Regular expression needed for specific string in PHP - regex

I need a regular expression to validate a string with the following conditions
String might contain any of digits space + - () / .
If string contain anything else then it should be invalid
If there is any + in the string then it should be at the beginning and there should at most one + , otherwise it would be invalid, if there are more than one + then it is invalid
String should be 7 to 20 character long
It is not compulsory to have all these digits space + - () / .
But it is compulsory to contain at least 7 digit

I think you are validating phone numbers with E.164 format. Phone number can contain many other format. It can contain . too. Multiple spaces in a number is not uncommon. So its better to format all the numbers to a common format and store that format in db. If that common format is wrong you can throw error.
I validate those phone numbers like this.
function validate_phone($phone){
// replace anything non-digit and add + at beginning
$e164 = "+". preg_replace('/\D+/', '', $phone);
// check validity by length;
return (strlen($e164)>6 && strlen($e164)<21);
}
Here I store $e164 in Db if its valid.
Even after that you can not validate a phone number. A valid phone number format does not mean its a valid number. For this an sms or call is generated against the number and activation code is sent. Once the user inputs the code phone number is fully validated.

You can do this in one regex:
/^(?=(?:.*\d){7})[0-9 ()\/+-][0-9 ()\/-]{6,19}$/
However I would personally do something like:
/^[0-9 ()\/+-][0-9 ()\/-]{6,19}$/
And then strip any non-digit and see if the remaining string is 7 or longer.

Let's try ...
preg_match('/^(?=(?:.*\d){7})[+\d\s()\/\-\.][\d\s()\/\-\.]{6,19}$/', $text);
Breaking this down:
We start with a positive look-ahead that requires a digit at least 7 times.
Then we match all the valid characters, including the plus.
Followed by matching all the valid characters without plus between 6 and 20 times.

A little more concise:
^\+?(?=(.*\d){7})[()/\d-]{7,19}$
'Course, why would you even use regular expressions?
function is_valid($string) {
$digits = 0;
$length = strlen($string);
if($length < 7 || $length > 20) {
return false;
}
for($i = 0; $i < $length; $i++) {
if(ctype_digit($string[$i])) {
$digits++;
} elseif(strpos('+-() ', $string[$i]) === false && ($string[$i] !== '+' || $i !== 0)) {
return false;
}
}
return $digits >= 7;
}

Related

how to write regular expression to see if at least some statements in expression passed

I am new to regular expressions and was wondering how to write a regular expression that would pass if some of the statements in the expression pass.
For example I have this regex
^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*(_|[^\w])).+$
which matches if the string has
1 lowercase character,
1 uppercase character,
1 digit,
1 symbol.
Is it possible to have a regular expression that would pass if at least 3 of the 4 conditions were true in any order?
(i.e. would pass if string had 1 lower, 1 upper, 1 symbol, or 1 upper, 1 digit, 1 symbol, etc.)
Any help is appreciated!
The best approach I can propose you is to capture every type of characters in a capture group and make sure that at least 3/4 capture groups have a value (if the group can't match anything, it should be an empty string).
^(?:([a-z])|([A-Z])|(\d)|(_|[^\w]))+$
You can also add a positive lookahead to make sure that the password have the required length (by example 8 to 32 characters).
^(?=.{8,32}$)(?:([a-z])|([A-Z])|(\d)|(_|[^\w]))+$
Edit: ([\W_]) is the equivalent of (_|[^\w]). Putting the "W" in upper case reverse it sense (match all the non-word characters). Moreover, using a single character class is faster than alternation (more details here)
If you are willing to use javascript, I adapted a function presented in "Regular Expression cookbook second edition" for the needs of my website:
var PASSWORD_RANKING = {
TOO_SHORT: 0,
WEAK: 1,
MEDIUM: 2,
STRONG: 3,
VERY_STRONG: 4
};
/**
* Take a password and returns it's ranking
* based of the strength of the password (length, numeric character, alphabetic character, special character, etc.)
*
* #param password String
* #param minLength Int
*
* #return Int
*/
function rankPassword(password, minLength){
var rank = PASSWORD_RANKING.TOO_SHORT;
var score = 0;
if (typeof minLength !== 'number' || minLength < 6){
minLength = 8;
}
if (typeof password === 'string' && password.length >= minLength){
if (/[A-Z]/.test(password)){ score++;}
if (/[a-z]/.test(password)){ score++;}
if (/[0-9]/.test(password)){ score++;}
if (/[_~!#.#$%^&]/.test(password)){ score++;}
score += Math.floor((password.length - minLength) / 2);
if (score < 3){
rank = PASSWORD_RANKING.WEAK;
}
else if (score < 4){
rank = PASSWORD_RANKING.MEDIUM;
}
else if (score < 6){
rank = PASSWORD_RANKING.STRONG;
}
else {
rank = PASSWORD_RANKING.VERY_STRONG;
}
}
return rank;
}
The section 4.19 present many regexes to enforce password strength. You can see all the code samples online :
http://examples.oreilly.com/0636920023630/Regex_Cookbook_2_Code_Samples.html

