I am making form and there is only one more thing which I cant figure it out :(
I need regular expression for password which must be at least 7 characters long. There can be small and big letters and must contain at least one number.
I tried
[0-9]+[a-zA-Z]){7}$
You can use lookahead:
^(?=.*\d)[a-zA-Z\d]{7,}$
(?=.*\d) is a lookahead which checks for a digit in the string. Basically, .* matches the whole string and then backtracks 1 by 1 to match a digit. If it matches a digit, the regex engine comes back to its position before match. So, it just checks for a pattern.
{7,} is a quantifier which matches previous pattern 7 to many times
^ is the beginning of a string
Related
I need to validate password with: At least one uppercase, at least one lowercase, at least one number OR symbol, at least 8 characters.
I have this regex:
/^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[^a-zA-Z0-9]).{8,20}$/
This works fine except of > it checks string on number AND symbol, but not on number OR symbol. And also character length 8-20, not at least 8 but gives range. I want it to check number OR symbol. Any ideas? Thanks and have a good day!
The (?=.*\d) positive lookahead requires a digit in the string AND (?=.*[^a-zA-Z0-9]) requires a char other than ASCII letter or digit.
To make the regex require a digit OR a char other than ASCII letter or digit, merge the two lookaheads as
(?=.*[^A-Za-z])
Basically, you need to remove 0-9 from the second lookahead and it will require any char but an ASCII letter.
Result:
/^(?=.*[^A-Za-z])(?=.*[a-z])(?=.*[A-Z]).{8,20}$/
Or, a much more efficient version based on the contrast principle:
/^(?=[A-Za-z]*[^A-Za-z])(?=[^a-z]*[a-z])(?=[^A-Z]*[A-Z]).{8,20}$/
See the regex demo.
If a space is not special, add it to the lookahead:
/^(?=[A-Za-z ]*[^A-Za-z ])(?=[^a-z]*[a-z])(?=[^A-Z]*[A-Z]).{8,20}$/
^ ^
Thanks for taking a look.
My goal is to come up with a regexp that will match input that contains no digits, whitespace or the symbols !#£$%^&*()+= or any other symbol I may choose.
I am however struggling to grasp precisely how regular expressions work.
I started out with the simple pattern /\D/, which from my understanding will match the first non-digit character it can find. This would match the string 'James' which is correct but also 'James1' which I don't want.
So, my understanding is that if I want to ensure that a pattern is not found anywhere in a given string, I use the ^ and $ characters, as in /^\D$/. Now because this will only match a single character that is not a digit, I needed to use + to specify that 1 or more digits should not be founds in the entire string, giving me the expression /^\D+$/. Brilliant, it no longer matches 'James1'.
Question 1
Is my reasoning up to this point correct?
The next requirement was to ensure no whitespace is in the given string. \s will match a single whitespace and [^\s] will match the first non-whitespace character. So, from my understanding I just had to add this to what I have already to match strings that contain no digits and no whitespace. Again, because [^\s] will only match a single non-white space character, I used + to match one or more whitespace characters, giving the new regexp of /^\D+[^\s]+$/.
This is where I got lost, as the expression now matches 'James1' or even 'James Smith25'. What? Massively confused at this point.
Question 2
Why is /^\D+[^\s]+$/ matching strings that contain spaces?
Question 3
How would I go about writing the regular expression I'm trying to solve?
While I am keen to solve the problem I am more interested in figuring where my understanding of regular expressions is lacking, so any explanations would be helpful.
Not quite; ^ and $ are actually "anchors" - they mean "start" and "end", it's actually a little more complicated, but you can consider them to mean the start and end of a line for now - look up the various modifiers on regular expressions if you're interested in learning more about this. Unfortunately ^ has an overloaded meaning; if used inside square brackets it means "not", which is the meaning you are already acquainted with. It's very important that you understand the difference between these two meanings and that the definition in your head actually applies only to character range matching!
