Develop Reference

extract number using regex

I have a string:
name:demo;morestuff.nbvideo:3;morestuff_here:45
from which I need to extract the nbvideo number. I managed it with 2 regexes, but I'm sure it can be done in just one regex.
Here's what I have now:
// get the nbvideo:XX part
videoPart = sink.tag.match(/nbvideo:([0-9]+)/gi);
// get the number from the video part
videoCount = videoPart[0].match(/([0-9]+)/gi)[0];
How can I extract the number behind 'nbvideo:' with one single regex?

Remove g from the modifiers and access the first capture group value like this:
var sink_tag = "name:demo;morestuff.nbvideo:3;morestuff_here:45";
var m = sink_tag.match(/nbvideo:([0-9]+)/i);
if (m) {
videoPart = m[1];
document.body.innerHTML = videoPart; // demo
}
The thing is that string#match does not keep captures if a global modifier is used with a regex, and it seems you just have one nbvideo:<NUMBER> in the input. So, removing /g seems to be enough. Else, use RegExp#exec() in a loop.

How to find in javascript with regular expression string from url?

Good evening, How can I find in javascript with regular expression string from url address for example i have url: http://www.odsavacky.cz/blog/wpcproduct/mikronebulizer/ and I need only string between last slashes (/ /) http://something.cz/something/string/ in this example word that i need is mikronebulizer. Thank you very much for you help.

You could use a regex match with a group.
Use this:
/([\w\-]+)\/$/.exec("http://www.odsavacky.cz/blog/wpcproduct/mikronebulizer/")[1];
Here's a jsfiddle showing it in action
This part: ([\w\-]+)
Means at least 1 or more of the set of alphanumeric, underscore and hyphen and use it as the first match group.
Followed by a /
And then finally the: $
Which means the line should end with this
The .exec() returns an array where the first value is the full match (IE: "mikronebulizer/") and then each match group after that.
So .exec()[1] returns your value: mikronebulizer

Simply:
url.match(/([^\/]*)\/$/);
Should do it.
If you want to match (optionally) without a trailing slash, use:
url.match(/([^\/]*)\/?$/);
See it in action here: http://regex101.com/r/cL3qG3

If you have the url provided, then you can do it this way:
var url = 'http://www.odsavacky.cz/blog/wpcproduct/mikronebulizer/';
var urlsplit = url.split('/');
var urlEnd = urlsplit[urlsplit.length- (urlsplit[urlsplit.length-1] == '' ? 2 : 1)];
This will match either everything after the last slash, if there's any content there, and otherwise, it will match the part between the second-last and the last slash.

Something else to consider - yes a pure RegEx approach might be easier (heck, and faster), but I wanted to include this simply to point out window.location.pathName.
function getLast(){
// Strip trailing slash if present
var path = window.location.pathname.replace(/\/$?/, '');
return path.split('/').pop();
}

Alternatively you could get using split:
var pieces = "http://www.odsavacky.cz/blog/wpcproduct/mikronebulizer/".split("/");
var lastSegment = pieces[pieces.length - 2];
// lastSegment == mikronebulizer

var url = 'http://www.odsavacky.cz/blog/wpcproduct/mikronebulizer/';
if (url.slice(-1)=="/") {
url = url.substr(0,url.length-1);
}
var lastSegment = url.split('/').pop();
document.write(lastSegment+"<br>");

JS / RegEx to remove characters grouped within square braces

I hope I can explain myself clearly here and that this is not too much of a specific issue.
I am working on some javascript that needs to take a string, find instances of chars between square brackets, store any returned results and then remove them from the original string.
My code so far is as follows:
parseLine : function(raw)
{
var arr = [];
var regex = /\[(.*?)]/g;
var arr;
while((arr = regex.exec(raw)) !== null)
{
console.log(" ", arr);
arr.push(arr[1]);
raw = raw.replace(/\[(.*?)]/, "");
console.log(" ", raw);
}
return {results:arr, text:raw};
}
This seems to work in most cases. If I pass in the string [id1]It [someChar]found [a#]an [id2]excellent [aa]match then it returns all the chars from within the square brackets and the original string with the bracketed groups removed.
The problem arises when I use the string [id1]It [someChar]found [a#]a [aa]match.
It seems to fail when only a single letter (and space?) follows a bracketed group and starts missing groups as you can see in the log if you try it out. It also freaks out if i use groups back to back like [a][b] which I will need to do.
I'm guessing this is my RegEx - begged and borrowed from various posts here as I know nothing about it really - but I've had no luck fixing it and could use some help if anyone has any to offer. A fix would be great but more than that an explanation of what is actually going on behind the scenes would be awesome.
Thanks in advance all.

You could use the replace method with a function to simplify the code and run the regexp only once:
function parseLine(raw) {
var results = [];
var parsed = raw.replace(/\[(.*?)\]/g, function(match,capture) {
results.push(capture);
return '';
});
return { results : results, text : parsed };
}

The problem is due to the lastIndex property of the regex /\[(.*?)]/g; not resetting, since the regex is declared as global. When the regex has global flag g on, lastIndex property of RegExp is used to mark the position to start the next attempt to search for a match, and it is expected that the same string is fed to the RegExp.exec() function (explicitly, or implicitly via RegExp.test() for example) until no more match can be found. Either that, or you reset the lastIndex to 0 before feeding in a new input.
Since your code is reassigning the variable raw on every loop, you are using the wrong lastIndex to attempt the next match.
The problem will be solved when you remove g flag from your regex. Or you could use the solution proposed by Tibos where you supply a function to String.replace() function to do replacement and extract the capturing group at the same time.

