I was wondering if there is a way having this
var string = "foo::bar"
To get the last part of the string: "bar" using just regex.
I was trying to do look-aheads but couldn't master them enough to do this.
--
UPDATE
Perhaps some examples will make the question clearer.
var st1 = "foo::bar::0"
match should be 0
var st2 = "foo::bar::0-3aab"
match should be 0-3aab
var st3 = "foo"
no match should be found
You can use a negative lookahead:
/::(?!.*::)(.*)$/
The result will then be in the capture.
Another approach:
/^.*::(.*)$/
This should work because the .* matches greedily, so the :: will match the last occurence of that string.
Simply,
/::(.+)$/
You can't use lookaheads unless you know exactly how long a string you're trying to match. Fortunately, this isn't an issue, because you're only looking at the end of the string $.
I wouldn't use regular expressions for this (although you certainly can); I'd split the string on ::, since that's conceptually what you want to do.
function lastToken(str) {
var xs = str.split('::');
return xs.length > 1 ? xs.pop() : null;
}
If you really want just a regular expression, you can use /::((?:[^:]|:(?!:))*)$/. First, it matches a literal ::. Then, we use parentheses to put the desired thing in capturing group 1. The desired thing is one or more copies of a (?:...)-bracketed string; this bracketing groups without capturing. We then look for either [^:], a non-colon character, or :(?!:), a colon followed by a non-colon. The (?!...) is a negative lookahead, which matches only if the next token doesn't match the contained pattern. Since JavaScript doesn't support negative lookbehinds, I can't see a good way to avoid capturing the :: as well, but you can wrap this in a function:
function lastTokenRegex(str) {
var m = str.match(/::((?:[^:]|:(?!:))*)$/);
return m && m[1];
}
var string2 = string.replace(/.*::/, "");
though perhaps string isn't the best choice of name for your string?
Related
I need a regular expression for :
<<12.txt>> <<45.txt>
I have created a regular expression :
<<.+.txt>>
But this found one match in whole string but here is 2 matches:
<<12.txt>>
<<45.txt>>
if anyone have solution for this problem please help me out there
Part of the issue is that the string you've specified wouldn't match because the second > is missing in <<45.txt>.
Also, you're using the . (dot) selector, and also trying to find a period. It works, but now how you think it is.
Here's the regex you want:
var regex = /<<\d+\.txt>>/g
\d matches only numbers
\. matches an actual period
/g means global, so it won't stop at the first match
Practice Regular Expressions
https://regexr.com/43bs4
Demo
var string = "<<12.txt>> <<45.txt>>";
var regex = /<<\d+\.txt>>/g;
var matches = string.match(regex);
console.log(matches);
P.S., if you actually want to match with 1 > or 2 >>, you can with:
var regex = /<<\d+\.txt>>?/g
? optionally matches the character right before it
/<<.+.txt>>/gm
g is for global (will search through entire source)
m is for multi line search support
I'm trying to create an analog for php's isset ($_GET[param]) but for JavaScript.
So long, I get this
[?&]param[&=]
But if param is at the end of URL (example.com?param) this regex won't work.
You can play with it here: https://regex101.com/r/fFeWPW/1
If you want to make sure your match ends with &, = or end of string, you may replace the [&=] character class with a (?:[&=]|$) alternation group that will match &, = or end of string (note that $ cannot be placed inside the character class as it would lose its special meaning there and will be treated as a $ symbol), or you may use a negative lookahead (?![^&=]) that fails the match if there is no & or = immediately after the current location, which might be a bit more efficient than an alternation group.
So, in your case, it will look like
[?&]param(?:[&=]|$)
or
[?&]param(?![^&=])
See a regex demo
JS demo:
var strs = ['http://example.com?param', 'http://example.com?param=123', 'http://example.com?param&another','http://example.com?params'];
var rx = /[?&]param(?![^&=])/;
for (var s of strs) {
console.log(s, "=>", rx.test(s))
}
My string can be something like A01, B02, C03, possibly AA18 in the future as well. I thought I could use a regex to get just the letters and work on my regex since I haven't done much with it. I wrote this function:
function rowOffset(sequence) {
console.log(sequence);
var matches = /^[a-zA-Z]+$/.exec(sequence);
console.log(matches);
var letter = matches[0].toUpperCase();
return letter;
}
var x = "A01";
console.log(rowOffset(x));
My matches continue to be null. Am I doing this correctly? Looking at this post, I thought the regex was correct: Regular expression for only characters a-z, A-Z
You can use String#replace to remove all non letters from input string:
var r = 'AA18'.replace(/[^a-zA-Z]+/g, '');
//=> "AA"
Your main issue is the use of the ^ and $ characters in the regex pattern. ^ indicates the beginning of the string and $ indicates the end, so you pattern is looking for a string that is ONLY a group of one or more letters, from the beginning to the end of the string.