Contributing further to your confusion is that \d means "a numerical digit" and \D means "not a numerical digit". Similarly \s means "a whitespace (space/tab/newline/etc.) character" and \S means "not a whitespace character."
It's worth noting that \d is effectively a shortcut for [0-9] (note that - has a special meaning inside square brackets), and \D is a shortcut for [^0-9].
The reason it's matching strings that contain spaces is that you've asked for "1+ non-numerical digits followed by 1+ non-space characters" - so it'll match lots of strings! I think that perhaps you don't understand that regular expressions match bits of strings, you're not adding constraints as you go, but rather building up bots of matchers that will match bits of corresponding strings.
/^[^\d\s!#£$%^&*()+=]+$/ is the answer you're looking for - I'd look at it like this:
i. [] - match a range of characters
ii. []+ - match one or more of that range of characters
iii. [^\d\s]+ - match one or more characters that do not match \d (numerical digit) or \s (whitespace)
iv. [^\d\s!#£$%^&*()+=]+ - here's a bunch of other characters I don't want you to match
v. ^[^\d\s!#£$%^&*()+=]+$ - now there are anchors applied, so this matcher has to apply to the whole line otherwise it fails to match
A useful website to explore regexs is http://regexr.com/3b9h7 - which I supply with my suggested solution as an example. Edit: Pruthvi Raj's link to debuggerx is awesome!
Is my reasoning up to this point correct?
Almost. /\D/ matches any character other than a digit, but not just the first one (if you use g option).
and [^\s] will match the first non-whitespace character
Almost, [^\s] will match any non-whitespace character, not just the first one (if you use g option).
/^\D+[^\s]+$/ matching strings that contain spaces?
Yes, it does, because \D matches a space (space is not a digit).
Why is /^\D+[^\s]+$/ matching strings that contain spaces?
Because \D+ in /^\D+[^\s]+$/can match spaces.
Conclusion:
Use
^[^\d\s!#£$%^&*()+=]+$
It will match strings that have no digits and spaces, and the symbols you do not allow.
Mind that to match a literal -, ] or [ with a character class, you either need to escape them, or use at the start or end of the expression. To play it safe, escape them.
Just insert every character you don't want to include in a negated character class as follows:
^[^\s\d!#£$%^&*()+=]*$
DEMO
Debuggex Demo
^ - start of the string
[^...] - matches one character that is not in `...`
\s - matches a whitespace (space, newline,tab)
\d - matches a digit from 0 to 9
* - a quantifier that repeats immediately preceeding element by 0 or more times
so the regex matches any string that has
1. string that has a beginning
2. containing 0 or more number of characters that is not whitesapce, digit, and all the symbols included in the character class ( In this example !#£$%^&*()+=) i.e., characters that are not included in the character class `[...]`
3.that has ending
NOTE:
If the symbols you don't want it to have also includes - , a hyphen, don't put it in between some other characters because it is a metacharacter in character class, put it at last of character class
I am trying to get a regular expression to work but am stumped. What I want is to do the inverse of this:
/(\w)\1{5,}/
This regex does the exact opposite of what I'm trying to do. I would like to get everything but a string that has 6 repeating numbers i.e. 111111 or 999999.
Is there a way to use a negative look-around or something with this regex?
You can use this rgex:
/^(?!.*?(\w)\1{5}).*$/gm
RegEx Demo
(?!.*?(\w)\1{5}) is a negative lookaahead that will fail the match if there are 6 consecutive same word characters in it.
I'd rather go with the \d shorthand class for digits since \w also allows letters and an underscore.
^(?!.*(\d)\1{5}).*$
Regex explanation:
^ - Start of string/line anchor
(?!.*(\d)\1{5}) - The negative lookahead checking if after an optional number of characters (.*) we have a digit ((\d)) that is immediately followed with 5 identical digits (\1{5}).
.* - Match 0 or more characters up to the
$ - End of string/line.
