I am intercepting messages which contain the following characters:
*_-
However, whenever any one of these characters comes through, it will always be preceded by a \. The \ is just for formatting though and I want to remove it before sending it off to my server. I know how to easily create a regex which would remove this backslash from a single letter:
'omg\_bbq\_everywhere'.replace(/\\_/g, '')
And I recognize I could just do this operation 3 times: once for each character I want to remove the preceding backslash for. But how can I create a single regex which would detect all three characters and remove the preceding backslash in all 3 cases?
You can use a character class like [*_-].
To remove only the backslash before these characters:
document.body.innerHTML =
"omg\\-bbq\\*everywhere\\-".replace(/\\([*_-])/g, '$1');
When you place a subpattern into a capturing group ((...)), you capture that subtext into a numbered buffer, and then you can reference it with a $1 backreference (1 because there is only one (...) in the pattern.)
This is a good time to use atomic matching. Specifically you want to check for the slash and then positive lookahead for any of those characters.
Ignoring the code, the raw regex you want is:
\\(?=[*_-])
A literal backslash, with one of these characters in front of it: *_-
So now you are matching the slash. The atomic match is a 0 length match, so it doesn't match anything, but sets a requirement that "for this to be a valid match, it needs to be followed by [*_-]"
Atomic groups: http://www.regular-expressions.info/atomic.html
Lookaround statements: http://www.regular-expressions.info/lookaround.html
Positive and negative lookahead and lookbehind matches are available.
Related
I want to match every word separated by comma, but it must not include a quote like ' or ".
I was using this regex:
^[a-zA-Z0-9][\!\[\#\\\:\;a-zA-Z0-9`_\s,]+[a-zA-Z0-9]$
However, it only matches a character and number and not a symbol.
The output should be:
example,example //true
exaplle,examp#3 //true, with symbol or number
example, //false, because there is no word after comma
,example //false, because there is no word before comma
##example&$123,&example& //true, with all character and symbol except quote
You can match 1+ times what is present in the character class. Then repeat 1+ times in a non capturing group (?: what is present in the character class, preceded by a comma.
^[!\[#\\:;a-zA-Z0-9`_ &$#]+(?:,[!\[#\\:;a-zA-Z0-9`_ &$#]+)+$
Regex demo
Note that you don't have to escape \!, \#, \: and \; in the character class, and that \s might also possibly match a newline.
I'm assuming you want the whole string to match perfectly with your conditions and return true then and then only.
These are the conditions-
Each word should be separated by a comma, said comma should have 2 valid words on each side
Words can contain anything except the 2 kinds of quotes (' and ") and whitespace characters (spaces and newlines).
The regex you would use is this- ^(?:[^,'"\s]+,[^,'"\s]+)+$, with the global flag (g) on.
Check out the demo here
Edit: As per request of being able to match only a single word.
This is the regex you would use for that- ^(?:(?:[^,'"\s]+,[^,'"\s]+)+|[^,'"\s]+)$
This will match words separated by a , as well as match just a single word.
The conditions for what qualifies as a word remains the same as aforementioned.
Quick explanation:-
^[^,'"\s]+,[^,'"\s]+$
This part matches 2 words separated by a comma, [^,'"\s]+ denotes a word
Wrapping that whole thing in ^(?:[^,'"\s]+,[^,'"\s]+)+$ simply makes it repeat, so it'll match N number of words separated by a comma, not just 2
Then adding another alternative using | and wrapping the whole thing in a group (non-capturing), we get ^(?:(?:[^,'"\s]+,[^,'"\s]+)+|[^,'"\s]+)$
This simply just adds the alternative [^,'"\s]+ - which matches a singular word.
Check out the updated demo here
I want to write regex for following
students/ad34567-06c1-498c-9b15-cdbac695c1f2/data/sessions
Where students, data and sessions should be exact match.
i have tried this
[students]\[a-z]\[a-z]\[a-z]
You can try this regex, although your question is not clear to me.
^students\/([\w\-\d]+)\/data\/sessions$
Check here https://regex101.com/r/xnxwCX/1
you can grab the data in between students/, /data/session.
