For text validation for chars I am using like [a-zA-z] and for numbers like [0-9] ..if I need to add special symbols I am adding slash like [a-zA-z/-/].
While including lot of symbols its getting difficult and my javascript is getting extremely big. Is there an easy way to do it ?
Regards
A.Collins
You can take a look at this cheat sheet. for instance, [0-9] can be reduced to \d.
For the general case of "a lot of characters" — no.
\w for alphanumerics and underscores
\d for digits
\s for whitespace
You can mix them, resulting in stuff like, for example, [\d.] (for matching numbers & dots).
In a character class, x-y means "all characters between x and y". If you just have one additional character, in your case / then you don't need to use the x-y format, you can just drop the character in:
[a-zA-z/]
That's not the correct way to escape characters. \ is the correct escape character to be used:
[a-zA-Z\/]
You can use character classes, [a-zA-Z0-9] can be replaced by [\w]. The only characters that needs to be escaped are \ and -. ^ should be escaped too when it's the first character in the character class.
Related
I'm looking for a Regular Expression for JavaScript that will identify word boundaries in English, while accepting hyphens and apostrophes that appear inside words, but excluding those that appear alone or at the beginning or end of a word.
For example, for the sentence ...
She said - 'That'll be all, Two-Fry.'
... I want the characters shown in grey below to be detected:
Shesaid- 'That'llbeall,Two-Fry.'
If I use the regex /[^A-Za-z'-]/g, then "loose" hyphens and apostrophes are not detected.
Shesaid-'That'llbeall,Two-Fry.'
How can I alter my regex so that it detects apostrophes and hyphens that don't have a word character on both sides?
You can test my regex here: https://regex101.com/r/bR8sV1/2
Note: the text I will be working on may contain other writing scripts, like руский and ไทอ so it will not be feasible to simply include all the characters that are not part of any English word.
You can organize your word-boundary characters into two groups.
Characters that cannot be alone.
Characters that can be alone.
A regex that works with your example would be:
[\s.,'-]{2,}|[\s.]
Regex101 Demo
Now all that's left is to keep adding all non-word characters into those two groups until it fits all of your needs. So you might start adding symbols and more punctuation to those character classes.
You could write something like that:
(\s|[!-/]|[:-#]|[\[-`]|[\{-~])*\s(\s|[!-/]|[:-#]|[\[-`]|[\{-~])*
Or the compact version:
(\s|[!-/:-#\[-`\{-~])*\s(\s|[!-/:-#\[-`\{-~])*
The RegExp requires one \s (Space character) and selects als spaces and non alphanumeric chars before and after it.
https://regex101.com/r/bR8sV1/4
\s matches all spaces
!-/ every char from ! to /
:-# every char from : to #
\[-`` every char from [ to ``
\{-~ every char from { to ~
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 !#£$%^&*()+=) i.e., characters that are not included in the character class `[...]`
3.that has ending
NOTE:
If the symbols you don't want it to have also includes - , a hyphen, don't put it in between some other characters because it is a metacharacter in character class, put it at last of character class
Two quick questions:
What would be a RegEx string for three letters and two numbers with space before and after them (i.e. " LET 12 ")?
Would you happen to know any good RegEx resources/tools?
For a good resource, try this website and the program RegexBuddy. You may even be able to figure out the answer to your question yourself using these sites.
To start you off you want something like this:
/^[a-zA-Z]{3}\s+[0-9]{2}$/
But the exact details depend on your requirements. It's probably a better idea that you learn how to use regular expressions yourself and then write the regular expression instead of just copying the answers here. The small details make a big difference. Examples:
What is a "letter"? Just A-Z or also foreign letters? What about lower case?
What is a "number"? Just 0-9 or also foreign numerals? Only integers? Only positive integers? Can there be leading zeros?
Should there be a single space between the letters and numbers? Or any amount of any whitespace? Even none?
Do you want to search for this string in a larger text? Or match a line exactly?
etc..
The answers to these questions will change the regular expression. It would be much faster for you in the long run to learn how to create the regular expression than to completely specify your requirements and wait for other people to reply.
I forgot to mention that there will be a space before and after. How do I include that?
Again you need to consider the questions:
Do you mean just one space or any amount of spaces? Possibly not always a space but only sometimes?
Do you mean literally a space character or any whitespace characters?
