I need to validate a password with these rules:
6 to 20 characters
Must contain at least one digit;
Must contain at least one letter (case insensitive);
Can contain the following characters: ! # # $ % & *
The following expression matches all but the last requirement. What can I do with the last one?
((?=.*\d)(?=.*[A-z]).{6,20})
I'm not completely sure I have this right, but since your last requirement is "Can contain the following characters: !##$%&*" I am assuming that other special characters are not allowed. In other words, the only allowed characters are letters, digits, and the special characters !##$%&*.
If this is the correct interpretation, the following regex should work:
^((?=.*\d)(?=.*[a-zA-Z])[a-zA-Z0-9!##$%&*]{6,20})$
Note that I changed your character class [A-z] to [a-zA-Z], because [A-z] will also include the following characters: [\]^_`
I also added beginning and end of string anchors to make sure you don't get a partial match.
^(?=.*\d)(?=.*[a-zA-Z])[a-zA-Z0-9!##$%&*]{6,20}$
Regex could be:-
^(?=.*\d)(?=.*[a-zA-Z])[a-zA-Z0-9!##$%&*]{6,20}$
How about this in Javascript:-
function checkPwd(str) {
if (str.length < 6) {
return("too_short");
} else if (str.length > 20) {
return("too_long");
} else if (str.search(/\d/) == -1) {
return("no_num");
} else if (str.search(/[a-zA-Z]/) == -1) {
return("no_letter");
} else if (str.search(/[^a-zA-Z0-9\!\#\#\$\%\^\&\*\(\)\_\+]/) != -1) {
return("bad_char");
}
return("ok");
}
Also check out this
Related
I have the following rules for password validation:
at least 6 characters
at least 1 capital letter
How can I validate this with a RegEx?
Here is my pattern: ^(?=.*[0-9]+.*)(?=.*[a-zA-Z]+.*)[0-9a-zA-Z]{6,}$
The pattern above enforces also numbers ... which I don't need. However user can enter any other characters he/she wishes except that must contain one capital and be longer or equal to 6 characters.
You can try this:
^(?=.*?[A-Z]).{6,}$
DEMO
If you want to allow special characters as well then change it like
^(?=.*?[A-Z])(?=.*?[#?!#$%^&*-]).{6,}$
^(?=.*[A-Z])[\w-[_\]]{6,}$
This force to contains a capital letter with (?=.*[A-Z]) and allow alphanumeric : [\w-[_\]] (\w is [a-zA-Z0-9_] in which we remove the _
You can use this :
(?=.*[A-Z]).{8}
or If you want to full rule, can try this :
(?=[a-z]*[A-Z])(?=\D*\d)(?=.*[#?!#$%^&*-]).{8}
It might be better to evaluate the rules separately to provide feedback to user as to which rule failed to validate.
I suggest doing something like this instead: https://jsfiddle.net/hejecr9d/
// will store list of rules
var rules = []
// at least one capital letter rule
rules.one_capital_letter = {
fail_message: 'password needs at least one capital letter',
validate: function(password){
return /[A-Z]/.test(password)
}
}
// at least six letters rule
rules.at_least_six_letters = {
fail_message: 'password needs to have at least 6 characters',
validate: function(password){
return password.length >= 6
}
}
function validate_password(password) {
// goes through all the rules
for(rule in rules) {
if(!rules[rule].validate(password)) {
return {success: false, message: rules[rule].fail_message}
}
}
// return success after all rules are checked
return {success: true, message: 'password validated successfully'}
}
// Test if it works
var test_strings = ['abcd', 'Abcd', 'Abcdef']
for(i in test_strings) {
$('body').append(test_strings[i] + ': ' + validate_password(test_strings[i]).message + '<br>')
}
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'.
Is it possible to do a find/replace using regular expressions on a string of dna such that it only considers every 3 characters (a codon of dna) at a time.
for example I would like the regular expression to see this:
dna="AAACCCTTTGGG"
as this:
AAA CCC TTT GGG
If I use the regular expressions right now and the expression was
Regex.Replace(dna,"ACC","AAA") it would find a match, but in this case of looking at 3 characters at a time there would be no match.
Is this possible?
