I was trying to install joomla on my website. While installing joomla, I was asked to create a MYSQl user. But I couldn't because, everytime I type as password, it gives a message saying the paswword doesn't meet the reqular expression requirement. Given below is the regular expression
'(?=^.{8,}$)((?=.*\d)|(?=.*\W+))(?![.\n])(?=.*[A-Z])(?=.*[a-z]).*$'
What does this mean? What password can I give? Give an example of a password that will pass this regular expression test. Please help me
(?=^.{8,}$)
This part means has 8 more more characters, and the match starts at the start of input.
((?=.*\d)
Means contains a digit.
|(?=.*\W+))
Or contains something that is neither a letter or a digit
(?![.\n])
not starting with a dot or UNIX newline.
(?=.*[A-Z])
Contains at least one capital letter.
(?=.*[a-z])
Contains at least one lowercase letter
.*$
Consists entirely of non-newline characters and the matched group will contain the entire string.
Password should be 8 symbols or more, atleast one digit or a non-character , atleast one lower alpha and atleast one upper alpha and not beginning with . or newline ( seriously?)
Example: Manojlds9
Related
I need to create a regular expression to check if a password has at least 1 uppercase letter, at least 2 numbers, and ends with a $ (dollar sign).
I've been trying to figure it out, but I can only get as far as checking if there's at least 1 uppercase and one number, rather than two.
These should be valid:
4hg5Fjkjk$
fh##Y5fFF5$
hgH5Hu6$
These should not be valid:
45tyghisu$ (No capital)
5THygfhy$ (Only one number)
Gh%hF45$h (No dollar sign at the end)
Here's what I have so far (checks for at least one number, one capital and dollar sign at the end)
/(?=.*[A-Z])(?=.*\d).*\$/
Any help would be greatly appreciated!
ps. I've looked on SO, and can't find anything relating to more than one required character.
In your pattern you have to repeat asserting a digit twice instead of one time using for example (?=(?:[^\d\r\n]*\d){2}) using contrast.
If you don't want to allow spaces in the password, you could use \S+ to match 1+ times a non whitespace char.
You could use:
^(?=[^A-Z\r\n]*[A-Z])(?=(?:[^\d\r\n]*\d){2})\S+\$$
Regex demo
According to the given answer by the OP, the number of characters should be 9-15:
^(?=[^A-Z\r\n]*[A-Z])(?=(?:[^\d\r\n]*\d){2})\S{9,15}\$$
Regex demo
This RegEx is simple and makes no other assumptions about what characters may be in a password other than what was specified by the OP.
^(?=.*?[A-Z])(?=.*?\d.*?\d).*\$$
See Demo (Click ON "RUN TESTS")
thanks for all the answers. Looked through them, and I figured out a pretty simple way to do it using the regular expression below. I edited it to allow for setting a length on the password, just change the 9 and 15 to your desired lengths.
/^(?=.*[A-Z])(?=.*\d.*\d)[^\s]{9,15}\$$/
So I've got /.+[^\x20-\x2A\x2C\x2F\x3A-\x40\x5B-\x5E\x60\x7B-\xFF]\#[\w+-?]+(.{1})\w{2,}/ pattern I want to use for email validation on client-side, which doesn't work as expected.
I know that my pattern is simple and doesn't cover every standard possibility, but it's part of my regex training.
Local part of address should be valid only when it has at least one digit [0-9] or letter [a-zA-Z] and can be mixed with comma or plus sign or underscore (or all at once) and then # sign, then domain part, but no IP address literals, only domain names with at least one letter or digit, followed by one dot and at least two letters or two digits.
In test string form it doesn't validate a#b.com and does validate baz_bar.test+private#e-mail-testing-service..com, which is wrong - it should be vice versa - validate a#b.com and not validate baz_bar.test+private#e-mail-testing-service..com
What specific error I've got there and where?
I can't locate this, sorry..
You need to change your regex
From: .+[^\x20-\x2A\x2C\x2F\x3A-\x40\x5B-\x5E\x60\x7B-\xFF]\#[\w+-?]+(\.{1})\w{2,}
To: .+[^\x20-\x2A\x2C\x2F\x3A-\x40\x5B-\x5E\x60\x7B-\xFF]?\#[\w+-]+(\.{1})\w{2,}
Notice that I added a ? before the # sign and removed the ? from the first "group" after the # sign. Adding that ? will make your regex to know that hole "group" is not mandatory.
See it working here: https://regex101.com/r/iX5zB5/2
You're requiring the local part (before #) to be at least two characters with the .+ followed by the character class [^...]. It's looking for any character followed by another character not in the list of exclusions you specify. That explains why "a#b.com" doesn't match.
The second problem is partly caused by the character class range +-? which includes the . character. I think you wanted [-\w+?]+. (Do you really want question marks?) And then later I think you wanted to look for a literal . character but it really ends up matching the first character that didn't match the previous block.
