i have this function to check my phone number:
function isValid( $what, $data ) {
switch( $what ) {
// validate a phone number
case 'phone_number':
$pattern = "/^[0-9-+]+$/";
break;
default:
return false;
break;
}
return preg_match($pattern, $data) ? true : false;
}
i want to change that regex to accept the following: the ) ( chars like (800) and the space.
So for example this number will pass the validation, right now is not passing:
+1 (201) 223-3213
Let us construct the regular expression step by step. Consider also that spaces are trimmed before matching.
at the beginning there might or might not be a + sign. This also needs to be escaped. \+?
then comes one or more digits, before the part with parenthesis [0-9]+ You might want to write [0-9]* if the number can begin directly with a group in parenthesis
then, optionally comes a group of digits in parenthesis: (\[0-9]+\)?. Suppose that only one such group is allowed
then comes the local phone number, hyphens also allowed: [0-9-]*
the final character must be a digit [0-9], hyphen is not allowed here
^\+?[0-9]+(\([0-9]+\))?[0-9-]*[0-9]$
See the result here. Trimming spaces looks like $trimmed = str_replace(' ', '', $pattern);.
How about this regexp:
/^[0-9-+()\s]+$/
See it in action here
'/\(?\b[0-9]{3}\)?[-. ]?[0-9]{3,5}[-. ]?[0-9]{4,8}\b/'
Since you seem to be using this for validation you can use str_replace('[\s\+\-\(\)]', '', $data) to get a string that should (if the phone number is valid) contain only digits. You can then test this assumption easily by running preg_match('\d{11}', $data) (the {11} means 11 digits, if there's a range allowed, use min, max like this {min,max}, e.g. \d{10,11}).
It's worth noting that this isn't as thorough as Lorlin's answer in that you're ignoring any invalid use of brackets, +s or -s. You may want to use a combination of the two, or whatever suits your needs the best.
Related
I've asked and it was answered but now, after years, it doesn't work.
I've even tried online regex validators. Not sure what is going on.
Version: PHP 7.0.30 on 64Bit OS
The string should only allow digits with commas.
No commas in the beginning or end.
Spaces between commas is ok but I'd rather not allow it.
The following isn't passing
My regex is:
$DateInvoicedIDs = "1031,453,808,387,111,342,962,706,251,442,362,858,950,738,310,288,99,665,1023,30,894,112,132,148,347,895,382,94,766,683,276,1104,658,34,348,235,786,769,2";
$reg = '/[0-9\s]+(,[0-9\s]+)*[0-9]$/';
if ( preg_match($reg, $DateInvoicedIDs) ) {
echo = $DateInvoicedIDs;
} else { echo "false"; }
I'm using preg_match and getting false.
Any idea?
Test your string and pattern # https://regex101.com/r/3TVmOv/1
When that loads, you will see that there is no match highlighted.
Then add a digit to the end of your string and Whalla! This is because (,[0-9\s]+)* is matching the final 2 and [0-9]$ cannot be satisfied because another digit is required.
If I understand your logic/requirements, I think I'd use ~^\d+(?:\s*,\s*\d+)*$~
This improves the validation because it doesn't allow a mixture of digits and spaces between commas like: 2, 3 4 56, 72 I don't think you want spaces in your comma-separated numerical values.
Pattern Demo
Code: (Demo)
$DateInvoicedIDs = "1031,453,808,387,111,342,962,706,251,442,362,858,950,738,310,288,99,665,1023,30,894,112,132,148,347,895,382,94,766,683,276,1104,658,34,348,235,786,769,2";
$reg = '/^\d+(?:\s*,\s*\d+)*$/';
if (preg_match($reg, $DateInvoicedIDs)) {
echo $DateInvoicedIDs;
} else {
echo "false";
}
It is not matching because of the last [0-9] in your regex. The * in (,[0-9\s]+)* is a greedy match which means that it is consuming all commas followed by digits in your string. There is nothing left after to match against the last [0-9].
So you probably want to reduce your regex to '/[0-9\s]+(,[0-9\s]+)*$/.
