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Get substring between substring and first occurrence of another string

I have URL pathnames that look similar to this: /service-area/i-need-this/but-not-this/. The /service-area/ part never changes, and the rest of the path is dynamic.
I need to get the part of the URL saying i-need-this.
Here was my attempt:
location.pathname.match(new RegExp('/service-area/' + "(.*)" + '/'));.
The goal was to get everything between /service-area/ and / but it's actually going up to the last occurrence of /, not the first occurrance. So the output from this is actually i-need-this/but-not-this.
I'm not so good with regex, is there a way it can be tweaked to get the desired result?

You need a lazy regex rather than a greedy one - so (.*?) instead of (.*). See also: What do 'lazy' and 'greedy' mean in the context of regular expressions?

You can do this without a regex too using replace and split:
var path = '/service-area/i-need-this/but-not-this/';
var res = path.replace('/service-area/', '').split('/')[0];
console.log(res);

Cut out, and replace a part of a string

There is a part in my string from, to which I would like to replace to an another string replace_string. My code should work, but what if there is an another part like the returned substring?
var from=10, to=17;
//...
str = str.replace(str.substring(from, to), replace_string);
For example:
from=4,to=6
str = "abceabxy"
replace_string = "zz"
the str should be "abcezzxy"

What you want to do is simple! Cut out and replace the string. Here is the basic tool, you need scissor and glue! Oops I mean string.Split() and string.Replace().
How to use?
Well I am not sure if you want to use string.Split() but you have used string.Replace() so here goes.
String.Replace uses two parameters, like this ("one", "two") what you need to make sure is that you are not replacing a char with a string or a string with a char. They are used as:
var str="Visit Microsoft!";
var n=str.replace("Microsoft","W3Schools");
Your code:
var from=10, to=17;
//...
var stringGot = str.replace(str.substring(from, to), replace_string);
What you should do will be to split the code first, and then replace the second a letter! As you want one in your example. Thats one way!
First, split the string! And then replaced the second a letter with z.
For String.Replace refer this: http://www.w3schools.com/jsref/jsref_replace.asp
For String.SubString: http://www.w3schools.com/jsref/jsref_substring.asp
For String.Split: http://www.w3schools.com/jsref/jsref_split.asp

Strings are immutable. This means they do not change after they are first instantiated. Every method to manipulate a string actually returns a new instance of a string. So you have to assign your result back to the variable like this:
str = str.replace(str.substring(from, to), replace_string);
Update: However, the more efficient way of doing this in the first place would be the following. it is also less prone to errors:
str = str.substring(0, from) + replace_string + str.substring(to);
See this fiddle: http://jsfiddle.net/cFtKL/
It runs both of the commands through a loop 100,000 times. The first takes about 75ms whereas the latter takes 20ms.

Finding image url via using Regex

Any working Regex to find image url ?
Example :
var reg = /^url\(|url\(".*"\)|\)$/;
var string = 'url("http://domain.com/randompath/random4509324041123213.jpg")';
var string2 = 'url(http://domain.com/randompath/random4509324041123213.jpg)';
console.log(string.match(reg));
console.log(string2.match(reg));
I tied but fail with this reg
pattern will look like this, I just want image url between url(" ") or url( )
I just want to get output like http://domain.com/randompath/random4509324041123213.jpg
http://jsbin.com/ahewaq/1/edit

I'd simply use this expression:
/url.*\("?([^")]+)/
This returns an array, where the first index (0) contains the entire match, the second will be the url itself, like so:
'url("http://domain.com/randompath/random4509324041123213.jpg")'.match(/url.*\("?([^")]+)/)[1];
//returns "http://domain.com/randompath/random4509324041123213.jpg"
//or without the quotes, same return, same expression
'url(http://domain.com/randompath/random4509324041123213.jpg)'.match(/url.*\("?([^")]+)/)[1];
If there is a change that single and double quotes are used, you can simply replace all " by either '" or ['"], in this case:
/url.*\(["']?([^"')]+)/

Try this regexp:
var regex = /\burl\(\"?(.*?)\"?\)/;
var match = regex.exec(string);
console.log(match[1]);
The URL is captured in the first subgroup.

If the string will always be consistent, one option would be simply to remove the first 4 characters url(" and the last two "):
var string = 'url("http://domain.com/randompath/random4509324041123213.jpg")';
// Remove last two characters
string = string.substr(0, string.length - 2);
// Remove first five characters
string = string.substr(5, string.length);
Here's a working fiddle.
Benefit of this approach: You can edit it yourself, without asking StackOverflow to do it for you. RegEx is great, but if you don't know it, peppering your code with it makes for a frustrating refactor.

