Here is my string:
Organization 2
info#something.org.au more#something.com market#gmail.com single#noidea.com
Organization 3
headmistress#money.com head#skull.com
Also this is my pattern:
/^.*?#[^ ]+|^.*$/gm
As you see in the demo, the pattern matches this:
Organization 2
info#something.org.au
Organization 3
headmistress#money.com
My question: How can I make it inverse? I mean I want to match this:
more#something.com market#gmail.com single#noidea.com
head#skull.com
How can I do that? Actually I can write a new (and completely different) pattern to grab expected result, but I want to know, Is "inverting the result of a pattern" possible?
No, I don't believe there is a way to directly inverse a Regular Expression but keeping it the same otherwise.
However, you could achieve something close to what you're after by using your existing RegExp to replace its matches with an empty string:
var everythingThatDidntMatchStr = str.replace(/^.*?#[^ ]+|^.*$/gm, '');
You can replace the matches from first RegExp by using Array.prototype.forEach() to replace matched RegExp with empty string using `String.ptototype.replace();
var re = str.match(/^.*?#[^ ]+|^.*$/gm);
var res = str;
re.forEach(val => res = res.replace(new RegExp(val), ""));
Related
I have a dilemma in using regex as I am very new in using this:
I have the URL below:
var url = https://website.com/something-here/page.html?p=null#confirmation?order=123
My expected result is:
/something-here/page.html #confirmation
It could be a space or a comma or simply combine the two(/something-here/page.html#confirmation)
I can do this using two regex below:
var a= url.match(/som([^#]+).html/)[0];
var b= url.match(/#([^#]+).tion/)[0];
console.log(a,b);
But I would like to have it done as a single regex with the same result.
You can use RegExp's group system to your advantage. Here's a snippet:
var matches = url.match(/(som[^#]+.html).*?(#[^#]+.tion)/);
console.log(matches[1] + " " + matches[2]); // prints /something-here/page.html #confirmation
I combined your two RegExp conditions into one, while enclosing them with parenthesis in the correct areas to create two groups.
That way, you can get the specified group and add a space in between.
Aside the fact that your example url is malformed (you have two search params), therefore not very suitable to work with - I have e proposition:
Why not use the URL object and its properties?
url = new URL("https://website.com/something-here/page.html?p=null#confirmation?order=123");
and precisely grab the properties with explicit syntax as in:
url.pathname; >> "something-here/page.html"
url.hash; >> "#confirmation?order=123"
But in case you explicitly need a RegExp variant
here is one
var url = "https://website.com/something-here/page.html?p=null#confirmation?order=123";
var match = url.match(/\/som.*?html|\#.*?tion/g);
console.log(match.join(" "));
Use each your condition in scope "( )" More details answer try find here
I have a url like http://www.somedotcom.com/all/~childrens-day/pr?sid=all.
I want to extract childrens-day. How to get that? Right now I am doing it like this
url = "http://www.somedotcom.com/all/~childrens-day/pr?sid=all"
url.match('~.+\/');
But what I am getting is ["~childrens-day/"].
Is there a (definitely there would be) short and sweet way to get the above text without ["~ and /"] i.e just childrens-day.
Thanks
You could use a negated character class and a capture group ( ) and refer to capture group #1. The caret (^) inside of a character class [ ] is considered the negation operator.
var url = "http://www.somedotcom.com/all/~childrens-day/pr?sid=all";
var result = url.match(/~([^~]+)\//);
console.log(result[1]); // "childrens-day"
See Working demo
Note: If you have many url's inside of a string you may want to add the ? quantifier for a non greedy match.
var result = url.match(/~([^~]+?)\//);
Like so:
var url = "http://www.somedotcom.com/all/~childrens-day/pr?sid=all"
var matches = url.match(/~(.+?)\//);
console.log(matches[1]);
Working example: http://regex101.com/r/xU4nZ6
Note that your regular expression wasn't actually properly delimited either, not sure how you got the result you did.
Use non-capturing groups with a captured group then access the [1] element of the matches array:
(?:~)(.+)(?:/)
Keep in mind that you will need to escape your / if using it also as your RegEx delimiter.
Yes, it is.
url = "http://www.somedotcom.com/all/~childrens-day/pr?sid=all";
url.match('~(.+)\/')[1];
Just wrap what you need into parenteses group. No more modifications into your code is needed.
References: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp
You could just do a string replace.
url.replace('~', '');
url.replace('/', '');
http://www.w3schools.com/jsref/jsref_replace.asp
If I have a String in JavaScript
key=value
How do I make a RegEx that matches key excluding =?