My guess is:
/^\s+[a-zA-Z]{3}\s+[0-9]{2}\s+$/
/[a-z]{3} [0-9]{2}/i will match 3 letters followed by a whitespace character, and then 2 numbers. [a-z] is a character class containing the letters a through z, and the {3} means that you want exactly 3 members of that class. The space character matches a literal space (alternately, you could use \s, which is a "shorthand" character class that matches any whitespace character). The i at the end is a pattern modifier specifying that your pattern is case-insenstive.
If you want the entire string to only be that, you need to anchor it with ^ and $:
/^[a-z]{3} [0-9]{2}$/i
Regular expression resources:
http://www.regular-expressions.info - great tutorial with a lot of information
http://rexv.org/ - online regular expression tester that supports a variety of engines.
^([A-Za-z]{3}) ([0-9]{2})$ assuming one space between the letters/numbers, as in your example. This will capture the letters and numbers separately.
I use http://gskinner.com/RegExr/ - it allows you to build a regex and test it with your own text.
As you can probably tell from the wide variety of answers, RegEx is a complex subject with a wide variety of opinions and preferences, and often more than one way of doing things. Here's my preferred solution.
^[a-zA-Z]{3}\s*\d{2}$
I used [a-zA-Z] instead of \w because \w sometimes includes underscores.
The \s* is to allow zero or more spaces.
I try to use character classes wherever possible, which is why I went with \d.
\w{3}\s{1}\d{2}
And I like this site.
EDIT:[a-zA-Z]{3}\s{1}\d{2} - The \w supports numeric characters too.
try this regularexpression
[^"\r\n]{3,}
I'm trying to write a regular that will check for numbers, spaces, parentheses, + and -
this is what I have so far:
/\d|\s|\-|\)|\(|\+/g
but im getting this error: unmatched ) in regular expression
any suggestions will help.
Thanks
Use a character class:
/[\d\s()+-]/g
This matches a single character if it's a digit \d, whitespace \s, literal (, literal ), literal + or literal -. Putting - last in a character class is an easy way to make it a literal -; otherwise it may become a range definition metacharacter (e.g. [A-Z]).
Generally speaking, instead of matching one character at a time as alternates (e.g. a|e|i|o|u), it's much more readable to use a character class instead (e.g. [aeiou]). It's more concise, more readable, and it naturally groups the characters together, so you can do e.g. [aeiou]+ to match a sequence of vowels.
References
regular-expressions.info/Character Class
Caveat
Beginners sometimes mistake character class to match [a|e|i|o|u], or worse, [this|that]. This is wrong. A character class by itself matches one and exactly one character from the input.
Related questions
Regex: why doesn’t [01-12] range work as expected?
Here is an awesome Online Regular Expression Editor / Tester! Here is your [\d\s()+-] there.
/^[\d\s\(\)\-]+$/
This expression matches only digits, parentheses, white spaces, and minus signs.
example:
888-111-2222
888 111 2222
8881112222
(888)111-2222
...
You need to escape your parenthesis, because parenthesis are used as special syntax in regular expressions:
instead of '(':
\(
instead of ')':
\)
Also, this won't work with '+' for the same reason:
\+
Edit: you may want to use a character class instead of the 'or' notation with '|' because it is more readable:
[\s\d()+-]
Try this:
[\d\s-+()]
I am learning javascript and I am analyzing existing codes.
In my JS reference book, it says to search on a single space use "\s"?
But I have came across the code
obj.match(/Kobe Bryant/);
Instead of using \s, it uses the actual space?
Why doesn't this generate an error?
The character class \s does not just contain the space character but also other Unicode white space characters. \s is equivalent to this character class:
[\t\n\v\f\r \u00a0\u2000\u2001\u2002\u2003\u2004\u2005\u2006\u2007\u2008\u2009\u200a\u200b\u2028\u2029\u3000]
No. It is perfectly legal to include a literal space in a regex.
However, it's not equivalent - \s will include any whitespace character, including tabs, non-breaking spaces, half-width spaces and other characters, whereas a literal space will only match the regular space character.
\s matches any whitespace character, including tabs etc. Sure you can use a literal space also without problems. Just like you can use [0-9] instead of \d to denote any digit. However, keep in mind that [0-9] is equivalent to \d whereas the literal space is a subset of \s.
In addition to normal spaces, \s matches different kinds of white space characters, including tabs (and possibly newline characters, according to configuration). That said, matching with a normal space is certainly valid, especially in your case where it seems you want to match a name, which is normally separated by a normal space.