I need to a regres that find all single backslashes followed by an alphabet.
So I want to find backslashes that exist in patterns like these:
\a
\f
\test
and not in these patterns:
\\a
\"
Thanks
Updated:
As #Amadan points out in the comments below, JavaScript does not implement lookbehind, which basically breaks my original answer.
There is an approach suggested in this stackoverflow post that may be a reasonable path to take for this problem.
Basically the poster suggests reversing the string and using lookahead to match. If we were to do that, then we would want to match a string of alphabetic characters followed by a single backslash, but not followed by multiple backslashes. The regex for that would look like this:
/[a-zA-Z]+\\(?![\\]+)/g
[a-zA-Z]+ - match one or more alphabetic characters
\\ - followed by a single backslash
(?![\\]+) - not followed by one or more backslashes
g - match it globally (more than one occurrence)
The downside of this approach (aside from having to reverse your string) is that you can't match only the backslash, but will have to also match the alphabetic characters that come before it (since JS doesn't have lookbehind).
Original Answer (using lookbehind):
/(?<!\\)\\[a-zA-Z]+/g (using negative lookbehind) will match a single backslash followed by one or more letters of the alphabet, regardless of case. This regular expression breaks down as follows:
(?<!\\)\\ - use negative lookbehind to match a \ that is not preceded by a \
[a-zA-Z]+ - match one or more letters of the alphabet, regardless of case
g - match it globally
If you only want to match the \ and not the alphabetic characters, then you can use positive lookahead. The regex for that would look like: /(?!>\\)\\(?=[a-zA-Z]+)/g and would break down like this:
(?<!\\)\\ - use negative lookbehind to match a \ that is not preceded by a \
(?=[a-zA-Z]+) - and is followed by one or more alphabetic characters
g - match it globally
If you only want the regex to match backslashes at the beginning of a line, prepend a ^ to it.
You can use a tool like Rubular to test and play with regular expressions.
Related
I have a few strings:
some-text-123123#####abcdefg/
some-STRING-413123#####qwer123t/
some-STRING-413123#####456zxcv/
I would like to receive:
abcdefg
qwer123t
456zxcv
I have tried regexp:
/[^#####]*[^\/]/
But this not working...
To get whatever comes after five #s and before the last /, you can use
/#####(.*)\//
and pick up the first group.
Demo:
const regex = /#####(.*)\//;
console.log('some-text-123123#####abcdefg/'.match(regex)[1]);
console.log('some-STRING-413123#####qwer123t/'.match(regex)[1]);
console.log('some-STRING-413123#####456zxcv/'.match(regex)[1]);
assumptions:
the desired part of the string sample will always:
start after 5 #'s
end before a single /
suggestion: /(?<=#{5})\w*(?=\/)/
So (?<=#{5}) is a lookbehind assertion which will check to see if any matching string has the provided assertion immediately behind it (in this case, 5 #'s).
(?=\/) is a lookahead assertion, which will check ahead of a matching string segment to see if it matches the provided assertion (in this case, a single /).
The actual text the regex will return as a match is \w*, consisting of a character class and a quantifier. The character class \w matches any alphanumeric character ([A-Za-z0-9_]). The * quantifier matches the preceding item 0 or more times.
successful matches:
'some-text-123123#####abcdefg/'
'some-STRING-413123#####qwer123t/'
'some-STRING-413123#####456zxcv/'
I would highly recommend learning Regular Expressions in-depth, as it's a very powerful tool when fully utilised.
MDN, as with most things web-dev, is a fantastic resource for regex. Everything from my answer here can be learned on MDN's Regular expression syntax cheatsheet.
Also, an interactive tool can be very helpful when putting together a complex regular expression. Regex 101 is typically what I use, but there are many similar web-tools online that can be found from a google search.
You pattern does not work because you are using negated character classes [^
The pattern [^#####]*[^\/] can be written as [^#]*[^\/] and matches optional chars other than # and then a single char other than /
Here are some examples of other patterns that can give the same match.
At least 5 leading # chars and then matching 1+ word chars in a group and the / at the end of the string using an anchor $, or omit the anchor if that is not the case:
#####(\w+)\/$
Regex demo
If there should be a preceding character other than #
[^#]#####(\w+)\/$
(?<!#)#####(\w+)\/$
Regex demo
Matching at least 5 # chars and no # or / in between using a negated character class in this case:
#####([^#\/]+)\/
Or with lookarounds:
(?<=(?<!#)#####)[^#\/]+(?=\/)
Regex demo
I have a string like this:
///////AB?\a\b\c\d\d\e\\f\a\a\b\cd\ed\fmnopqrstuvwxy\z\a\a\a\a\a\a\a\a\a///imgy
it started with /// and ended with ///imgy (i and/or m and/or g and/or y), and between the beginning and end are the character are normal character like a or escaped character like \a.