See demo. This regex will allow
Saw a challenge on Twitter so I've been working my way through it, granted I am not the best with Regular Expressions. This is what I have so far:
var pass_regex = new RegExp(/^[a-z][A-Z][0-9]|[!##$%^&*()_]+$/);
I am trying to match a password input that contains:
1 Lowercase Letter
1 Uppercase Letter
1 Digit OR Special Character
Where I am getting stuck is on the 'OR' part, I thought the pipe separator between [0-9] and my set of special characters would work but it doesn't seem to. Trying to better understand how you would use regular expressions to to check for 1 Digit OR 1 Special Character. Thank you in advance for any help provided.
Atleast one:
You need to use a positive lookahead based regex for checking multiple conditions.
^(?=.*?[A-Z])(?=.*?[a-z]).*?[\W\d].*
OR
^(?=.*?[A-Z])(?=.*?[a-z]).*?[!##$%^&*()_\d].*
(?=.*?[A-Z]) Asserts that there must be atleast one uppercase letter.
(?=.*?[a-z]) Atleast one lowercase letter.
.*? non-greedy match of any character zero or more times.
If the above conditions are satisfied then match that corresponding string and also the string must contain atleast a single character from the given list [!##$%^&*()_\d] . \d in this list matches any digit character.
.* matches the following zero or more characters.
DEMO
I need to build a JavaScript regular expression with the following constraints:
The input string needs to be at least 6 characters long
The input string needs to contain at least 1 alphabetical character
The input string needs to contain at least 1 non-alphabetical character
I'm seriously lacking a lookback feature in JavaScript. The thing I came up with:
((([a-zA-Z][^a-zA-Z])|([^a-zA-Z][a-zA-Z]))....)|
(.(([a-zA-Z][^a-zA-Z])|([^a-zA-Z][a-zA-Z]))...)|
(..(([a-zA-Z][^a-zA-Z])|([^a-zA-Z][a-zA-Z]))..)|
(...(([a-zA-Z][^a-zA-Z])|([^a-zA-Z][a-zA-Z])).)|
(....(([a-zA-Z][^a-zA-Z])|([^a-zA-Z][a-zA-Z])))
This looks pretty long. Is there a better way?
How I came to this:
Regex for alphabetical character is [a-zA-Z]
Regex for non-alphabetical character is [^a-zA-Z]
So I need to look for a [a-zA-Z][^a-zA-Z] or [^a-zA-Z][a-zA-Z] so (([a-zA-Z][^a-zA-Z])|([^a-zA-Z][a-zA-Z])).
I need to check for n preceding characters and 6-n succeeding characters.
/^(?=.{6})(?=.*[a-zA-Z])(?=.*[^a-zA-Z])/
This means:
^ - start of the string
(?= ... ) - followed by (i.e. an independent submatch; it won't move the current match position)
.{6} - six characters ("start of string followed by six characters" implements the "must be at least six characters long" rule)
.* - 0 or more of any character (except newline - may need to fix this?)
[a-zA-Z] - a letter (.*[a-zA-Z] therefore finds any string with a letter anywhere in it (technically it finds the last letter in it))
[^a-zA-Z] - a non-letter character
In summary: Starting from the beginning of the string, we try to match each of the following in turn:
6 characters (if we find those, the string must be 6 characters long (or more))
an arbitrary string followed by a letter
an arbitrary string followed by a non-letter
Use this regex...
/^(?=.{6,})(?=.*[a-zA-Z])(?=.*[^a-zA-Z]).*$/
-------- ------------- --------------
^ ^ ^
| | |->checks for a single non-alphabet
| |->checks for a single alphabet
|->checks for 6 to many characters
(?=) is a zero width look ahead which checks for a match.It doesn't consume characters.This is the reason why we can use multiple lookaheads back to back
Similar answer to others, thus this doesn't need much explanation, I think the best way is to do
/^(?=.*[a-zA-Z])(?=.*[^a-zA-Z]).{6,}$/
This starts at the beginning of the string, looks ahead for an alphabetical character, looks ahead for a non-alphabetical character and, in the end, it finds a string of 6+ chars, I think there's no need for lookaheads about length