How do I remove a substring from a string?

[Unnel]

Hi,

I've developed an application that reads URLs from a file. I would like to know how I can chop off a part of a string I'm not interested in. Below is an example of what I mean.

http://www.aghdncbyd.com/ghjjsyk

I would like to remove the 'ghjjsyk' part after the last foreslash so that I can store the rest.

Thank you,

[JosAH]

The lastIndexOf( ... ) and substring( ... ) methods can help you out; read the API documentation for the String class.

kind regards,

Jos

[Unnel]

The lastIndexOf( ... ) and substring( ... ) methods can help you out; read the API documentation for the String class.

kind regards,

Jos

Thanks Jos,

But what about long URLS such as http://www.aghdtfsj.com/hsgdtt/ghdtuii/?

How do I go about only retrieving http://www.aghdtfs.com/ and chopping off the rest? I'm looking for a generic way of doing this because my URLs are differing from one another.

Thank you

[JosAH]

Thanks Jos,

But what about long URLS such as http://www.aghdtfsj.com/hsgdtt/ghdtuii/?

How do I go about only retrieving http://www.aghdtfs.com/ and chopping off the rest? I'm looking for a generic way of doing this because my URLs are differing from one another.

Thank you

Start your search for a / from the front of the String; you want to stop ,and chop off, the String starting at the third / from the start of the String. Read the API documentation for the two argument indefOf( ... ) method in the String class; or else, use the URL class; it can do such things such as chopping file, host and protocol parts.

kind regards,

Jos

[fakepics500]

i tried to code your problem

import java.lang.String;
class Str
{
	public static void main(String ar[])
	{
		String s1="/";
		int i;
		i=ar[0].indexOf(s1,8);
		String ns;
		ns=ar[0].substring(0,i);
		System.out.println("String ="+ns);
	}
}

url entered may be

http://google.com/kkk  
http://google.com/kkk/lklj
https://google.com/kkk

[Junky]

One thing you can do to make the code a bit more robust in case the http:// bit is not present is to do an indexOf "//". If it is greater than 0 then indexOf "/" starting from that position (2 parameter method). Else indexOf "/" from the start of the URL (1 parameter method).

[DarrylBurke]

Wot no regex? ;)

public class DomainOnly {

  public static void main(String[] args) {
    String[] inputs = {
      "http://www.aghdncbyd.com/ghjjsyk",
      "http://www.aghdtfsj.com/hsgdtt/ghdtuii/?",
      "http://google.com/kkk",
      "http://google.com/kkk/lklj",
      "https://google.com/kkk"
    };
    String regex = "(^[^/]+(//)?[^/]*).*";
    for (String input : inputs) {
      System.out.println("[" + input.replaceAll(regex, "$1") + "]");
    }
  }
}

db

[JosAH]

Wot no regex? ;)

    String regex = "(^[^/]+(//)?[^/]*).*";

How come I didn't think of that gibberish; /*&(^^*?//?\// to you too ;-)

kind regards,

Jos

[DarrylBurke]

Feast your eyes:
How to mask string not conforming to a regular expression pattern (Java in General forum at JavaRanch)

db

[JosAH]

Feast your eyes:
How to mask string not conforming to a regular expression pattern (Java in General forum at JavaRanch)

Don't you ever do that again; I didn't have my espresso yet nor did I smoke my first tobacco and you're showing me that mess? I'm a very vulnerable little puppy at this hour of the day you know, *yuck!*

kind regards,

Jos (<--- goes to take a shower)

[DarrylBurke]

The regex is actually quite fundamental.

(         start capturing group 1
^         start of input
[^/]+     not-slash, one or more times
(//)?     double-slash, 0 or 1 time
[^/]*     not-slash, 0 or more times
)         end capturing group 1
.*        any character, 0 or more times

"$1"      the content of capturing group 1: 
      start of input
      + not-slash, one or more times
      + double-slash, 0 or 1 time
      + not-slash, 0 or more times

Just noticed that OP wants to retain the slash after the domain, so add that to the end of the capturing group.

String regex = "(^[^/]+(//)?[^/]*/).*";

db

[JosAH]

The regex is actually quite fundamental.

Yes I know; type 2 (context free) grammars are also fundamental and so much more powerful; I really don't understand the popularity of regular expressions. There's nothing a RE can do that a type 2 grammar parser can't do ... and those REs are ugly to.

kind regards,

Jos

[fakepics500]

The regex is actually quite fundamental.

(         start capturing group 1
^         start of input
[^/]+     not-slash, one or more times
(//)?     double-slash, 0 or 1 time
[^/]*     not-slash, 0 or more times
)         end capturing group 1
.*        any character, 0 or more times

"$1"      the content of capturing group 1: 
      start of input
      + not-slash, one or more times
      + double-slash, 0 or 1 time
      + not-slash, 0 or more times

Just noticed that OP wants to retain the slash after the domain, so add that to the end of the capturing group.

String regex = "(^[^/]+(//)?[^/]*/).*";

db

i am still not able understand about regex

[Joel]

You should read up on regular expressions.

[Norm]

i am still not able understand about regex

Not many people do. They are very hard to debug. They are a "black-box" type of program that either works or not.

Is there a way to debug them? A way to trace the logic flow as they are scanned and rescanned etc???

They are great when they work and a head ache when they don't.

[JosAH]

Not many people do. They are very hard to debug. They are a "black-box" type of program that either works or not.

Is there a way to debug them? A way to trace the logic flow as they are scanned and rescanned etc???

They are great when they work and a head ache when they don't.

Regular expressions are a PITA but they don't need to be rescanned all the time. Most (all?) RE engines build a NFA (Non-deterministic Finit Automaton) out of the RE which is often turned into a DFA (Deterministic Finit Automaton) which can be interpreted in linear time (O(n)) given an input String. The size of the DFA can be O(2^n) w.r.t. to original RE though. I don't think Sun's/Oracle's implementation can be easily debuged ... but I agree: they can be (and often are) a head ache.

kind regards,

Jos

[Unnel]

Hi guys,

Thanks a lot for your much appreciated and valuable input. I managed to get my piece of code to work thanks to y'all.

Have a good one.

Cheers