You need to escape the last bracket: \[(.*?)\].

javascript regex to extract the first character after the last specified character

I am trying to extract the first character after the last underscore in a string with an unknown number of '_' in the string but in my case there will always be one, because I added it in another step of the process.
What I tried is this. I also tried the regex by itself to extract from the name, but my result was empty.
var s = "XXXX-XXXX_XX_DigitalF.pdf"
var string = match(/[^_]*$/)[1]
string.charAt(0)
So the final desired result is 'D'. If the RegEx can only get me what is behind the last '_' that is fine because I know I can use the charAt like currently shown. However, if the regex can do the whole thing, even better.

If you know there will always be at least one underscore you can do this:
var s = "XXXX-XXXX_XX_DigitalF.pdf"
var firstCharAfterUnderscore = s.charAt(s.lastIndexOf("_") + 1);
// OR, with regex
var firstCharAfterUnderscore = s.match(/_([^_])[^_]*$/)[1]
With the regex, you can extract just the one letter by using parentheses to capture that part of the match. But I think the .lastIndexOf() version is easier to read.
Either way if there's a possibility of no underscores in the input you'd need to add some additional logic.

Regex to extract substring, returning 2 results for some reason

I need to do a lot of regex things in javascript but am having some issues with the syntax and I can't seem to find a definitive resource on this.. for some reason when I do:
var tesst = "afskfsd33j"
var test = tesst.match(/a(.*)j/);
alert (test)
it shows
"afskfsd33j, fskfsd33"
I'm not sure why its giving this output of original and the matched string, I am wondering how I can get it to just give the match (essentially extracting the part I want from the original string)
Thanks for any advice

match returns an array.
The default string representation of an array in JavaScript is the elements of the array separated by commas. In this case the desired result is in the second element of the array:
var tesst = "afskfsd33j"
var test = tesst.match(/a(.*)j/);
alert (test[1]);

Each group defined by parenthesis () is captured during processing and each captured group content is pushed into result array in same order as groups within pattern starts. See more on http://www.regular-expressions.info/brackets.html and http://www.regular-expressions.info/refcapture.html (choose right language to see supported features)
var source = "afskfsd33j"
var result = source.match(/a(.*)j/);
result: ["afskfsd33j", "fskfsd33"]
The reason why you received this exact result is following:
First value in array is the first found string which confirms the entire pattern. So it should definitely start with "a" followed by any number of any characters and ends with first "j" char after starting "a".
Second value in array is captured group defined by parenthesis. In your case group contain entire pattern match without content defined outside parenthesis, so exactly "fskfsd33".
If you want to get rid of second value in array you may define pattern like this:
/a(?:.*)j/
where "?:" means that group of chars which match the content in parenthesis will not be part of resulting array.
Other options might be in this simple case to write pattern without any group because it is not necessary to use group at all:
/a.*j/
If you want to just check whether source text matches the pattern and does not care about which text it found than you may try:
var result = /a.*j/.test(source);
The result should return then only true|false values. For more info see http://www.javascriptkit.com/javatutors/re3.shtml

I think your problem is that the match method is returning an array. The 0th item in the array is the original string, the 1st thru nth items correspond to the 1st through nth matched parenthesised items. Your "alert()" call is showing the entire array.

Just get rid of the parenthesis and that will give you an array with one element and:
Change this line
var test = tesst.match(/a(.*)j/);
To this
var test = tesst.match(/a.*j/);
If you add parenthesis the match() function will find two match for you one for whole expression and one for the expression inside the parenthesis
Also according to developer.mozilla.org docs :
If you only want the first match found, you might want to use
RegExp.exec() instead.
You can use the below code:
RegExp(/a.*j/).exec("afskfsd33j")

I've just had the same problem.
You only get the text twice in your result if you include a match group (in brackets) and the 'g' (global) modifier.
The first item always is the first result, normally OK when using match(reg) on a short string, however when using a construct like:
while ((result = reg.exec(string)) !== null){
console.log(result);
}
the results are a little different.
Try the following code:
var regEx = new RegExp('([0-9]+ (cat|fish))','g'), sampleString="1 cat and 2 fish";
var result = sample_string.match(regEx);
console.log(JSON.stringify(result));
// ["1 cat","2 fish"]
var reg = new RegExp('[0-9]+ (cat|fish)','g'), sampleString="1 cat and 2 fish";
while ((result = reg.exec(sampleString)) !== null) {
console.dir(JSON.stringify(result))
};
// '["1 cat","cat"]'
// '["2 fish","fish"]'
var reg = new RegExp('([0-9]+ (cat|fish))','g'), sampleString="1 cat and 2 fish";
while ((result = reg.exec(sampleString)) !== null){
console.dir(JSON.stringify(result))
};
// '["1 cat","1 cat","cat"]'
// '["2 fish","2 fish","fish"]'
(tested on recent V8 - Chrome, Node.js)
The best answer is currently a comment which I can't upvote, so credit to #Mic.