I'm working on a JavaScript to extract a URL from a Google search URL, like so:
http://www.google.com/search?client=safari&rls=en&q=thisisthepartiwanttofind.org&ie=UTF-8&oe=UTF-8
Right now, my code looks like this:
var checkForURL = /[\w\d](.org)/i;
var findTheURL = checkForURL.exec(theURL);
I've ran this through a couple regex testers and it seems to work, but in practice the string I get returned looks like this:
thisisthepartiwanttofind.org,.org
So where's that trailing ,.org coming from?
I know my pattern isn't super robust but please don't suggest better patterns to use. I'd really just like advice on what in particular I did wrong with this one. Thanks!
Remove the parentheses in the regex if you do not process the .org (unlikely since it is a literal). As per #Mark comment, add a + to match one or more characters of the class [\w\d]. Also, I would escape the dot:
var checkForURL = /[\w\d]+\.org/i;
What you're actually getting is an array of 2 results, the first being the whole match, the second - the group you defined by using parens (.org).
Compare with:
/([\w\d]+)\.org/.exec('thisistheurl.org')
→ ["thisistheurl.org", "thisistheurl"]
/[\w\d]+\.org/.exec('thisistheurl.org')
→ ["thisistheurl.org"]
/([\w\d]+)(\.org)/.exec('thisistheurl.org')
→ ["thisistheurl.org", "thisistheurl", ".org"]
The result of an .exec of a JS regex is an Array of strings, the first being the whole match and the subsequent representing groups that you defined by using parens. If there are no parens in the regex, there will only be one element in this array - the whole match.
You should escape .(DOT) in (.org) regex group or it matches any character. So your regex would become:
/[\w\d]+(\.org)/
To match the url in your example you can use something like this:
https?://([0-9a-zA-Z_.?=&\-]+/?)+
or something more accurate like this (you should choose the right regex according to your needs):
^https?://([0-9a-zA-Z_\-]+\.)+(com|org|net|WhatEverYouWant)(/[0-9a-zA-Z_\-?=&.]+)$
Related
I was a bit surprised, that actually no one had the exact same issue in javascript...
I tried several different solutions none of them parse the content correctly.
The closest one I tried : (I stole its regex query from a PHP solution)
const test = `abc?aaa.abcd?.aabbccc!`;
const sentencesList = test.split("/(\?|\.|!)/");
But result just going to be
["abc?aaa.abcd?.aabbccc!"]
What I want to get is
['abc?', 'aaa.', 'abcd?','.', 'aabbccc!']
I am so confused.. what exactly is wrong?
/[a-z]*[?!.]/g) will do what you want:
const test = `abc?aaa.abcd?.aabbccc!`;
console.log(test.match(/[a-z]*[?!.]/g))
To help you out, what you write is not a regex. test.split("/(\?|\.|!)/"); is simply an 11 character string. A regex would be, for example, test.split(/(\?|\.|!)/);. This still would not be the regex you're looking for.
The problem with this regex is that it's looking for a ?, ., or ! character only, and capturing that lone character. What you want to do is find any number of characters, followed by one of those three characters.
Next, String.split does not accept regexes as arguments. You'll want to use a function that does accept them (such as String.match).
Putting this all together, you'll want to start out your regex with something like this: /.*?/. The dot means any character matches, the asterisk means 0 or more, and the questionmark means "non-greedy", or try to match as few characters as possible, while keeping a valid match.
To search for your three characters, you would follow this up with /[?!.]/ to indicate you want one of these three characters (so far we have /.*?[?!.]/). Lastly, you want to add the g flag so it searches for every instance, rather than only the first. /.*?[?!.]/g. Now we can use it in match:
const rawText = `abc?aaa.abcd?.aabbccc!`;
const matchedArray = rawText.match(/.*?[?!.]/g);
console.log(matchedArray);
The following code works, I do not think we need pattern match. I take that back, I have been answering in Java.
final String S = "An sentence may end with period. Does it end any other way? Ofcourse!";
final String[] simpleSentences = S.split("[?!.]");
//now simpleSentences array has three elements in it.
I am trying to edit a DateTime string in typescript file.
The string in question is 02T13:18:43.000Z.
I want to trim the first three characters including the letter T from the beginning of a string AND also all 5 characters from the end of the string, that is Z000., including the dot character. Essentialy I want the result to look like this: 13:18:43.
From what I found the following pattern (^(.*?)T) can accomplish only the first part of the trim I require, that leaves the initial result like this: 13:18:43.000Z.
What kind of Regex pattern must I use to include the second part of the trim I have mentioned? I have tried to include the following block in the same pattern (Z000.)$ but of course it failed.
Thanks.
Any help would be appreciated.
There is no need to use regular expression in order to achieve that. You can simply use:
let value = '02T13:18:43.000Z';
let newValue = value.slice(3, -5);
console.log(newValue);
it will return 13:18:43, assumming that your string will always have the same pattern. According to the documentation slice method will substring from beginIndex to endIndex. endIndex is optional.
as I see you only need regex solution so does this pattern work?
