that will contain only digits and
white space character like space and tab
group 3: optional(?) group that will contain anything but digit
Demo
Source (run)
$re = '/^(\D+)?([\d \t]+)(\D+)?$/m';
$str = 'from 8 000 packs
432534534
from 344454 packs
45054 packs
04 555
434654
54 564 packs
';
preg_match_all($re, $str, $matches, PREG_SET_ORDER, 0);
foreach ($matches as $matchgroup)
{
echo "before: ".$matchgroup[1]."\n";
echo "number:".preg_replace('/\D/m','',$matchgroup[2])."\n";
echo "after:".$matchgroup[3]."";
echo "\n\n\n";
}
I corrected your regex and added groups, the regex looks like this:
^(?<before>[a-zA-Z]+)?\s?(?<number>[0-9].*?)\s?(?<after>[a-zA-Z]+)?$`
Test regex here: https://regex101.com/r/QLEC9g/2
By using groups you can easily separate the words and numbers, and handle them any way you want.
Your pattern does not match because there are 4 required parts that all expect 1 character to be present:
(^[0-9|a-zA-Z].*?)\s([0-9].*?)\s([a-zA-Z]*$)
^^^^^^^^^^^^ ^^ ^^^^^ ^^
The other thing to note is that the first character class [0-9|a-zA-Z] can also match digits (you can omit the | as it would match a literal pipe char)
If you would allow all other chars than digits on the left and right, and there should be at least a single digit present, you can use a negated character class [^\d\r\n]* optionally matching any character except a digit or a newline:
^([^\d\r\n]*)\h*(\d+(?:\h+\d+)*)\h*([^\d\r\n]*)$
^ Start of string
([^\d\r\n]*) Capture group 1, match any char except a digit or a newline
\h* Match optional horizontal whitespace chars
(\d+(?:\h+\d+)*) Capture group 2, match 1+ digits and optionally repeat matching spaces and 1+ digits
\h* Match optional horizontal whitespace chars
([^\d\r\n]*) Capture group 3, match any char except a digit or a newline
$ End of string
See a regex demo and a PHP demo.
For example
$re = '/^([^\d\r\n]*)\h*(\d+(?:\h+\d+)*)\h*([^\d\r\n]*)$/m';
$str = 'from 8 000 packs
test from 8 000 packs test
432534534
from 344454 packs
45054 packs
04 555
434654
54 564 packs';
preg_match_all($re, $str, $matches, PREG_SET_ORDER, 0);
foreach($matches as $match) {
list(,$before, $num, $after) = $match;
echo sprintf(
"before: %s\nnum:%s\nafter:%s\n--------------------\n",
$before, preg_replace("/\h+/", "", $num), $after
);
}
Output
before: from
num:8000
after:packs
--------------------
before: test from
num:8000
after:packs test
--------------------
before:
num:432534534
after:
--------------------
before: from
num:344454
after:packs
--------------------
before:
num:45054
after:packs
--------------------
before:
num:04555
after:
--------------------
before:
num:434654
after:
--------------------
before:
num:54564
after:packs
--------------------
If there should be at least a single digit present, and the only allowed characters are a-z for the word(s), you can use a case insensitive pattern:
(?i)^((?:[a-z]+(?:\h+[a-z]+)*)?)\h*(\d+(?:\h+\d+)*)\h*((?:[a-z]+(?:\h+[a-z]+)*)?)?$
See another regex demo and a php demo.
How to remove all numbers exept alphanumeric, for example if i have string like this:
Abs_1234abcd_636950806858590746.lands
to become it like this
Abs_1234abcd_.lands
It is probably done like this
Find (?i)(?<![a-z\d])\d+(?![a-z\d])
Replace with nothing.
Explained:
It's important to note that in the class [a-z\d] within assertions,
there exists a digit, without which could let "abc901234def" match.
(?i) # Case insensitive
(?<! [a-z\d] ) # Behind, not a letter nor digit
\d+ # Many digits
(?! [a-z\d] ) # Ahead, not a letter nor digit
Note - a speedier version exists (?i)\d(?<!\d[a-z\d])\d*(?![a-z\d])
Regex1: (?i)\d(?<!\d[a-z\d])\d*(?![a-z\d])
Completed iterations: 50 / 50 ( x 1000 )
Matches found per iteration: 2
Elapsed Time: 0.53 s, 530.56 ms, 530564 µs
Matches per sec: 188,478
Regex2: (?i)(?<![a-z\d])\d+(?![a-z\d])
Completed iterations: 50 / 50 ( x 1000 )
Matches found per iteration: 2
Elapsed Time: 0.91 s, 909.58 ms, 909577 µs
Matches per sec: 109,941
In this specific example, we can simply use _ as a left boundary and . as the right boundary, collect our digits, and replace:
Test
$re = '/(.+[_])[0-9]+(\..+)/m';
$str = 'Abs_1234abcd_636950806858590746.lands';
$subst = '$1$2';
$result = preg_replace($re, $subst, $str);
echo $result;
Demo
For your example data, you could also match not a word character or an underscore [\W_] using a character class. Then forget what is matched using \K.
Match 1+ digits that you want to replace with a empty string and assert what is on the right is again not a word character or an underscore.
[\W_]\K\d+(?=[\W_])
Regex demo
I split a string '3(1-5)' like this:
$pattern = '/^(\d+)\((\d+)\-(\d+)\)$/';
preg_match($pattern, $string, $matches);
But I need to do the same thing for decimals, i.e. '3.5(1.5-4.5)'.
And what do I have to do, if the user writes '3,5(1,5-4,5)'?
Output of '3.5(1.5-4.5)' should be:
$matches[1] = 3.5
$matches[2] = 1.5
$matches[3] = 4.5
You can use the following regular expression.