Regex to match carriage return in Javascript

I am trying to write some Javascript to hide some elements that contain only carriage returns. I appreciate that the correct way to solve this problem would be to stop these elements being created, but unfortunately that is not possible in this instance. I am trying to user a regular expression to search for the unwanted elements but am not having much luck. The function I have written is as follows:
function HideEmptyP()
{
var patt = (\\r)
for(var i = 0;i<desc[i].length;i++);
{
var desc[i] = document.getElementsByClassName('sitspagedesc');
var result[i] = patt.test(desc[i]);
if (result[i] == true)
{
desc[i].style.display='none';
}
else
{
alert("No Match!");
}
}
The error I'm getting in the Web Console is 'Syntax Error: Illegal Character'.
Grateful for any ideas on how to solve this.
Thanks in advance.

I am trying to write some Javascript to hide some elements that contain only carriage returns.
There's no need for a regular expression for that, just compare the element's innerHTML property to "\\r", e.g.:
if (demo[i].innerHTML === "\\r") {
// Remove it
}
But beware that some browsers may transform a single carriage return. You might want to check for "\\r", "\\n", and just a space. To do that, you might want to use a regular expression.
Your regular expression literal ((\\r)) is just completely invalid, it's worth reading up on them to learn the correct syntax. To write a regular expression literal in JavaScript, you use / as the delimiter. So: /\\r/. To test that a string contains only \r, \n, or space, you can use /^[\r\n ]+$/ (which requires there be at least one character that matches, and uses ^ to indicate start-of-string, and $ to indicate end-of-string):
if (demo[i].innerHTML.match(/^[\r\n ]+$/) {
// Remove it
}

The reason you are getting Syntax error is because the declaration
var patt = (\r)
is incorrect it should be somethign like var patt = '\r';
Also the whole for loop is wrong.
You should define demo before you start the for loop not inside it, and result need not be an array but just a normal variable

Your litteral seems odd.
Try var patt = /\r/;

var patt=/\n/gi
should work.
extra i flag to denote case insensitive.
g for global search.

what's wrong with this regular expression? getting the hash part of an url

I´m trying to get the first part of a hash from a url (the part between the # and a /, a ? or the end of the string
So far now I came out with this:
r = /#(.*)[\?|\/|$]/
// OK
r.exec('http://localhost/item.html#hash/sub')
["#hash/", "hash"]
// OK
r.exec('http://localhost/item.html#hash?sub')
["#hash?", "hash"]
// WAT?
r.exec('http://localhost/item.html#hash')
null
I was expeting to receive "hash"
I tracked down the problem to
/#(.*)[$]/
r2.exec('http://localhost/item.html#hash')
null
any idea what could be wrong?

r = /#(.*)[\?|\/|$]/
When $ appears in [] (character class, it's the literal "$" character, not the end of input/line. In fact, your [\?|\/|$] part is equivalent to just [?/$|], which matches the 4 specific characters (including pipe).
Use this instead (JSFiddle)
r = /#(.+?)(\?|\/|$)/

You aren't supposed to write [$] (within a character class) unless you want to match the $ literally and not the end of line.
/#(.*)$/
Code:
var regex = /\#(.*)$/;
regex.exec('http://localhost/item.html#hash');
Output:
["#hash", "hash"]
Your regex: /#(.*)[\?|\/|$]/
//<problem>-----^ ^-----<problem>
| operator won't work within [], but within ()
$ will be treated literally within []
.* will match as much as possible. .*? will be non-greedy
On making the above changes,
you end up with /#(.*?)(\?|\/|$)/

I use http://regexpal.com/ to test my regular expressions.
Your problem here is that your regular expression wants a /. So it don't works with http://localhost/item.html#hash but it works with http://localhost/item.html#hash/
Try this one :
r = /#([^\?|\/|$]*)/

You can't use the $ end-of-string marker in a character class. You're probably better off just matching characaters that aren't / or ?, like this:
/#([^\?\/]*)/

Why Regex? Do it like this (nearly no regex):
var a = document.createElement('a');
a.href = 'http://localhost/item.html#hash/foo?bar';
console.log(a.hash.split(/[\/\?]/)[0]); // #hash
Just for the sake, if it is node.js you are working with:
var hash = require('url').parse('http://localhost/item.html#hash').hash;

I found this regular expression that seems to work
r = /#([^\/\?]*)/
r.exec('http://localhost/item.html#hash/sub')
["#hash", "hash"]
r.exec('http://localhost/item.html#hash?sub')
["#hash", "hash"]
r.exec('http://localhost/item.html#hash')
["#hash", "hash"]
Anyway, I still don't get why the original one isn't working