Is there anyway to pick up back-references?
var name = "HELLO WORLD"
var patt = /\S+\s(.+)/;
alert(name.match(patt));
This is just a simple example to get every word after the first.
But, if I alert $1, nothing pops up and I'm not sure why. I'd appreciate any help!
Just a quick solution to this if anyone is actually looking for this.
You can use RegExp.$X where X is the digit of the reference you seek. match returns an array as well, which you can use to see whether things matched as well as the backreferences in the following indices.
Related
I wrote a fairly simple regular expression to detect when a string looks like it could be an email:
var looksLikeEmail = /^\S+#\S+\.\S+$/gi;
I'm using Knockout and the string being tested is the value of a textarea.
Essentially, say we have the value of the textarea in a variable text. This value was, for example, the typed in value abc#example.com.
What's odd, is it seems like, even though text === text.trim(), looksLikeEmail.test(text) returns true, but looksLikeEmail.test(text.trim()) returns false.
On the other hand, if I manually create the string var test2 = 'abc#example.com', it does not have this issue.
This seems to indicate to me that the textarea is inserting some odd characters or something... that .trim() is doing something weird with. But test.length === test2.length and test.length === test.trim().length
Does anyone know how to make this behave correctly?
I've written up a jsfiddle to quickly demonstrate the behavior...
If you go to the fiddle and try typing in an email... you will see the problem. another weird behavior: add a space after the email, then remove it. /confused
Any help is much appreciated. Thanks.
.test(), just like .exec() will remember the last index of a match when using a global regex, and try to match from it onward, failing on the second call. Just remove the /g option from your regex - it doesn't make sense to have /g in a non-multiline regex which matches beginning and end.
I need to split a string to one or more substrings each of which contains no more or less than two dots. For example, if the string is foo.boo.coo.too" then what would be the regex to get the following array?: ["foo.boo.coo", "boo.coo.too"]. I hope there will be someone to answer this question - I will really admire you, as I've been programming for several years and have not still be used to regular expressions well enough to solve this particular problem by myself. Thank you very much in advance. Let me know your identity so that I can credit you as a contributor of the program I am creating.
RegEx is for this Problem not the best solution a similar problem was discussed here: split-a-sting-every-3-characters-from-back-javascript
A good javascript solution would be a javascript function like this
function splitter(text){
var parts = text.split(".");
var times = parts.length - 2;
var values = [];
for(var index = 0; index<times;index++)
{
values.push(parts.slice(index,index+3).join("."));
}
return values;
}
splitter("too.boo.coo.too")
//=> Result tested on Chrome 25+ ["too.boo.coo", "boo.coo.too"]
I hope this helps
If you want to Use Regex try the Lookhead Stuff, this could help http://www.regular-expressions.info/lookaround.html
Regex by its nature will return non-intersecting results, so if you want "all matches" from a single regex - it's not possible.
So basically you will need to find first match, and then start from next position to find next match and so on; something like this technique described here regex matches with intersection in C# (it's not JavaScript but idea is the same)
You can use the following regex for example:
(?<=^|\.)((?:[^.]*\.){2}[^.]*?)(?=$|\.)
It ensures that it starts and ends with dot, or at begin/end of line, and contains exactly two dots inside, and captures result in first capture. You can replace * with + to make sure at least one symbol exists between dots, if it is required.
But you need to understand that such approach has really bad performance for the task you are solving, so may be using other way (like split + for) will be better solution.
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_\-?=&.]+)$
Hi need to extract ONE letter from a string.
The string i have is a big block of html, but the part where i need to search in is this text:
Vahvistustunnus M :
And I need to get the M inside the nbsp's
So, who is the quickest regex-guru out there? :)
Ok, according to this page in the molybdenum api docs, the results will be all of the groups concatenated together. Given that you just want the char between the two 's then it's not good enough to match the whole thing and then pull out the group. Instead you'll need to do something like this:
(?<=Vahvistustunnus )[a-zA-Z](?= )
Warning
This might not work for you because lookbehinds (?<=pattern) are not available in all regex flavors. Specifically, i think that because molybdenum is a firefox extension, then it's likely using ECMA (javascript) regex flavor. And ECMA doesn't support lookbehinds.
If that's the case, then i'm gonna have to ask someone else to answer your question as my regex ninja (amateur) skills don't go much further than that. If you were using the regex in javascript code, then there are ways around this limitation, but based on your description, it sounds like you have to solve this problem with nothing but a raw regex?
Looks like it uses JavaScript and if so
var str = "Vahvistustunnus M :";
var patt = "Vahvistustunnus ([A-Z]) :";
var result = str.match(patt)[1];
should work.
I have IDs that look like: 185-51-671 but they can also have letters at the end, 175-1-7b
All I want to do is remove the hyphens, as a pre-processing step. Show me some cool ways to do this in javascript? I figure there are probably quite a few questions like this one, but I'm interested to see what optimizations people will come up with for "just hyphens"
Thanks!
edit: I am using jQuery, so I guess .replace(a,b) does the trick (replacing a with b)
numberNoHyphens = number.replace("-","");
any other alternatives?
edit #2:
So, just in case anyone is wondering, the correct answer was
numberNoHyphens = number.replace(/-/g,"");
and you need the "g" which is the pattern switch or "global flag" because
numberNoHyphens = number.replace(/-/,"");
will only match and replace the first hyphen
You need to include the global flag:
var str="185-51-671";
var newStr = str.replace(/-/g, "");
This is not faster, but
str.split('-').join('');
should also work.
I set up a jsperf test if anyone wants to add and compare their methods, but it's unlikely anything will be faster than the replace method.
http://jsperf.com/remove-hyphens-from-string
var str='185-51-671';
str=str.replace(/-/g,'');
Gets much easier in String.prototype.replaceAll(). Check out the browser support for the built-in method.
const str = '185-51-671';
console.log(str.replaceAll('-', ''));
Som of these answers, prior to edits, did not remove all of the hyphens. You would need to use .replaceAll("-","")
In tidyverse, there are multiple functions that could suit your needs. Specifically, I would use str_remove, which will replace in a string, the giver character by an empty string (""), effectively removing it (check here the documentation). Example of its usage:
str_remove(x, '-')