Additionally, if you want to get each individual instance of the letters, you want to include the "global" indicator (g) at the end of your regex pattern: /[a-zA-Z]+/g. Leaving that out means that it will only find the first instance of the pattern and then stop searching . . . adding it will match all instances.
Those two updates should get you going.
EDIT:
Also, you may want to use match() rather than exec(). If you have a string of multiple values (e.g., "A01, B02, C03, AA18"), match() will return them all in an array, whereas, exec() will only match the first one. If it is only ever one value, then exec() will be fine (and you also wouldn't need the "global" flag).
If you want to use match(), you need to change your code order just a bit to:
var matches = sequence.match(/[a-zA-Z]+/g);
To return an array of separate letters remove +:
var matches = sequence.match(/[a-zA-Z]/g);
You're confused about what the goal of the other question was: he wanted to check that there were only letters in his string.
You need to remove the anchors ^$, who match respectively the beginning and end of the string:
[a-zA-Z]+
This will match the first of letters in your input string.
If there might be more (ie you want multiple matches in your single string), use
sequence.match(/[a-zA-Z]+/g)
This /[^a-z]/g solves the problem. Look at the example below.
function pangram(str) {
let regExp = /[^a-z]/g;
let letters = str.toLowerCase().replace(regExp, '');
document.getElementById('letters').innerHTML = letters;
}
pangram('GHV 2## %hfr efg uor7 489(*&^% knt lhtkjj ngnm!##$%^&*()_');
<h4 id="letters"></h4>
You can do this:
var r = 'AA18'.replace(/[\W\d_]/g, ''); // AA
Also can be done by String.prototype.split(regex).
'AA12BB34'.split(/(\d+)/); // ["AA", "12", "BB", "34", ""]
'AA12BB34'.split(/(\d+)/)[0]; // "AA"
Here regex divides the giving string by digits (\d+)
It's my first time using Java Script....
What does this do?
var INTEGER_SINGLE = /\d+/;
What does the forward slashes tell you? How about the backslash? d means for digit?
Thanks!
That creates a regular expression that matches one or more digits.
Anything inside / / is a regular expression. \d matches a digit, and + is the positive closure, which means one or more.
Having said that, depending on what this regex is supposed to do, you may want to change it to:
var INTEGER_SINGLE = /^\d+$/;
^ matches the beginning of the string, and $ the end. The end result would be that any strings you try to match against the regex would have to satisfy it in the string's entirety.
var INTEGER_SINGLE = /^\d+$/;
console.log(INTEGER_SINGLE.test(12)); //true
console.log(INTEGER_SINGLE.test(12.5)); //false
Of course if the regex is supposed to only match a single integer anywhere in the string, then of course it's perfect just the way it is.
I need to replace any occurrence of a sequence of integers followed by a dash and then another sequence of integers, with only the first sequence of integers. For example:
THIS IS A STRING 2387263-1111 STRING CONTINUES
Will become:
THIS IS A STRING 2387263 STRING CONTINUES
Can I use that with Javascript and replace()?
You can do:
str = str.replace(/(\d+)-\d+/,'$1');
See it
Which replaces a group of digits followed by a hyphen followed by a group of digits with the first group of digits.
If you want to replace multiple occurrences of such pattern just use the g modifier as:
str = str.replace(/(\d+)-\d+/g,'$1');
NEW ANSWER -
Yes, in your case according to me, first you need to match that whole string "2387263-1111" using a regex and then remove that part followed by '-' and then replace the result in the original string.
Check the answer from codaddict. Mine would've almost been same but his answer seems more appropriate.
OLD ANSWER: -
Why replace? Just use split and get the first value.
var str = "2387263-1111";
var output = str.split("-")[0];
User RegExp function of javascript
str = str.replace(new RegExp("-[0-9]+"), " ");
Not really a new answer, just an addition to codaddict. Don't know JScript's regex
all that well, but I asume it uses extended regular expressions (if not, forget this).
If you need validation on boundry conditions you could do something like this:
str = str.replace( /((^|\s)\d+)-\d+(?=\s|$)/g, '$1' );
That would prevent matching this type of thing:
A STRING 2387263-1111STRING CONTINUES
A STRING2387263-1111 STRING CONTINUES
A STRING2387263-1111STRING CONTINUES