(\d{2}:)+\d{2} or simply \d{2}:\d{2}:\d{2}
it searches much times for digit-digit-doubleDot combos and digit-digit-doubleDot at the end
the only disadvange is that it doesn't check whether say there are no minutes>59 and etc.
The main reason why I didn't include checking just because I kept in mind that you get your dates from sources where data that are stored are already valid, ex. database.
Solution
This should suffice to remove both the prefix from beginning to T and postfix from . to end:
/^.*T|\..*$/g
console.log(new Date().toISOString().replace(/^.*T|\..*$/g, ''))
See the visualization on debuggex
Explanation
The section ^.*T removes all characters up to and including the last encountered T in the string.
The section \..*$ removes all characters from the first encountered . to the end of the string.
The | in between coupled with the global g flag allows the regular expression to match both sections in the string, allowing .replace(..., '') to trim both simultaneously.
Say I have a string "&something=variable&something_else=var2"
I want to match between &something= and &, so I'll write a regular expression that looks like:
/(&something=).*?(&)/
And the result of .match() will be an array:
["&something=variable&", "&something=", "&"]
I've always solved this by just replacing the start and end elements manually but is there a way to not include them in the match results at all?
You're using the wrong capturing groups. You should be using this:
/&something=(.*?)&/
This means that instead of capturing the stuff you don't want (the delimiters), you capture what you do want (the data).
You can't avoid them showing up in your match results at all, but you can change how they show up and make it more useful for you.
If you change your match pattern to /&something=(.+?)&/ then using your test string of "&something=variable&something_else=var2" the match result array is ["&something=variable&", "variable"]
The first element is always the entire match, but the second one, will be the captured portion from the parentheses, which is much more useful, generally.
I hope this helps.
If you are trying to get variable out of the string, using replace with backreferences will get you what you want:
"&something=variable&something_else=var2".replace(/^.*&something=(.*?)&.*$/, '$1')
gives you
"variable"
I have a string of text I'm trying to parse with RegExp in JavaScript. Let's say it looks like this:
var myString = "This is a string of text (item item item) and more text here.";
I need to match the first occurrence of the word 'item' based on the fact that it is the first item inside the parentheses. I can't figure out how to write a pattern that will match ONLY the first item inside a set of parentheses.
Some of you may want to think of it like this: Pretend I'm parsing a string of Lisp and want to match all cars.
Thanks in advance.
If you can avoid checking for an opening parenthesis you could use a look ahead for the closing parenthesis.
item(?=[^\)]*\))
You can also use capturing groups with:
\(.*?(item).*?\)
EDIT
For a word that is repeated somewhere in the parentheses at least once:
\(.*?(\b\w+\b).*?\1.*?\)
EDIT 2
For just the first alphanumeric set of characters in a parentheses:
\([^\w]*(\b\w+\b).*?\)
Or a simpler alternative:
\(.*?\b(\w+)\b.*?\)
Use something like this:
/\((\w*?)(?:\s|\))/
Live example: http://tinkerbin.com/EJyjEX30 (click run to run the code))
You cannot make a pattern that doesn't match the opening parenthesis, since Javascript regexes don't support lookback.
You probably need to update your lexer code to understand capturing groups (or general functions or something convenient...) or you need to restructure you code so that it can deal with that extra paren you can't get rid of.
If I am correct, the following code will only match a URL that is exactly as presented.
However, what would it look like if you wanted to identify subdomains as well as urls that contain various different query strings - in other words, any address that contains this domain:
var url = /test.com/
if (window.location.href.match(url)){
alert("match!");
}
If you want this regex to match "test.com" you need to escape the "." and both of the "/" that means any character in regex syntax.
Escaped : \/test\.com\/
Take a look for here for more info
No, your pattern will actually match on all strings containing test.com.
The regular expresssion /test.com/ says to match for test[ANY CHARACTER]com anywhere in the string
Better to use example.com for example links. So I replaces test with example.
Some example matches could be
http://example.com
http://examplexcom.xyz
http://example!com.xyz
http://example.com?q=123
http://sub.example.com
http://fooexample.com
http://example.com/asdf/123
http://stackoverflow.com/?site=example.com
I think you need to use /g. /g enables "global" matching. When using the replace() method, specify this modifier to replace all matches, rather than only the first one:
var /test.com/g;
If you want to test if an URL is valid this is the one I use. Fairly complex, because it takes care also of numeric domain & a few other peculiarities :
var urlMatcher = /(([\w]+:)?\/\/)?(([\d\w]|%[a-fA-f\d]{2,2})+(:([\d\w]|%[a-fA-f\d]{2,2})+)?#)?([\d\w][-\d\w]{0,253}[\d\w]\.)+[\w]{2,4}(:[\d]+)?(\/([-+_~.\d\w]|%[a-fA-f\d]{2,2})*)*(\?(&?([-+_~.\d\w]|%[a-fA-f\d]{2,2})=?)*)?(#([-+_~.\d\w]|%[a-fA-f\d]{2,2})*)?/;
Takes care of parameters and anchors etc... dont ask me to explain the details pls.