Why use a regex? Try this instead, which is probably more efficient to boot:
public string DnaReplaceCodon(string input, string match, string replace) {
if (match.Length != 3 || replace.Length != 3)
throw new ArgumentOutOfRangeException();
var output = new StringBuilder(input.Length);
int i = 0;
while (i + 2 < input.Length) {
if (input[i] == match[0] && input[i+1] == match[1] && input[i+2] == match[2]) {
output.Append(replace);
} else {
output.Append(input[i]);
output.Append(input[i]+1);
output.Append(input[i]+2);
}
i += 3;
}
// pick up trailing letters.
while (i < input.Length) output.Append(input[i]);
return output.ToString();
}
Solution
It is possible to do this with regex. Assuming the input is valid (contains only A, T, G, C):
Regex.Replace(input, #"\G((?:.{3})*?)" + codon, "$1" + replacement);
DEMO
If the input is not guaranteed to be valid, you can just do a check with the regex ^[ATCG]*$ (allow non-multiple of 3) or ^([ATCG]{3})*$ (sequence must be multiple of 3). It doesn't make sense to operate on invalid input anyway.
Explanation
The construction above works for any codon. For the sake of explanation, let the codon be AAA. The regex will be \G((?:.{3})*?)AAA.
The whole regex actually matches the shortest substring that ends with the codon to be replaced.
\G # Must be at beginning of the string, or where last match left off
((?:.{3})*?) # Match any number of codon, lazily. The text is also captured.
AAA # The codon we want to replace
We make sure the matches only starts from positions whose index is multiple of 3 with:
\G which asserts that the match starts from where the previous match left off (or the beginning of the string)
And the fact that the pattern ((?:.{3})*?)AAA can only match a sequence whose length is multiple of 3.
Due to the lazy quantifier, we can be sure that in each match, the part before the codon to be replaced (matched by ((?:.{3})*?) part) does not contain the codon.
In the replacement, we put back the part before the codon (which is captured in capturing group 1 and can be referred to with $1), follows by the replacement codon.
NOTE
As explained in the comment, the following is not a good solution! I leave it in so that others will not fall for the same mistake
You can usually find out where a match starts and ends via m.start() and m.end(). If m.start() % 3 == 0 you found a relevant match.
I am looking for a perl regex which will validate a string containing only the letters ACGT. For example "AACGGGTTA" should be valid while "AAYYGGTTA" should be invalid, since the second string has "YY" which is not one of A,C,G,T letters. I have the following code, but it validates both the above strings
if($userinput =~/[A|C|G|T]/i)
{
$validEntry = 1;
print "Valid\n";
}
Thanks
Use a character class, and make sure you check the whole string by using the start of string token, \A, and end of string token, \z.
You should also use * or + to indicate how many characters you want to match -- * means "zero or more" and + means "one or more."
Thus, the regex below is saying "between the start and the end of the (case insensitive) string, there should be one or more of the following characters only: a, c, g, t"
if($userinput =~ /\A[acgt]+\z/i)
{
$validEntry = 1;
print "Valid\n";
}
Using the character-counting tr operator:
if( $userinput !~ tr/ACGT//c )
{
$validEntry = 1;
print "Valid\n";
}
tr/characterset// counts how many characters in the string are in characterset; with the /c flag, it counts how many are not in the characterset. Using !~ instead of =~ negates the result, so it will be true if there are no characters not in characterset or false if there are characters not in characterset.
Your character class [A|C|G|T] contains |. | does not stand for alternation in a character class, it only stands for itself. Therefore, the character class would include the | character, which is not what you want.
Your pattern is not anchored. The pattern /[ACGT]+/ would match any string that contains one or more of any of those characters. Instead, you need to anchor your pattern, so that only strings that contain just those characters from beginning to end are matched.
$ can match a newline. To avoid that, use \z to anchor at the end. \A anchors at the beginning (although it doesn't make a difference whether you use that or ^ in this case, using \A provides a nice symmetry.
So, you check should be written:
if ($userinput =~ /\A [ACGT]+ \z/ix)
{
$validEntry = 1;
print "Valid\n";
}
I need a regular expression to validate a string with the following conditions
String might contain any of digits space + - () / .
If string contain anything else then it should be invalid
If there is any + in the string then it should be at the beginning and there should at most one + , otherwise it would be invalid, if there are more than one + then it is invalid
String should be 7 to 20 character long
It is not compulsory to have all these digits space + - () / .
But it is compulsory to contain at least 7 digit
I think you are validating phone numbers with E.164 format. Phone number can contain many other format. It can contain . too. Multiple spaces in a number is not uncommon. So its better to format all the numbers to a common format and store that format in db. If that common format is wrong you can throw error.
I validate those phone numbers like this.
function validate_phone($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;
}