Between the regex provided and the explanatory text I'm not sure what rules you intend to implement though. And since this is an exercise it's probably better to just give hints anyway.
You will also want to use the ^ and $ anchors to makes sure the entire string matches.
I have the following regular expression:
^.*(?=^.{8,}$)(?=.*\d)(?=.*[!##$%^&*-])(?=.*[A-Z])(?=.*[a-z]).*$
I am using it to validate for
At least one letter
least one capital letter
least one number
least one special characters
least 8 characters
But along with this I need to restrict the underscore (_).
If I enter password Pa$sw0rd, this is validating correctly, which is true.
If I enter Pa$_sw0rd this is also validating correctly, which is wrong.
The thing is the regex is passing when all the rules are satisfied. I want a rule to restrict underscore along with above.
Any help will be very appreciable.
I think you can use a negated character class [^_]* to add this restriction (also, remove the initial .*, it is redundant, and the first look-ahead is already at the beginning of the pattern, no need to duplicate ^, and it is totally redundant since the total length limit can be checked at the end):
^(?=.*\d)(?=.*[!##$%^&*-])(?=.*[A-Z])(?=.*[a-z])[^_]{8,}$
See demo
^(?=.*?\d)(?=.*?[!##$%^&*-])(?=.*?[A-Z])(?=.*?[a-z])(?!.*_).{8,}$
You can try this..* at start is of no use.See demo.
https://regex101.com/r/pG1kU1/34
I've got this RegEx example: http://regexr.com?34hihsvn
I'm wondering if there's a more elegant way of writing it, or perhaps a more optimised way?
Here are the rules:
Digits and dashes only.
Must not contain more than 10 digits.
Must have two hyphens.
Must have at least one digit between each hyphen.
Last number must only be one digit.
I'm new to this so would appreciate any hints or tips.
In case the link expires, the text to search is
----------
22-22-1
22-22-22
333-333-1
333-4444-1
4444-4444-1
4444-55555-1
55555-4444-1
666666-7777777-1
88888888-88888888-1
1-1-1
88888888-88888888-22
22-333-
333-22
----------
My regex is: \b((\d{1,4}-\d{1,5})|(\d{1,5}-\d{1,4}))-\d{1}\b
I'm using this site for testing: http://gskinner.com/RegExr/
Thanks for any help,
Nick
Here is a regex I came up with:
(?=\b[\d-]{3,10}-\d\b)\b\d+-\d+-\d\b
This uses a look-ahead to validate the information before attempting the match. So it looks for between 3-10 characters in the class of [\d-] followed by a dash and a digit. And then after that you have the actual match to confirm that the format of your string is actually digit(dash)digit(dash)digit.
From your sample strings this regex matches:
22-22-1
333-333-1
333-4444-1
4444-4444-1
4444-55555-1
55555-4444-1
1-1-1
It also matches the following strings:
22-7777777-1
1-88888888-1
Your regexp only allows a first and second group of digits with a maximum length of 5. Therefore, valid strings like 1-12345678-1 or 123456-1-1 won't be matched.
This regexp works for the given requirements:
\b(?:\d\-\d{1,8}|\d{2}\-\d{1,7}|\d{3}\-\d{1,6}|\d{4}\-\d{1,5}|\d{5}\-\d{1,4}|\d{6}\-\d{1,3}|\d{7}\-\d{1,2}|\d{8}\-\d)\-\d\b
(RegExr)
You can use this with the m modifier (switch the multiline mode on):
^\d(?!.{12})\d*-\d+-\d$
or this one without the m modifier:
\b\d(?!.{12})\d*-\d+-\d\b
By design these two patterns match at least three digits separated by hyphens (so no need to put a {5,n} quantifier somewhere, it's useless).
Patterns are also build to fail faster:
I have chosen to start them with a digit \d, this way each beginning of a line or word-boundary not followed by a digit is immediately discarded. Other thing, using only one digit, I know the remaining string length.
Then I test the upper limit of the string length with a negative lookahead that test if there is one more character than the maximum length (if there are 12 characters at this position, there are 13 characters at least in the string). No need to use more descriptive that the dot meta-character here, the goal is to quickly test the length.
finally, I describe the end of string without doing something particular. That is probably the slower part of the pattern, but it doesn't matter since the overwhelming majority of unnecessary positions have already been discarded.
I've looked on here for some ideas but I still seem to be struggling with coming up with a regular expression to meet my requirements.
I need a regular expression to check a password format, the criteria are:
At least 1 uppercase letter
At least 1 number
Only alphanumeric characters (no special characters)
At least 8 characters long
The regular expression I'm using is:
^(?=.*[a-z])(?=.*[A-Z])(?=.*\d).{8,}$
However this is also allowing characters like !$&.
Is there a modification I need to make to this to get it to stop these special characters being accepted?
Change the last part .{8,} to [a-zA-Z\d]{8,}
^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)[a-zA-Z\d]{8,}$