In other words:
var regex = //Regular Expression goes here
regex.exec("key=value")[0]//Should be "key"
How do I make a RegEx that matches value excluding =?
I am using this code to define a language for the Prism syntax highlighter so I do not control the JavaScript code doing the Regular Expression matching nor can I use split.
Well, you could do this:
/^[^=]*/ // anything not containing = at the start of a line
/[^=]*$/ // anything not containing = at the end of a line
It might be better to look into Prism's lookbehind property, and use something like this:
{
'pattern': /(=).*$/,
'lookbehind': true
}
According to the documentation this would cause the = character not to be part of the token this pattern matches.
use this regex (^.+?)=(.+?$)
group 1 contain key
group 2 contain value
but split is better solution
.*=(.*)
This will match anything after =
(.*)=.*
This will match anything before =
Look into greedy vs ungreedy quantifiers if you expect more than one = character.
Edit: as OP has clarified they're using javascript:
var str = "key=value";
var n=str.match(/(.*)=/i)[1]; // before =
var n=str.match(/=(.*)/i)[1]; // after =
var regex = /^[^=]*/;
regex.exec("key=value");
I have several Javascript strings (using jQuery). All of them follow the same pattern, starting with 'ajax-', and ending with a name. For instance 'ajax-first', 'ajax-last', 'ajax-email', etc.
How can I make a regex to only grab the string after 'ajax-'?
So instead of 'ajax-email', I want just 'email'.
You don't need RegEx for this. If your prefix is always "ajax-" then you just can do this:
var name = string.substring(5);
Given a comment you made on another user's post, try the following:
var $li = jQuery(this).parents('li').get(0);
var ajaxName = $li.className.match(/(?:^|\s)ajax-(.*?)(?:$|\s)/)[1];
Demo can be found here
Below kept for reference only
var ajaxName = 'ajax-first'.match(/(\w+)$/)[0];
alert(ajaxName);
Use the \w (word) pattern and bind it to the end of the string. This will force a grab of everything past the last hyphen (assuming the value consists of only [upper/lower]case letters, numbers or an underscore).
The non-regex approach could also use the String.split method, coupled with Array.pop.
var parts = 'ajax-first'.split('-');
var ajaxName = parts.pop();
alert(ajaxName);
you can try to replace ajax- with ""
I like the split method #Brad Christie mentions, but I would just do
function getLastPart(str,delimiter) {
return str.split(delimiter)[1];
}
This works if you will always have only two-part strings separated by a hyphen. If you wanted to generalize it for any particular piece of a multiple-hyphenated string, you would need to write a more involved function that included an index, but then you'd have to check for out of bounds errors, etc.
I was wondering how to use a regexp to match a phrase that comes after a certain match. Like:
var phrase = "yesthisismyphrase=thisiswhatIwantmatched";
var match = /phrase=.*/;
That will match from the phrase= to the end of the string, but is it possible to get everything after the phrase= without having to modify a string?
You use capture groups (denoted by parenthesis).
When you execute the regex via match or exec function, the return an array consisting of the substrings captured by capture groups. You can then access what got captured via that array. E.g.:
var phrase = "yesthisismyphrase=thisiswhatIwantmatched";
var myRegexp = /phrase=(.*)/;
var match = myRegexp.exec(phrase);
alert(match[1]);
or
var arr = phrase.match(/phrase=(.*)/);
if (arr != null) { // Did it match?
alert(arr[1]);
}
phrase.match(/phrase=(.*)/)[1]
returns
"thisiswhatIwantmatched"
The brackets specify a so-called capture group. Contents of capture groups get put into the resulting array, starting from 1 (0 is the whole match).
It is not so hard, Just assume your context is :
const context = "https://example.com/pa/GIx89GdmkABJEAAA+AAAA";
And we wanna have the pattern after pa/, so use this code:
const pattern = context.match(/pa\/(.*)/)[1];
The first item include pa/, but for the grouping second item is without pa/, you can use each what you want.
Let try this, I hope it work
var p = /\b([\w|\W]+)\1+(\=)([\w|\W]+)\1+\b/;
console.log(p.test('case1 or AA=AA ilkjoi'));
console.log(p.test('case2 or AA=AB'));
console.log(p.test('case3 or 12=14'));
If you want to get value after the regex excluding the test phrase, use this:
/(?:phrase=)(.*)/
the result will be
0: "phrase=thisiswhatIwantmatched" //full match
1: "thisiswhatIwantmatched" //matching group