Regular expression for comma seperated name search with wild card

Right now I am using multiple if conditions to valid the input for search by name with wildcard(*). Since I have multiple 'if' with inner 'if' statements I am trying to use regular expression to validate my input. I want to use this expression in both front end and back end.
Appreciate if anyone can help.
Validating rules are follow
Input is last name, first name i.e. separated by comma.
Must have at least two characters while using wild card search.
Valid wildcard character is '*' only.
At most two wildcard characters can be used.
No consecutive wild cards.
If no wild card used no constraint on length of characters in both last and first name.
Some of the valid inputs are:
- hopkins, johns
- h, j
- ho*, jp*
- *ins, johns
- *op*, john*
Some of the invalid inputs are:
- hopkins johns
- h*, johns
- hop**, joh*
- h*pk*n*
If regular expression not going to be complex we can consider this as valid otherwise it OK to consider this as invalid
- ho*in*, jo*
In short general name format is
[*]XX[*], [*]XX[*]
where [] ==> Optional
X ==> A-Z, a-z
XX ==> length 2 or more if wild card used
You can use this regex
\*?[a-zA-Z]{2,}\*?, \*?[a-zA-Z]{2,}\*?
The before doing validation with the above regex, just do something like match the number of * with the regex /\*/g and make sure it's length is between 0 to 2.
With the help of #Amit_Joki answer I wrote the following code and its working fine.
var nameArray = [...];
var re = /\*?[a-zA-Z]{2,}\*?, \*?[a-zA-Z]{2,}\*?/;
for (var i = 0; i < nameArray.length; i++) {
if(nameArray[i].indexOf(',') < 0 ||
(nameArray[i].indexOf('*') >= 0 && !re.test(nameArray[i]))) {
console.log(nameArray[i] + ": Invalid");
} else {
console.log(nameArray[i] + ": Valid");
}
}

as3 regular expression to validate a password

I have a form that asks for a password and I want to validate if the password has at least eight characters, and of these eight characters at least two must be numbers and two must be letters in any order. I'm trying with this:
function validatePassword():void
{
var passVal:String = pass.text;
if(validPass(passVal))
{
trace("Password Ok");
sendForm();
}
else
{
trace("You have entered an invalid password");
}
function validPass(passVal:String):Boolean{
var pw:RegExp = /^?=.{8,}[A-Za-z]{2,}[0-9]{2,}/;
return(pw.test(passVal));
}
}
But it doesn't work. What I'm doing wrong?
Any help would be really appreciated!
use this pattern ^(?=.{8})(?=(.*\d){2})(?=(.*[A-Za-z]){2}).*$
^ anchor
(?=.{8}) look ahead for at least 8 characters
(?=(.*\d){2}) look ahead for at least 2 digits in any order
(?=(.*[A-Za-z]){2}) look ahead for at least 2 letters in any order
.*$ catch everything to the end if passed previous conditions
The problem is that your regex is forcing the numbers to follow the letters ([A-Za-z]{2,}[0-9]{2,}). While it is possible to write such a regex, I suggest using a simple length check and two regexes:
function validPass(passVal:String):Boolean{
if (passVal.length < 8)
return False;
var letterRegex:RegExp = /^.*?[A-Za-z].*?[A-Za-z].*?$/;
var numberRegex:RegExp = /^.*?\d.*?\d.*?$/;
return letterRegex.test(passVal) && numberRegex.test(passVal);
}

Reg expression validate / \ # & characters

I've been learning how Regular expressions work, which is very tricky for me. I would like to validate this chars below from input field. Basically if string contains any of these characters, alert('bad chars')
/
\
#
&
I found this code, but when I change it around doesn't seem to work. How can I alter this code to meet my needs?
var str = $(this).val();
if(/^[a-zA-Z0-9- ]*$/.test(str) == false) {
alert('bad');
return false;
} else {
alert('good');
}
/^[a-zA-Z0-9- ]*$/ means the following:
^ the string MUST start here
[a-zA-Z0-9- ] a letter between a and z upper or lower case, a number between 0 and 9, dashes (-) and spaces.
* repeated 0 or more times
$ the string must end here.
In the case of "any character but" you can use ^ like so: /^[^\/\\#&]*$/. If this matches true, then it doesn't have any of those characters. ^ right after a [ means match anything that isn't the following.
.
You could just try the following:
if("/[\\/#&]/".test(str) == true) {
alert('bad');
return false;
} else {
alert('good');
}
NOTE: I'm not 100% on what characters need to be escaped in JavaScript vs. .NET regular expressions, but basically, I'm saying if your string contains any of the characters \, /, # or &, then alert 'bad'.

