I want to use following code for password validation of at lest one character,number, upper case and lower case character
function fd3() {
alert("hii");
var v1 = document.getElementById("h3").value,
pass = /^([a-zA-Z0-9_##$%^&*]+$)/;
if(!pass.test(v1)) {
alert("Wrong Password");
}
}
Strictly speaking it does match your example of Abc#123.
Follow this link to see it matching Abc#123
However it doesn't do anything like what you think it does - it will match one or more character in that group - so simply "A" alone will match it.
You need something MUCH more complex like this:
^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=]).*$
Debuggex Demo of the working example
I would also note that you did not have a requirement for a minimal number of characters so I did not include this in the regex. My list of special characters in this regex is also different from yours, but that can easily be changed by you if that is important.
Related
I have to set some rules on not accepting wrong url for my project. I am using regex for this.
My Url is "http ://some/resource/location".
This url should not allow space in beginning or middle or in end.
For example these spaces are invalid:
"https ://some/(space here in middle) resource/location"
"https ://some/resource/location (space in end)"
"(space in starting) https ://some/resource/location"
"https ://(space here) some/resource/location"
Also these scenario's are invalid.
"httpshttp ://some/resource/location"
"https ://some/resource/location,https ://some/resource/location"
Currently I am using a regex
var regexp = /(ftp|http|https):\/\/(\w+:{0,1}\w*#)?(\S+)(:[0-9]+)?(\/|\/([\w#!:.?+=&%#!\-\/]))?/;
This regex accepts all those invalid scenarios. I am unable to find the correct matching regex which will accept only if the url is valid. Can anyone help me out on this?
We need to validate n number of scenarios for URL validation. If your particular about your given pattern then above regex expression from other answer looks good.
Or
If you want to take care of all the URL validation scenarios please refer In search of the perfect URL validation regex
/(ftp|http|https){1}:\/\/(?:.(?! ))+$/
is this regex OK ?
use this
^\?([\w-]+(=[\w-]*)?(&[\w-]+(=[\w-]*)?)*)?$
See live demo
This considers each "pair" as a key followed by an optional value (which maybe blank), and has a first pair, followed by an optional & then another pair,and the whole expression (except for the leading?) is optional. Doing it this way prevents matching ?&abc=def
Also note that hyphen doesn't need escaping when last in the character class, allowing a slight simplification.
You seem to want to allow hyphens anywhere in keys or values. If keys need to be hyphen free:
^\?(\w+(=[\w-]*)?(&\w+(=[\w-]*)?)*)?$
I have this code:
if(address.length<=0)
{
msg.setAttribute("style", "color:red");
msg.innerHTML='Please enter address';
return false;
}
I would like to change so it checks whether the webform contains BOTH numbers and letters. Can you help me?
Thank you so much,
Jones
p.s.: So I want to make sure they also enter street name AND house number as well (example: 24 Sunshine street would be good, but if they forget house number, they would get the message).
That doesn't look like PHP at all. More like JavaScript...
Here's one way to do it in JavaScript:
var re = /^\d+\s+\D+$/;
if (re.test(address)) {
//We get here if the address is correctly formated
}
else {
//We get here if the string is badly formated
}
The regex works like this:
\d+ matches to one or more numbers
\s+ matches to one or more spaces
\D+ matches to one or more letters
If you want to accept both "24 Sunshine" and "Sunshine 24" you could instead use this:
/^(\d+\s+\D+)|(\D+\s+\d+)$/
And if we want to be extra safe and protect from the case that the user might enter an extra trailing or leading space we could either trail the string or use this ReGex:
/^\s*(\d+\s+\D+)|(\D+\s+\d+)\s*$/
Apart from regular expression which is a very nice and clear solution you can use these php functions:
first the ctype_alnum () in order to check if your string contains letters and digits and then
this on ctype_alpha() in case the above is true to check if user forgot to enter number.
In case you are interested there is also this one ctype_digit() for checking if user missed the address but gave the number.
Or if you want just a regex this it will do the job:
^[a-zA-Z]([a-zA-Z-]+\s)+\d{1,4}
I am trying and failing hard in validating a phone number within jQuery validation. All I want is to allow a number like (01660) 888999. Looking around the net I find a million examples but nothing seems to work. Here is my current effort
$.validator.addMethod("phonenumber", function(value) {
var re = new RegExp("/[\d\s()+-]/g");
return re.test(value);
//return value.match("/[\d\s]*$");
}, "Please enter a valid phone number");
Bergi is correct that the way you are constructing the regular expression is wrong.
Another problem is that you are missing anchors and a +:
var re = /^[\d\s()+-]+$/;
Note though that a regular expression based solution will still allow some inputs that aren't valid phone numbers. You can improve your regular expression in many ways, for example you might want to require that there are at least x digits, for example.