In your regex [students]\\[a-z]\\[a-z]\\[a-z] you are trying to match with word students in a character class [students] which will match one of the specified characters instead of matching the whole word.
To match a forward slash you have to use \/ instead of //. [a-z] is specified without a quantifier and will match 1 character from a-z.
To match your example string you might use
^students\/[a-z0-9]+(?:-[a-z0-9]+)+\/data\/sessions$
Regex demo
This part [a-z0-9]+(?:-[a-z0-9]+)+ matches one or more times a lowercase character or a digit [a-z0-9]+
Following a non capturing group repeated one or more times that will match a hyphen followed by matching one or more times a lowercase character or a digit (?:-[a-z0-9]+)+
You might also use [a-f0-9] if your characters are a -f
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 want to extract a string from another using JavaScript / RegExp.
Here is what I got:
var string = "wp-button wp-image-45 wp-label";
string.match(/(?:(?:.*)?\s+)?(wp-image-([0-9]+))(:?\s(?:.*)?)?/);
// returnes: ["wp-button ", "wp-image-45", "45", undefined]
I just want to have "wp-image-45", so:
(Optional) any character
(Optional) followed by whitespace
(Required) followed by "wp-image-"
(Required) followed by any number
(Optional) followed by whitespacy
(Optional) followed by any character
What is missing here? Is it just some kind of bracketing or more?
I also tried
string.match(/(?:(?:.*)?\s+)?(?=(wp-image-([0-9]+)))(?=(:?\s(?:.*)?)?)/)
Edit: In the end I just want to have the number. But I'd also make this step in between.
Regexps are not required to start matching at the beginning of the string, so your attempts to match whitespace and any character aren't necessary. Also, "any character" includes whitespace (except newlines in certain modes).
This should be all you need:
string.match(/\bwp-image-(\d+)\b/)
This will capture, for example, "wp-image-123" into matching group 0, and "123" into matching group 1.
\b means "word boundary", which ensures that you won't match "abcwp-image-123def". A word boundary is defined as any place where a non-word character is followed by a word character, or vice versa. A word character is consists of a letter, a number or an underscore.
Also, I used \d instead of [0-9] simply out of convenience. They have slightly different meaning (\d also matches characters considered numbers in other languages), but that won't make a difference in your case.
If all of that surrounding stuff is optional and all you want is the number then there's no point to matching for any of that stuff except for that "wp-image-" prefix, just do:
var string = "wp-button wp-image-45 wp-label";
string.match(/wp-image-([0-9]+)/);
I am using this regexp - [^\s\da-zA-ZåäöÅÄÖ]+$ to filter out anything but A-Z, 0-9 plus the Swedish characters ÅÄÖ. It works as expected as long as the string isn't ending with whitespace and I am a bit confused on what I need correct to make it accept strings even if they end with whitespace. The \s is there but is apparently not enough.
What is wrong in my regexp?
"something #¤%&/()=?".replace(/[^\s\da-zA-ZåäöÅÄÖ]+$/, '') # => a string
"something ending with whitespace #¤%&/()=? ".replace(/[^\s\da-zA-ZåäöÅÄÖ]+$/, '')# => a string ending with space #¤%&/()=?
You're using a negated character class ("anything that is not a space, a digit, a letter etc."), therefore your regex fails to match.
Drop the \s from it, and also the $ (which ties the match to the end of the string), and it should work.
If you do want to keep spaces inside the string and only remove them at the end, use
"something with whitespace #¤%&/()=? ".replace(/[^\s\da-zA-ZåäöÅÄÖ]+|\s+$/g, '')
Result:
something with whitespace
Your regex says: "match one or more instances of the characters not in the following range, followed by end-of-string". This essentially means that your regex will match only sequences of not-allowed characters appearing at the end of the string. Since your test string ends with a whitespace, which is allowed by your logic, there's no 'sequence of not-allowed characters appearing at the end of the string' and so the regex doesn't match anything.
You can achieve your desired filtering if you remove the $ from the end of the regex and instead use the g flag to make it globally replace anything not in the specified character range with the empty string.
If you additionally want to trim trailing whitespace, it'd be better to do so using another regex, or a simpler trimRight call.