Regular Expression to find numbers with same digits in different order

I have been looking for a regular expression with Google for an hour or so now and can't seem to work this one out :(
If I have a number, say:
2345
and I want to find any other number with the same digits but in a different order, like this:
2345
For example, I match
3245 or 5432 (same digits but different order)
How would I write a regular expression for this?
There is an "elegant" way to do it with a single regex:
^(?:2()|3()|4()|5()){4}\1\2\3\4$
will match the digits 2, 3, 4 and 5 in any order. All four are required.
Explanation:
(?:2()|3()|4()|5()) matches one of the numbers 2, 3, 4, or 5. The trick is now that the capturing parentheses match an empty string after matching a number (which always succeeds).
{4} requires that this happens four times.
\1\2\3\4 then requires that all four backreferences have participated in the match - which they do if and only if each number has occurred once. Since \1\2\3\4 matches an empty string, it will always match as long as the previous condition is true.
For five digits, you'd need
^(?:2()|3()|4()|5()|6()){5}\1\2\3\4\5$
etc...
This will work in nearly any regex flavor except JavaScript.
I don't think a regex is appropriate. So here is an idea that is faster than a regex for this situation:
check string lengths, if they are different, return false
make a hash from the character (digits in your case) to integers for counting
loop through the characters of your first string:
increment the counter for that character: hash[character]++
loop through the characters of the second string:
decrement the counter for that character: hash[character]--
break if any count is negative (or nonexistent)
loop through the entries, making sure each is 0:
if all are 0, return true
else return false
EDIT: Java Code (I'm using Character for this example, not exactly Unicode friendly, but it's the idea that matters now):
import java.util.*;
public class Test
{
public boolean isSimilar(String first, String second)
{
if(first.length() != second.length())
return false;
HashMap<Character, Integer> hash = new HashMap<Character, Integer>();
for(char c : first.toCharArray())
{
if(hash.get(c) != null)
{
int count = hash.get(c);
count++;
hash.put(c, count);
}
else
{
hash.put(c, 1);
}
}
for(char c : second.toCharArray())
{
if(hash.get(c) != null)
{
int count = hash.get(c);
count--;
if(count < 0)
return false;
hash.put(c, count);
}
else
{
return false;
}
}
for(Integer i : hash.values())
{
if(i.intValue()!=0)
return false;
}
return true;
}
public static void main(String ... args)
{
//tested to print false
System.out.println(new Test().isSimilar("23445", "5432"));
//tested to print true
System.out.println(new Test().isSimilar("2345", "5432"));
}
}
This will also work for comparing letters or other character sequences, like "god" and "dog".
Put the digits of each number in two arrays, sort the arrays, find out if they hold the same digits at the same indices.
RegExes are not the right tool for this task.
You could do something like this to ensure the right characters and length
[2345]{4}
Ensuring they only exist once is trickier and why this is not suited to regexes
(?=.*2.*)(?=.*3.*)(?=.*4.*)(?=.*5.*)[2345]{4}
The simplest regular expression is just all 24 permutations added up via the or operator:
/2345|3245|5432|.../;
That said, you don't want to solve this with a regex if you can get away with it. A single pass through the two numbers as strings is probably better:
1. Check the string length of both strings - if they're different you're done.
2. Build a hash of all the digits from the number you're matching against.
3. Run through the digits in the number you're checking. If you hit a match in the hash, mark it as used. Keep going until you don't get an unused match in the hash or run out of items.
I think it's very simple to achieve if you're OK with matching a number that doesn't use all of the digits. E.g. if you have a number 1234 and you accept a match with the number of 1111 to return TRUE;
Let me use PHP for an example as you haven't specified what language you use.
$my_num = 1245;
$my_pattern = '/[' . $my_num . ']{4}/'; // this resolves to pattern: /[1245]{4}/
$my_pattern2 = '/[' . $my_num . ']+/'; // as above but numbers can by of any length
$number1 = 4521;
$match = preg_match($my_pattern, $number1); // will return TRUE
$number2 = 2222444111;
$match2 = preg_match($my_pattern2, $number2); // will return TRUE
$number3 = 888;
$match3 = preg_match($my_pattern, $number3); // will return FALSE
$match4 = preg_match($my_pattern2, $number3); // will return FALSE
Something similar will work in Perl as well.
Regular expressions are not appropriate for this purpose. Here is a Perl script:
#/usr/bin/perl
use strict;
use warnings;
my $src = '2345';
my #test = qw( 3245 5432 5542 1234 12345 );
my $canonical = canonicalize( $src );
for my $candidate ( #test ) {
next unless $canonical eq canonicalize( $candidate );
print "$src and $candidate consist of the same digits\n";
}
sub canonicalize { join '', sort split //, $_[0] }
Output:
C:\Temp> ks
2345 and 3245 consist of the same digits
2345 and 5432 consist of the same digits