This question already has answers here:
RegEx for Javascript to allow only alphanumeric
(22 answers)
Closed 8 years ago.
I am trying to interrupt a form submission to test field for alphanumeric characters. After googling i found a lot of this implementation for regex that everyone claims works great...
if( !jQuery('input[name=myUsername]').val().test(/^[a-zA-Z0-9]+/) ) {
alert('ERROR: Your username contains invalid characters. Please use numbers and letters only. No spaces, no punctuation.');
jQuery('input[name=myUsername]').focus();
return false;
}
However, this ONLY returns false and creates an alert if the value STARTS with a non-alphanumeric character. If i enter "bo$$" it allows it as alphanumeric even tho $ is not an alphanumeric character... If i enter "$$ob" it fails and alerts.
How do i fix this so that it will fail for ANY invalid character in the entire value? I've tried .match() instead of .test() but same issue im assuming it's something in the regex that is wrong
^[a-zA-Z0-9]+$
put $ the end anchor to limit that.Without $ your regex will make a partial match.
$ assert position at end of a line
bo$$ will make a partial match upto bo as $ is not there.
Use anchor $ to avoid matching unwanted character at the end:
/^[a-zA-Z0-9]+$/
Related
This question already has answers here:
How do I make part of a regex match optional?
(2 answers)
Closed 1 year ago.
I should allow 2 different input strings formats, with each their own validation.
So eg:
AA2222222222222222
and
2222222222222222
This means that if the first character is a letter, I should validate for ^[a-zA-Z]{2}\d{16}$. If the first character is numeric, I should validate for d{16}.
I tried to write it in an conditional regex:
^(([a-zA-Z])(?([a-zA-Z])^[a-zA-Z]{2}\d{16}$|d{16})
but I get a pattern error and can't figure out what exactly is wrong.
Any insight would be apreciated
I tried to write it in an conditional regex
JavaScript doesn't support regular expression conditional syntax, so (?ifthen|else) doesn't work in JavaScript.
This means that if the first character is a letter, I should validate for ^[a-zA-Z]{2}\d{16}$. If the first character is numeric, I should validate for d{16}.
Since the \d{16} part is the same, you can just make the [a-zA-Z]{2} part optional:
/^(?:[a-zA-Z]{2})?\d{16}$/
That uses a non-capturing group around the [a-zA-Z]{2} and makes the entire group optional via the ? after it.
If the validation were different (say, maybe the version with the letters at the start only does \d{14}), you could use an alternation:
/^(?:[a-zA-Z]{2}\d{14}|\d{16})$/
(Beware the gotcha: Without the non-capturing around around the alternation, the ^ would be part of the first alternative but not the second, and the $ would be part of the second alternative, but not the first.)
But in your specific case, you don't need that.
This question already has answers here:
How can I validate an email address using a regular expression?
(79 answers)
Closed 3 years ago.
I have following regex string to check for valid email formats
/^(([^<>()[\]\\.,;:\s#"]+(\.[^<>()[\]\\.,;:\s#"]+)*)|(".+"))#((\[[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}])|(([a-zA-Z\-0-9]+\.)+[a-zA-Z]{2,}))$/
At the very end of it I want to add unicode flag u so it will look like this
/^(([^<>()[\]\\.,;:\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,}))$/u
However I am getting error saying that regex becomes invalid with unicode flag. Is there any possibility to set it here?
There are multiple solutions in order to validate unicode characters, but this flag cannot be used like that. The \u flag is most used to be followed by a char code like \u00C0.
I think the most reliable solution is to specify the range of accepted unicode characters in the regex.
Something like this should work:
/^(?!\.)((?!.*\.{2})[a-zA-Z0-9\u00E0-\u00FC.!#$%&'*+-/=?^_`{|}~\-\d]+)#(?!\.)([a-zA-Z0-9\u00E0-\u00FC\-\.\d]+)((\.([a-zA-Z]){2,63})+)$/
The solution applied here is to support characters from à to ü.
Regex tester: https://www.regexpal.com/?fam=108260
Related question for mathching unicode characters: Matching accented characters with Javascript regexes
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Learning Regular Expressions [closed]
(1 answer)
Match empty string, comma, hyphen or underscore once using regex
(2 answers)
Closed 4 years ago.
I have a requirement to do a field validation using regEx. The requirement is not to have any special characters. It is working for me for a single white space, but not for multiple white spaces. It does not work if there is a white space at the beginning. Can you please help me with this?
This is what I used:
^(\w+ ?)*$
You might be looking for
^[\w ]*$
The ^ and $ are anchors for the start/end of the string, the [...] is called a character class and would allow only [A-Za-z0-9_ ]. The * is a quantifier and means zero or more times, thus the expression would also allow an empty string. If this is not what you want, change it to + instead of the *. Please note that this would also allow a string with only spaces (it really depends on what you want).
This question already has answers here:
Validate phone number with JavaScript
(30 answers)
Closed 5 years ago.
console.log(/\d+?\d+?\d+?-\d+?\d+?\d+?-\d+?\d+?\d+?\d+?$/.test("555-555-55539"));
Answer --> true
I was looking for false, i am validating phone numbers. e.g. 555-555-5555 is a correct response([0-9])
I am a newbie to regex, can anyone explain what i am doing wrong here?
How about this.
console.log(/\d{3}-\d{3}-\d{4}$/.test("555-555-55539"));
You used wrong quantifiers in your regex. You made them lazy (+?), but it will still match all characters until the next character from regex is found. In case of your last quantifier (just before $) it will match all digits until the end of string is found. Hence it matches not only one digit but all of them. Same thing happens before each hyphen (555555555-5555-555555555 is valid for your regex).
This question already has answers here:
regular expression for password with few rules
(3 answers)
Closed 8 years ago.
I took an online JavaScript test where the 3rd problem was as follows:
Complete the checkPassword function, which should check if the password parameter adheres to the following rules:
Must be longer than 6 characters.
Allowed characters are lower or uppercase Latin alphabetic characters (a-z), numbers (0-9), and special characters +, $, #, \, / only.
Must not have 3 or more consecutive numbers (e.g. "pass12p" is fine,
but "pass125p" is not, because it contains "125")
>
My solution was as follows:
function checkPassword(password) {
return /^[a-zA-Z0-9\+\$\\\/#]{6,}$/.test(password) && !/[0-9]{3,}/.test(password);
}
This solution gave me the correct outputs for the given inputs.
But the ultimate test result told that the solution is just 75% correct sadly.
I think the perfect answer is expected to be a solution just with a single regular expression. Is there anyone who can give me a single regular expression that gives the same result? I am not so good at regular expression so please advise.
I appreciate your help in advance.
Just use a negative lookahead at the start.
^(?!.*?\d{3})[a-zA-Z0-9\+\$\\\/#]{6,}$
(?!.*?\d{3}) at the start asserts that the match won't contain atleast three consecutive digits. Then the regex engine tries to match the pattern [a-zA-Z0-9\+\$\\\/#] against the input string 6 or more times only if this condition is satisfied.
DEMO
^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])[a-zA-Z\d]{6,}$
For consecutive number check
public boolean isValid(final String userName,final String password) {
for(int i=0;(i+2)<userName.length();i++) if(password.indexof(userName.substring(i,i+2))!=-1) return false; return true;
}