The last part of your regex [0-9]$ is what's causing it to fail:
[0-9\s]+ is matching the first number only 1031,
(,[0-9\s]+)* is covering everything until ,2 because it's a single number right after a comma which is what it's looking for
Then [0-9]$ is trying to find one more number but it can't
If the last number is a double-digit number, i.e. ,25 instead of 2, then the that second part (,[0-9\s]+)* would be satisfied because it found at least one number and [0-9]$ would match the next number which is 5 (https://regex101.com/r/0XbHsw/1)
Adding ? for that last part would solve the problem: [0-9\s]+(,[0-9\s]+)*[0-9]?$
I have many Strings looking like this:
QR-DF-6549-1 and QR-DF-6549
I want to get these parts of the strings:
DF-6549
Edit:
So I also want to get rid of the "-1" at the end, in case it exists.
How can I do this with php? I know substr but I am a bit lost at this one.
Thank you very much!
A regular expression is probably the best way given your sample data
// ^ start matching at start of string
// [A-Z]{2}- must start with two capital letters and a dash
// ( we want to capture everything that follows
// [A-Z]{2}- next part must start with two capital letters and a dash
// \d+ a sequence of one or more digits
// ) end the capture - this will be index 1 in the $match array allowed
if (preg_match('/^[A-Z]{2}-([A-Z]{2}-\d+)/', $str, $match)) {
$data=$match[1];
}
I'm trying to make a function to verify names on PHP using Regex, I want the names to be able to carry infinite amount of spaces and ' and -, and to allow only capital characters after spaces but to allow capital and none capitals after - and '.. Also the total length should be of 50 characters and the name should end with a lowercase, note that the uppercases are A to Z plus those characters :
ÙÒÌÈÀÁÉÍßÓÚÝÂÊÎÔÛÃÑÕÄÅÆŒÇÐØËÏÖÜŸ
and the lower cases are a to z plus those characters :
éçàèàèìòùáéíóúýâêîôûãñõäëïöüÿåæœçðøß
each word (between a space , ' or - and another) should count at least 2 characters the name should also start with an uppercase and finish with a lower case and in words (between a space , ' or - and another) no uppercases but that of the beginning is allowed
Examples of acceptable names are :
Adam Klsld
Adam'odskdl
Adam'Ddlsl
Ùdam-ddkkdk
Addssd-Ddsdsd
I've been trying a lot but here's my last try that I still keep in my php file, the others I've deleted in the chaos of non-successful attempts (using mb_ereg function to match, so this is a posix-ere):
([A-ZÙÒÌÈÀÁÉÍßÓÚÝÂÊÎÔÛÃÑÕÄÅÆŒÇÐØËÏÖÜŸ][a-zéçàèàèìòùáéíóúýâêîôûãñõäëïöüÿåæœçðøß]+){1}((^[\'\-\s])[A-ZÙÒÌÈÀÁÉÍßÓÚÝÂÊÎÔÛÃÑÕÄÅÆŒÇÐØËÏÖÜŸ][a-zéçàèàèìòùáéíóúýâêîôûãñõäëïöüÿåæœçðøß]+)*
(this does not necessarily mean it's the best attempt but I though it may help and give an idea on how much of a dork am I)
I wouldn't exactly suggest you use this... but I think this does what you want?
^([A-ZÙÒÌÈÀÁÉÍßÓÚÝÂÊÎÔÛÃÑÕÄÅÆŒÇÐØËÏÖÜŸ][a-zéçàèàèìòùáéíóúýâêîôûãñõäëïöüÿåæœçðøß]+){1}((([\s])[A-ZÙÒÌÈÀÁÉÍßÓÚÝÂÊÎÔÛÃÑÕÄÅÆŒÇÐØËÏÖÜŸ][a-zéçàèàèìòùáéíóúýâêîôûãñõäëïöüÿåæœçðøß]+)|((['\-])([A-ZÙÒÌÈÀÁÉÍßÓÚÝÂÊÎÔÛÃÑÕÄÅÆŒÇÐØËÏÖÜŸ]|[a-zéçàèàèìòùáéíóúýâêîôûãñõäëïöüÿåæœçðøß])[a-zéçàèàèìòùáéíóúýâêîôûãñõäëïöüÿåæœçðøß]+))*$
Here it is in a non-code block so you can see how insane it is... think it strips some characters here though:
^([A-ZÙÒÌÈÀÁÉÍßÓÚÝÂÊÎÔÛÃÑÕÄÅÆŒÇÐØËÏÖÜŸ][a-zéçàèàèìòùáéíóúýâêîôûãñõäëïöüÿåæœçðøß]+){1}((([\s])[A-ZÙÒÌÈÀÁÉÍßÓÚÝÂÊÎÔÛÃÑÕÄÅÆŒÇÐØËÏÖÜŸ][a-zéçàèàèìòùáéíóúýâêîôûãñõäëïöüÿåæœçðøß]+)|((['-])([A-ZÙÒÌÈÀÁÉÍßÓÚÝÂÊÎÔÛÃÑÕÄÅÆŒÇÐØËÏÖÜŸ]|[a-zéçàèàèìòùáéíóúýâêîôûãñõäëïöüÿåæœçðøß])[a-zéçàèàèìòùáéíóúýâêîôûãñõäëïöüÿåæœçðøß]+))*$
Is this Regex answering what you need to check ?