Here is my regex:
/^\/{3}((?:\\?[\s\S])+?)\/{3}([imgy]{0,4})(?!\w)/
But the problem is that it is reported as "vulnerable to denial-of-service attacks". The main part that has the problem is
(?:\\?[\s\S])+
How can I create a right one that can figure out both a and \a? Thank you!
Regex Demo
Update:
I just found to use the following regex:
(?:\\[\s\S]+?)|(?:(?<!\\)[\s\S]+?)|(?:(?<=\\\\)[\s\S]+?)
to replace the old problematic part (?:\\?[\s\S])+?, and in this way, it can avoid requires exponential time to match certain inputs, and avoid vulnerable to denial-of-service attacks.
The details:
(?:\\[\s\S]+?) match any \a
(?:(?<!\\)[\s\S]+?) match any a, but not following \.
(?:(?<=\\\\)[\s\S]+?) match any a, but much following \\. This to make sure f is matched that following \\.
So the whole regex will look like this:
^\/{3}((?:\\[\s\S]+?)|(?:(?<!\\)[\s\S]+?)|(?:(?<=\\\\)[\s\S]+?))\/{3}([imgy]{0,4})(?!\w)
You might list the characters that are allowed to a character class, and optionally repeat an escaped character [a-z]
^\/{3,}[A-Za-z?]+(?:\\[a-z\\][A-Za-z?]*)*\/\/\/[imgy]{0,4}$
The pattern matches:
^ Start of string
\/{3,}[A-Za-z?]+ Match 3 or more / and 1 or more times any of the listed allowed chars
(?: Non capture group
\\[a-z\\] Match an escaped char a-z or \\
[A-Za-z?]* Optionally match any of the listed
)* Close an optionally repeat the group
\/\/\/[imgy]{0,4} Match /// and 0-4 times any of i m g or y If there should be at least a single char, you can use {1,4}
$ End of string
Regex demo
Having the following regex: ([a-zA-Z0-9//._-]{3,12}[^//._-]) used like pattern="([a-zA-Z0-9/._-]{3,12}[^/._-])" to validate an HTML text input for username, I wonder if is there anyway of telling it to check that the string has only one of the following: ., -, _
By that I mean, that I'm in need of regex that would accomplish the following (if possible)
alex-how => Valid
alex-how. => Not valid, because finishing in .
alex.how => Valid
alex.how-ha => Not valid, contains already a .
alex-how_da => Not valid, contains already a -
The problem with my current regex, is that for some reason, accepts any character at the end of the string that is not ._-, and can't figure it out why.
The other problem, is that it doesn't check to see that it contains only of the allowed special characters.
Any ideas?
Try this one out:
^(?!(.*[.|_|-].*){2})(?!.*[.|_|-]$)[a-zA-Z0-9//._-]{3,12}$
Regexpal link. The regex above allow at max one of ., _ or -.
What you want is one or more strings containing all upper, lower and digit characters
followed by either one or none of the characters in "-", ".", or "_", followed by at least one character:
^[a-zA-Z0-9]+[-|_|\.]{0,1}[a-zA-Z0-9]+$
Hope this will work for you:-
It says starts with characters followed by (-,.,_) and followed and end with characters
^[\w\d]*[-_\.\w\d]*[\w\d]$
Seems to me you want:
^[A-Za-z0-9]+(?:[\._-][A-Za-z0-9]+)?$
Breaking it down:
^: beginning of line
[A-Za-z0-9]+: one or more alphanumeric characters
(?:[\._-][A-Za-z0-9]+)?: (optional, non-captured) one of your allowed special characters followed by one or more alphanumeric characters
$: end of line
It's unclear from your question if you wanted one of your special characters (., -, and _) to be optional or required (e.g., zero-or-one versus exactly-one). If you actually wanted to require one such special character, you would just get rid of the ? at the very end.
Here's a demonstration of this regular expression on your example inputs:
http://rubular.com/r/SQ4aKTIEF6
As for the length requirement (between 3 and 12 characters): This might be a cop-out, but personally I would argue that it would make more sense to validate this by just checking the length property directly in JavaScript, rather than over-complicating the regular expression.