$pattern = '/^(\d+(?:[.,]\d+)?)\(((?1))-((?1))\)$/';
The first capturing group ( ... ) matches the following pattern:
( # group and capture to \1:
\d+ # digits (0-9) (1 or more times)
(?: # group, but do not capture (optional):
[.,] # any character of: '.', ','
\d+ # digits (0-9) (1 or more times)
)? # end of grouping
) # end of \1
Afterwords we look for an opening parenthesis and then recurse (match/capture) the 1st subpattern followed by a hyphen (-) and then recurse (match/capture) the 1st subpattern again followed by a closing parenthesis.
Code Demo
This pattern should help:
^(\d+\.?\,?\d+)\((\d+\,?\.?\d+)\-(\d+\.?\,?\d+)\)$
I want to create a regular expression in PHP, which will allow to user to enter a phone number in either of the formats below.
345-234 898
345 234-898
235-123-456
548 812 346
The minimum length of number should be 7 and maximum length should be 12.
The problem is that, the regular expression doesn't care about the minimum and maximum length. I don't know what is the problem in it. Please help me to solve it. Here is the regular expression.
if (preg_match("/^([0-9]+((\s?|-?)[0-9]+)*){7,12}$/", $string)) {
echo "ok";
} else {
echo "not ok";
}
Thanks for reading my question. I will wait for responses.
You should use the start (^) and the end ($) sign on your pattern
$subject = "123456789";
$pattern = '/^[0-9]{7,9}$/i';
if(preg_match($pattern, $subject)){
echo 'matched';
}else{
echo 'not matched';
}
You can use preg_replace to strip out non-digit symbols and check length of resulting string.
$onlyDigits = preg_replace('/\\D/', '', $string);
$length = strlen($onlyDigits);
if ($length < 7 OR $length > 12)
echo "not ok";
else
echo "ok";
Simply do this:
if (preg_match("/^\d{3}[ -]\d{3}[ -]\d{3}$/", $string)) {
Here \d means any digits from 0-9. Also [ -] means either a space or a hyphen
You can check the length with a lookahead assertion (?=...) at the begining of the pattern:
/^(?=.{7,12}$)[0-9]+(?:[\s-]?[0-9]+)*$/
Breaking down your original regex, it can read like the following:
^ # start of input
(
[0-9]+ # any number, 1 or more times
(
(\s?|-?) # a space, or a dash.. maybe
[0-9]+ # any number, 1 or more times
)* # repeat group 0 or more times
)
{7,12} # repeat full group 7 to 12 times
$ # end of input
So, basically, you're allowing "any number, 1 or more times" followed by a group of "any number 1 or more times, 0 or more times" repeat "7 to 12 times" - which kind of kills your length check.
You could take a more restricted approach and write out each individual number block:
(
\d{3} # any 3 numbers
(?:[ ]+|-)? # any (optional) spaces or a hyphen
\d{3} # any 3 numbers
(?:[ ]+|-)? # any (optional) spaces or a hyphen
\d{3} # any 3 numbers
)
Simplified:
if (preg_match('/^(\d{3}(?:[ ]+|-)?\d{3}(?:[ ]+|-)?\d{3})$/', $string)) {
If you want to restrict the separators to be only a single space or a hyphen, you can update the regex to use [ -] instead of (?:[ ]+|-); if you want this to be "optional" (i.e. there can be no separator between number groups), add in a ? to the end of each.
if (preg_match('/^(\d{3}[ -]\d{3}[ -]\d{3})$/', $string)) {
may it help you out.
Validator::extend('price', function ($attribute, $value, $args) {
return preg_match('/^\d{0,8}(\.\d{1,2})?$/', $value);
});
What would be Regex to match the following 10-digit numbers:
0108889999 //can contain nothing except 10 digits
011 8889999 //can contain a whitespace at that place
012 888 9999 //can contain two whitespaces like that
013-8889999 // can contain one dash
014-888-9999 // can contain two dashes
If you're just looking for the regex itself, try this:
^(\d{3}(\s|\-)?){2}\d{4}$
Put slightly more legibly:
^ # start at the beginning of the line (or input)
(
\d{3} # find three digits
(
\s # followed by a space
| # OR
\- # a hyphen
)? # neither of which might actually be there
){2} # do this twice,
\d{4} # then find four more digits
$ # finish at the end of the line (or input)
EDIT: Oops! The above was correct, but it was also too lenient. It would match things like 01088899996 (one too many characters) because it liked the first (or the last) 10 of them. Now it's more strict (I added the ^ and $).
I'm assuming you want a single regex to match any of these examples:
if (preg_match('/(\d{3})[ \-]?(\d{3})[ \-]?(\d{4})/', $value, $matches)) {
$number = $matches[1] . $matches[2] . $matches[3];
}
preg_match('/\d{3}[\s-]?\d{3}[\s-]?\d{4}/', $string);
0108889999 // true
011 8889999 // true
012 888 9999 // true
013-8889999 // true
014-888-9999 // true
To match the specific parts:
preg_match('/(\d{3})[\s-]?(\d{3})[\s-]?(\d{4}/)', $string, $matches);
echo $matches[1]; // first 3 numbers
echo $matches[2]; // next 3 numbers
echo $matches[3]; // next 4 numbers
You can try this pattern. It satisfies your requirements.
[0-9]{3}[-\s]?[0-9]{3}[-\s]?[0-9]{4}
Also, you can add more conditions to the last character by appending [\s.,]+: (phone# ending with space, dot or comma)
[0-9]{3}[-\s]?[0-9]{3}[-\s]?[0-9]{4}[\s.,]+