(You'll have to add the weird characters inside each brackets of course).
You can use this to avoid accented characters issue:
$pattern = "~^[\p{Lu}ß]\p{Ll}*+(?>(?> [\p{Lu}ß]|['-]\p{L})\p{Ll}*+)*$~u";
if(preg_match($pattern, $name)) { ...
Or for a more specific set of characters:
$pattern = "~(?(DEFINE)(?<Up>[A-ZÙÒÌÈÀÁÉÍßÓÚÝÂÊÎÔÛÃÑÕÄÅÆŒÇÐØËÏÖÜŸ]))
(?(DEFINE)(?<Lo>[a-zéçàèàèìòùáéíóúýâêîôûãñõäëïöüÿåæœçðøß]))
^\g<Up>\g<Lo>*+(?>(?>\h\g<Up>|['-]\g<Up>?+\g<Lo>)\g<Lo>*+)*+$~ux";
if (preg_match($pattern, $name, $matches)) { ...
or the same in a shorter way:
$pattern = "~(?(DEFINE)(?<Up>[A-ZÀ-ÖØ-ݟߌ]))
(?(DEFINE)(?<Lo>[a-zà-öø-ýÿßœ]))
^\g<Up>\g<Lo>*+(?>(?>\h\g<Up>|['-]\g<Up>?+\g<Lo>)\g<Lo>*+)*+$~ux";
I want to make a bid system on a website. That means users can post their bid (natural number). I want to make sure users don't try to post characters, decimal numbers, etc.
I don't want to use is_numeric function because hexadecimal notation is allowed.
I was thinking to use preg_match for this. But in php.net the documentation for this function is little and I have no idea how to use preg_match.
So how should I check if a variable is a natural number with preg_match?
If you don't require decimal points: ctype_digit or filter_var($var, FILTER_VALIDATE_INT).
If you do: filter_var($var, FILTER_VALIDATE_FLOAT).
ctype_digit does what you want:
Checks if all of the characters in the provided string, text, are numerical.
(Before PHP 5.1.0, this function returned TRUE when text was an empty string.)
Either preg_match('/^[0-9]+$/', $var); or ctype_digit
I would generally caution against using regex for parsing numerics, as there are generally better solutions than regex for this, but since you're asking, I'll try to give you some assistance with it:
preg_match uses regular expressions (regex) for it's matching.
You can find out more about regex syntax at sites like http://www.regular-expressions.info/
If you want to match a digit in regex, you can either use [0-9] or \d.
If you want to match one or more of anything, you would use a plus sign.
Finally, regex strings need to be enclosed in a pair of characters. The character chosen is usually a slash (/) character, as some languages specifically require this character, but PHP also allows other characters to be used; tilde (~) is quite common.
So your regex string to match any number of digits would be "/\d+/". This can then be put into a preg_match call like so:
$isnumeric = preg_match("/\d+/",$input_string);
If you have more specific requirements, you can limit the number of characters allowed by replacing the plus sign with {max} or {min,max} where 'min' and 'max' are the number of times the preceding match is allowed. So to allow a number between two and six digits long, you would use this:
$isnumeric = preg_match("/\d{2,6}/",$input_string);
If you need to allow a decimal point, you need to know that the dot character is a special character in regex (it means 'match any character at all'), so you need to escape it with a back-slash.
Therefore, a regex to match a currency amount with two decimal places, and at least one digit before the point would be like this:
$isnumeric = preg_match("/\d+\.\d\d/",$input_string);
Finally, note that regex will return true in all the above if the string simply contains the matched value. To ensure it doesn't contain anything else, you would need to 'anchor' it to the front and end of the string, using the anchor characters: ^ for the start of the string, and $ for the end.