There are many rules for what phone numbers are valid and invalid. It is unlikely you could encode all those rules into a regular expression in a maintainable way, so you could try one of these approaches:
Find a library that is able to validate phone numbers (but possibly not regular expression based).
If you need a regular expression, aim for something that is a close approximation to the rules, but doesn't attempt to handle all the special cases. I would suggest trying to write an expression that accepts all valid phone numbers, but doesn't necessarily reject all invalid phone numbers.
You may also want to consider writing test cases for your solution. The tests will also double as a form of documentation of which inputs you wish to accept and reject.
You need to use either a regex literal or a string literal in the RegExp constructor:
var re = /[\d\s()+-]/g;
// or
var re = new RegExp("[\\d\\s()+-]", "g");
See also Creating a Regular Expression.
Apart from that, you would need to use start- and end-of-string anchors to make sure that the regex matches the whole string, not only a part of it, and some repetition modifier to allow more than one character:
var re = /^[\d\s()+-]+$/g;
Another approach may be:
function(value) {
return /^\d+$/.test(value.replace(/[()\s+-]/g,''));
}
and if you want to check for the length of the number too, say it has to be a string with 10 digits:
function(value) {
return /^\d{10}$/.test(value.replace(/[()\s+-]/g,''));
}
I am trying to validate a password string with javascript and need some help with a regex. I have tried some tutorials, but I think I have some problems understanding how to escape quantifiers and/or metacharacters.
I want to make sure that the password string only contains one or more (max 32) characters from the following spans:
"abcdefghijklmnopqrstuvwxyz"
"ABCDEFGHIJKLMNOPQRSTUVWXYZ"
"012345678901234567890123456789"
"!##%&/(){}[]=?+*^~-_.:,;"
The first three spans are pretty easy, but I can't figure out the last one. Basically my script looks something like this:
var password = "user_input_password";
if (/^[A-Za-z0-9!##$%...]{1,32}$/.test(password)) {
document.write('OK');
} else {
document.write('Not OK');
}
Any help or input is highly appreciated, thanks!
In general, you can escape a meta-character using a backslash \; however, inside a character class, the only ones you have to escape are ] , \ and - (the ^ only has a meaning at the very beginning). Something like [\w!##%&/(){}[\]=?+*^~\-.:,;] will do what you want.
The \w is equal to [A-Za-z0-9_].
So the full test would be something like:
/^[\w!##%&/(){}[\]=?+*^~\-.:,;]{1,32}$/.test(password)
/^[A-Za-z0-9!##%&\/(){}\[\]=?+*^~\-_\.:,;]{1,32}$/
You can also match all characters that are not considered white space (space, newline, tab)
/^[^\s]{1,32}$/.test(password);
To exclude quotes as well (I didn't see them in your example) you can add those in:
/^[^\s'"]{1,32}$/.test(password);
Regex fun again...
Take for example http://something.com/en/page
I want to test for an exact match on /en/ including the forward slashes, otherwise it could match 'en' from other parts of the string.
I'm sure this is easy, for someone other than me!
EDIT:
I'm using it for a string.match() in javascript
Well it really depends on what programming language will be executing the regex, but the actual regex is simply
/en/
For .Net the following code works properly:
string url = "http://something.com/en/page";
bool MatchFound = Regex.Match(url, "/en/").Success;
Here is the JavaScript version:
var url = 'http://something.com/en/page';
if (url.match(/\/en\//)) {
alert('match found');
}
else {
alert('no match');
}
DUH
Thank you to Welbog and Chris Ballance to making what should have been the most obvious point. This does not require Regular Expressions to solve. It simply is a contains statement. Regex should only be used where it is needed and that should have been my first consideration and not the last.
If you're trying to match /en/ specifically, you don't need a regular expression at all. Just use your language's equivalent of contains to test for that substring.
If you're trying to match any two-letter part of the URL between two slashes, you need an expression like this:
/../
If you want to capture the two-letter code, enclose the periods in parentheses:
/(..)/
Depending on your language, you may need to escape the slashes:
\/..\/
\/(..)\/
And if you want to make sure you match letters instead of any character (including numbers and symbols), you might want to use an expression like this instead:
/[a-z]{2}/
Which will be recognized by most regex variations.
Again, you can escape the slashes and add a capturing group this way:
\/([a-z]{2})\/
And if you don't need to escape them:
/([a-z]{2})/
This expression will match any string in the form /xy/ where x and y are letters. So it will match /en/, /fr/, /de/, etc.
In JavaScript, you'll need the escaped version: \/([a-z]{2})\/.
You may need to escape the forward-slashes...
/\/en\//
Any reason /en/ would not work?
/\/en\// or perhaps /http\w*:\/\/[^\/]*\/en\//
You don't need a regex for this:
location.pathname.substr(0, 4) === "/en/"
Of course, if you insist on using a regex, use this:
/^\/en\//.test(location.pathname)