How do i get the string from after the third "/"

Question

Totally new to programming so kindly excuse the silly question. I have this URL http://www.blahblah.com/some/thing/blah..

Now i need the string that appears after the third"/" ie., just "/some/thing".. I dont want to go from reverse as the URL is different for each page but the protocol and the host name is the same.. Is there a way i can extract "/some/thing" along using Javascript.. Appreciate any help.. Thanks!!

use the split method http://stackoverflow.com/questions/96428/how-do-i-split-this-string-with-javascript — Paddy Carroll, Jun 15 '12 at 01:36

odiszapc · Answer 1 · 2012-06-15T01:51:37.260

3

var part = string.split('/')[2]

Or more complicated:

var string = "/a/b/c/d/e/d"
var arr1 = string.split('/')
var arr2 = []
for (i in  arr1)
{
  if (arr1[i].length > 0)
     arr2.push(arr1[i])
}
var part = arr2[2]

edited Jun 15 '12 at 01:51

answered Jun 15 '12 at 01:34

odiszapc

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This will fail for `/a/b/c/d/e/d` – Jun 15 '12 at 01:45
Working now on the second snippet – odiszapc Jun 15 '12 at 01:52

Ryan · Answer 2 · 2012-06-15T02:15:03.260

2

using the string.split method: http://www.w3schools.com/jsref/jsref_split.asp

var str = "www.blah.com/blah/blah/thing/blah.php";
var arrayOfString = str.split("/");
alert(arrayOfString[2]); //coming back to do regex/substring for you

have fun! :)

By the way... if you are new to web development, www.w3schools.com was extremely good to me :)

so to respond to your edit you need to do something like this :

assuming above already:

alert(arrayOfString.slice(3).join("/"));

ok fixed it based on actual needs here.. but like i said, check out http://www.w3schools.com it really taught me some good basics.

to fix for the fail case in the comments do this before the split/slice/join:

if (str.substring(0,1) == "/"){ str = str.substring(1);}

edited Jun 15 '12 at 02:15

answered Jun 15 '12 at 01:34

Ryan

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This will fail for `/a/b/c/d/e` – Jun 15 '12 at 01:45
@pst, i think that fixes the fail case, lemme know if you think of any other failz :) – Ryan Jun 15 '12 at 02:15

score 1 · Accepted Answer · 2012-06-15T02:10:27.697

I would use a simple regular expression here.

One way to get stuff after the third "/" just throw it away with everything before it.

Basic idea:

"/a/b/c/d/e/f".replace(/.*?\/.*?\/s.*?\//, "")
// -> "c/d/e/f"

Alternative #1, to reduce repeat:

"/a/b/c/d/e/f".replace(/(?:.*?\/){3}/, "")

Alternative #2, if input can contain newlines:

"/a/b/c/d/e/f".replace(/(?:[^\/]*\/){3}/, "")

The *? qualifier means "don't be greedy" so . won't skip over /'s. It's not needed in the negative character class case because [^\/] will never match /.

This could of course be written in a "less generic" form because because the real intent is just to git rid of the protocol/domain portions of the URI (the number of /'s is only an observed relationship):

"http://foo/a/b/c/d/e/f".replace(/https?:\/\/[^\/]*\/?/i, "")
// -> "a/b/c/d/e/f"

Note that the last \/ has a ? after it so it will return "" for "http://foo" and the i is so it will match HTTP:// if that is used for some strange reason (completely optional, of course).

"HTTPS://foo".replace(/https?:\/\/[^\/]*\/?/i, "")
// -> ""

Happy coding!

Note: all of the above regular expressions will extract the content after the 3rd /, as stated in the question. They will need to be slightly modified to include the 3rd / in the result. Also, it helps to not ask X-Y questions.

This looks like some hard core coding.. God! i have a long way to go. Will try this out too..Thank you so much!! — vidya lakshmi, Jun 15 '12 at 02:13
I just understood these regular expressions and found how accurate they can be:) thanks tried yours and worked just fine.. — vidya lakshmi, Jun 15 '12 at 07:14

Shoaib · Answer 4 · 2012-06-15T01:53:55.813

var str = 'http://www.blahblah.com/some/thing/blah'
var afterThirdSlash = str.split('/').slice(3).join('/')

The split returns an array which is the string split on every / character. The slice selects the part of the array from the fourth element (since arrays are zero-based) onwards, since that corresponds to the part after the third slash. The join pieces these elements back together by re-inserting the /. Is this what you wanted?

score 0 · Answer 5 · answered Jun 15 '12 at 02:06

0

if you know the url begins with 'http://www.blahblah.com/', why not use url.substring(24)?

answered Jun 15 '12 at 02:06

kennebec

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I have two UAT servers and 1 production.. Boss wants generic;) – vidya lakshmi Jun 15 '12 at 02:13

How do i get the string from after the third "/"

5 Answers5