So for the previous example, if you want it to only contain a decimal number, and nothing else, you would need this:
$isnumeric = preg_match("/^\d+\.\d\d$/",$input_string);
Regex is a complex subject, but I hope that gives you a start.
I know this is very old but I wanted to share the next solution in case someone else comes up with this problem.
I'm assuming that by natural number you meant positive integer (which excludes the number 0).
function is_positive_integer( $value ) {
// Check if is integer and greater than zero
if( is_int( $value ) && $value > 0 ) {
return true;
}
// Is not a positive integer
else {
return false;
}
}
This kind of depends on your definition of natural numbers - according to different theories, the number zero (0) does or does not count as a natural number.
To answer your question on how to solve this with preg_match:
If you want to include zero, using preg_match is pretty easy preg_match('^[0-9]+$', $input).
Usage:
if (preg_match('^[0-9]+$', $input))
// $input represents a non-negative numeric value
else
// $input does not represent a non-negative numeric value
If you don't want to include the zero, use preg_match('^[1-9][0-9]*$', $input):
if (preg_match('^[1-9][0-9]*$', $input))
// $input represents a positive numeric value
else
// $input does not represent a positive numeric value
That said - for your particular problem, using ctype_digit is a faster solution, as others already pointed out (you'd have to do a second check if you don't want to allow the number zero).
in_array(str_replace(str_split('0123456789'), '', $s), array(',','.',''));
simple function:
function isnature($x){
$y = ceil($x)-floor($x);
return $y == 0 ? true : false;
}
From a mathematical point of view, a natural number is a positive integer, including zero, so you could check it like this:
is_int($bid) && $bid >= 0
Simplest and faster
if( is_numeric( $key ) && intval( $key ) == $key )
{
//key == number
}
I have a PHP array of strings. The strings are supposed to represent PIN codes which are of 6 digits like:
560095
Having a space after the first 3 digits is also considered valid e.g. 560 095.
Not all array elements are valid. I want to filter out all invalid PIN codes.
Yes you can make use of regex for this.
PHP has a function called preg_grep to which you pass your regular expression and it returns a new array with entries from the input array that match the pattern.
$new_array = preg_grep('/^\d{3} ?\d{3}$/',$array);
Explanation of the regex:
^ - Start anchor
\d{3} - 3 digits. Same as [0-9][0-9][0-9]
? - optional space (there is a space before ?)
If you want to allow any number of any whitespace between the groups
you can use \s* instead
\d{3} - 3 digits
$ - End anchor
Yes, you can use a regular expression to make sure there are 6 digits with or without a space.
A neat tool for playing with regular expressions is RegExr... here's what RegEx I came up with:
^[0-9]{3}\s?[0-9]{3}$
It matches the beginning of the string ^, then any three numbers [0-9]{3} followed by an optional space \s? followed by another three numbers [0-9]{3}, followed by the end of the string $.
Passing the array into the PHP function preg_grep along with the Regex will return a new array with only matching indeces.
If you just want to iterate over the valid responses (loop over them), you could always use a RegexIterator:
$regex = '/^\d{3}\s?\d{3}$/';
$it = new RegexIterator(new ArrayIterator($array), $regex);
foreach ($it as $valid) {
//Only matching items will be looped over, non-matching will be skipped
}
It has the benefit of not copying the entire array (it computes the next one when you want it). So it's much more memory efficient than doing something with preg_grep for large arrays. But it also will be slower if you iterate multiple times (but for a single iteration it should be faster due to the memory usage).
If you want to get an array of the valid PIN codes, use codaddict's answer.
You could also, at the same time as filtering only valid PINs, remove the optional space character so that all PINs become 6 digits by using preg_filter:
$new_array = preg_filter('/^(\d{3}) ?(\d{3})$/D', '$1$2', $array);
The best answer might depend on your situation, but if you wanted to do a simple and low cost check first...
$item = str_replace( " ", "", $var );
if ( strlen( $item ) !== 6 ){
echo 'fail early';
}
Following that, you could equally go on and do some type checking - as long as valid numbers did not start with a 0 in which case is might be more difficult.
If you don't fail early, then go on with the regex solutions already posted.