^(?=[a-zA-Z0-9/._-]{3,12}$)[a-zA-Z0-9]+(?:[/._-][a-zA-Z0-9]+)?$
or, as a JavaScript regex literal:
/^(?=[a-zA-Z0-9\/._-]{3,12})[a-zA-Z0-9]+(?:[\/._-][a-zA-Z0-9]+)?$/
The lookahead, (?=[a-zA-Z0-9/._-]{3,12}$), does the overall-length validation.
Then [a-zA-Z0-9]+ ensures that the name starts with at least one non-separator character.
If there is a separator, (?:[/._-][a-zA-Z0-9]+)? ensures that there's at least one non-separator following it.
Note that / has no special meaning in a regex. You only have to escape it if you're using a regex literal (because / is the regex delimiter), and you escape it by prefixing with a backslash, not another forward-slash. And inside a character class, you don't need to escape the dot (.) to make it match a literal dot.
The dot in regex has a special meaning: "any character here".
If you mean a literal dot, you should escape it to tell the regex parser so.
Escape dot in a regex range
I need to match these characters. This quote is from an API documentation (external to our company):
Valid characters: 0-9 A-Z a-z & # - . , ( ) / : ; ' # "
I used this Regex to match characters:
^[0-9a-z&#-\.,()/:;'""#]*$
However, this wrongly matches characters like %, $, and many other characters. What's wrong?
You can test this regular expression online using http://regexhero.net/tester/, and this regular expression is meant to work in both .NET and JavaScript.
You are not escaping the dash -, which is a reserved character. If you add replace the dash with \- then the regex no longer matches those characters between # and \
Move the literal - to the front of the character set:
^[-0-9a-z&#\.,()/:;'""#]*$
otherwise it is taken as specifying a range like when you use it in 0-9.
- sign, when not escaped, has special meaning in square brackets. #-\. is transformed into #-. (BTW, backslash before dot is not necessary in square brackets), which means "any character between # (ASCII 0x23) and . (ASCII 0x2E). The correct notation is
^[0-9a-z&#\-.,()/:;'"#]*$
The special characters in a character class are the closing bracket (]), the backslash (\), the caret (^) and the hyphen (-).
As such, you should either escape them with a backslash (\), or put them in a position where there is no ambiguity and they do not need escaping. In the case of a hyphen, this would be the first or last position.
You also do not need to escape the dot (.).
Your regex thus becomes:
^[-0-9a-z&#.,()/:;'"#]*$
As a side note, there are many available regex evaluators which provide code hinting. This way, you can simply hover your mouse over your regular expression and it can be explained in English words.
One such free one is RegExr.
Typing your original regex in it and hovering over the hyphen shows:
Matches characters in the range '#-\'
Try that
^[0-9a-zA-Z\&\#\-\.\,\(\)\/\:\;\'\"\#]*$
How can I write a regex to match strings following these rules?
1 letter followed by 4 letters or numbers, then
5 letters or numbers, then
3 letters or numbers followed by a number and one of the following signs: ! & # ?
I need to allow input as a 15-character string or as 3 groups of 5 chars separated by one space.
I'm implementing this in JavaScript.
I'm not going to write out the whole regex for you since this is homework, but here are some hints which should help you out:
Use character classes. [A-Z] matches all uppercase. [a-z] matches all lowercase. [0-9] matches numbers. You can combine them like so [A-Za-z0-9].
Use quantifiers like {n} so [A-Z]{3} gives you 3 uppercase letters.
You can put other characters in character classes. Let's say you wanted to match % or # or #, you could do [%##] which would match any of those characters.
Some meta-characters (characters which have special meaning in the context of regular expressions) will need to be escaped like so: \$ (since $ matches the end of a line)
^ and $ match the beginning and end of the line respectively.
\s matches white-space, but if you sanitize your input, you shouldn't need to use this.
Flags after the regex do special things. For example in /[a-z]/i, the i ignores case.
This should be it:
/^[a-z][a-z0-9]{4} ?[a-z0-9]{5} ?[a-z0-9]{3}[0-9][!&#?]$/i
Feel free to change 0-9 and [0-9] with \d if you see fit.
The regex is simple and readable enough. ^ and $ make sure this is a whole match, so there aren't extra characters before or after the code, and the /i flag allows upper or lower case letters.
I would start with a tutorial.
Pay attention to the quantifiers (like {N}) and character classes (like [a-zA-Z])
^[a-zA-Z][a-zA-Z0-9]{4} ?[a-zA-Z0-9]{5} ?[a-zA-Z0-9]{3}[\!\&\#\?]$