This question already has answers here:
Regex lazy vs greedy confusion
(2 answers)
Why does a simple .*? non-greedy regex greedily include additional characters before a match?
(3 answers)
Closed 3 years ago.
I'm trying this in javascript
/\/.*?$/.exec('foo/bar/tar')[0]
I was expecting to get /tar as result but getting /bar/tar. As far as I understand non-greed regex would take the smallest match.
I'm circumventing this with myvar.split('/').reverse()[0] but I couldn't understand what is going wrong with the regex.
There is nothing wrong with the regex but the pattern \/.*?$ matches from the first forward slash until the end of the string non greedy.
The dot matches any character except a newline and does not take a forward slash into account, so that will result in /bar/tar.
If you want to match /tar, you could match a forward slash, followed by not matching anymore forward slashes using a negated character class and then assert the end of the string.
\/[^\/]+$
Pattern demo
console.log(/\/[^\/]+$/.exec('foo/bar/tar')[0]);
Related
This question already has answers here:
Why this javascript regex doesn't work?
(1 answer)
Match exact string
(3 answers)
Closed 5 years ago.
Regex is the bane of my existence. I've done plenty tutorials, but the rules never stick, and when I look them up they seem to conflict. Anyways enough of my whining. Could someone tell me why this regex doesn't exclude hyphens or brackets:
/^[A-Za-z_][A-Za-z\d_]*/
The way I understand it (or at least what I'm trying to do), the ^ character dictates that the regex should start with the next thing on the list That means the regex should start with [A-Za-z_] or any character a-z and A-Z as well as and underscore _. Then the string can have anything that includes [A-Za-z\d_] which is any alphanumeric character and an underscore. Then I use the * to say that the string can have any number of what was presented previously (any alphanumeric character plus underscore). At no point to I specify a bracket [ or a hyphen -. Why does this expression not exclude these characters
Extra info
I'm verifying this with javascript:
function variableName(name) {
const reg = RegExp("^[A-Za-z_][A-Za-z\d_]*")
return reg.test(name)
}
function variableName("va[riable0") // returns true should be false
It's actually matching the first 2 letters("va"), that's why it's true.
To match the whole phrase, your reg expression should have "$" at the end:
"^[A-Za-z_][A-Za-z\d_]*$"
Your regex matches the part of the string that does not contain the bracket, because your're missing the $ anchor that would (together with ^) force it to match the whole string. Use
const reg = /^[A-Za-z_][A-Za-z\d_]*$/g
// ^
function variableName(name) {
return reg.test(name)
}
console.log(variableName("va[riable0"))
This question already has answers here:
How can I match a pipe character followed by whitespace and another pipe?
(5 answers)
What special characters must be escaped in regular expressions?
(13 answers)
Closed 5 years ago.
I'm using this JS regex text.match(/|(https:.+?path.+?)|/)[1] to get a regex of a URL that is in between pipe | characters but it's not working.
The text is ||https://url.com/path/value|| but I can't seem to extract the URL from it. I need to have path in the middle to identify this particular URL since there are other URLs in the file.
It doesn't have to be a URL that I'm extracting. I mainly would like to know how to extract something from between a pair of characters (| in this case).
You need to escape the pipe ("|") characters:
text.match(/\|(https:.+?path.+?)\|/)[1]
Pipe is a special character that basically means "or". https://www.regular-expressions.info/alternation.html
To grab everything between the two sets of || then you could use this regex:
text.match(/\|\|(.*)\|\|/)
The first part \|\| matches the characters || literally.
The next part (.*)matches any character zero or more and groups the result.
The last part \|\| matches the closing characters || literally.
This question already has answers here:
Regex to find not start and end with dot and allow some special character only not all
(3 answers)
Closed 2 years ago.
I am using following
ng-pattern="/^[a-zA-Z][a-zA-Z0-9._](.*[a-zA-Z0-9])?$/"
The matching String should
not start with a special character,
not end with special character, and
not include consecutive symbols except . (dot) and _ (underscore).
But it is not working.
Please, any suggestion.
Try using the word character class as a start ([\w] = [a-zA-Z0-9_]):
I'm not sure what you mean by consecutive symbols. But this might help:
/^[a-zA-Z]([\w.]*[a-zA-Z0-9])?$/
Maybe, have a look at the JavaScript RegExp Reference
This question already has answers here:
regex pattern to match a type of strings
(4 answers)
Closed 8 years ago.
I need to match the below type of strings using a regex pattern in javascript.
E.g. /this/<one or more than one word with hyphen>/<one or more than one word with hyphen>/<one or more than one word with hyphen>/<one or more than one word with hyphen>
So this single pattern should match both these strings:
1. /this/is/single-word
2. /this/is more-than/single/word-patterns/to-be-matched
Only the slash (/)and the 'this' in the beginning are consistent and contains only alphabets.
Try this -
^\/this(?:\/[\w\- ]+)+$
Demo here
There are some inconsistencies in your question, and it's not quite clear exactly what you want to match.
That being said, the following regex will provide a loose starting point for the exact strings that you want.
/this/(?:[\w|-]+/?){1,10}
This assumes the ' ' in your url was not intentional. This example will match a url with '/this/' + 1 to 10 additional '/' chunks.
(?:) -> non-matching group
[\w|-]+ -> one or more word characters or a hyphen
/? -> zero or one slashes
{1,10} -> 1 to 10 of the previous element, the non-matching group
This question already has answers here:
Reference - What does this regex mean?
(1 answer)
What is a non-capturing group in regular expressions?
(18 answers)
What does ?! mean?
(3 answers)
Closed 2 years ago.
Like the regexp in this one? What does it match?
document.getElementById("MyElement").className =
document.getElementById("MyElement").className.replace
( /(?:^|\s)MyClass(?!\S)/ , '' )
?: means make the capturing group a non capturing group, i.e. don't include its match as a back-reference. This is often done to increase performance and de-clutter the back-references when a capturing group is necessary to use the | operator.
In your example, it is being used to allow the or (|) of the start of the string ^ or whitespace (\s). Since the author of this code doesn't care about what it matched, they have made it a non capturing group.
?! is the negative lookahead. The regex will only match if the capturing group does not match.
In this example, the author wants to ensure the character after MyClass is not a whitespace character (\S).
It is somewhat possible the author of this code could have used word boundaries instead (\b).
The regular expression (?:^|\s) is a non-capturing group that matches either the start of the line or a whitespace character.
The regular expression (?!\S) is a negative lookahead assertion which succeeds either at the end of the string, or else when the next character is a whitespace character.