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04.21.2008 at 09:40PM PDT, ID: 23341847

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5.6

error java.net.MalformedURLException unknown protocol

Zones: Java Editors & IDEs, Eclipse, Domain Name Service (DNS)

Hi
I have a code which is monitor URL open by client machine.But Problem is that when client open a www.google.com then it will work open a page and show on serve side.But when client open a goole linke like http://picasaweb.google.co.in/home?hl=en&tab=wq' then give error like this.
error java.net.MalformedURLException: unknown protocol: www.google.com.

I am useing httprequest.java ,httpreply.java and some more java file. If any one need my code then i give link.

Regard's
Ddipro

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Question Stats

Zone: Programming

Question Asked By: ddipro

Solution Provided By: Bart_Cr

Participating Experts: 2

Solution Grade: B

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04.21.2008 at 11:15PM PDT, ID: 21408435

Bart_Cr:

It's indeed impossible to answer this question without the code.

04.22.2008 at 12:39AM PDT, ID: 21408773

ddipro:

ok i am sending code link.

04.22.2008 at 12:41AM PDT, ID: 21408777

ddipro:

http://www.dhruvasagar.com/Proxy.zip

04.22.2008 at 01:28AM PDT, ID: 21408989

Bart_Cr:

Parsing the request is not done correctly. First, you don't need to do a substring(1) on the token for the url. The url should be the entire token. Second, the url is not always prefixed with a protocol. The java URL class expects this. If the protocol is missing in the url, you need to make sure you prefix it correctly.

You can find a list of port numbers and the attached protocol on following address:
http://en.wikipedia.org/wiki/List_of_TCP_and_UDP_port_numbers

You'll need to make sure you map the needed ports to the correct protocol. Make sure you provide at least a mapping for http and https. If you can't map the port to a protocol you can default safely to http IMHO.

04.22.2008 at 02:33AM PDT, ID: 21409280

ddipro:

thxs for reply

but can you give me code after debug.. becouse i am new programmar..point will be adjust..

04.22.2008 at 03:34AM PDT, ID: 21409540

Bart_Cr:

url = getToken(tz).substring(1);

should become

url = getToken(tz);

This will make the proxy server work for the requests where the protocol is correctly specified. I don't have the time or ambition to create a working proxy server, but I'll be glad to review solutions and to help find causes for errors if you encounter them.

04.22.2008 at 04:26AM PDT, ID: 21409807

ddipro:

thxs for reply

i did this but still " error java.net.MalformedURLException: unknown protocol: www.google.com is found "

04.22.2008 at 04:41AM PDT, ID: 21409886

Bart_Cr:

That's a case where the request does not contain a protocol in the URL. If it does not contain a protocol you should prefix the url string before creating a java.net.URL object with the string. In this case it should be prefixed with "http://".

Checking the string for occurence of ":/" should be sufficient to find out if a protocol is specified.

url = getToken(tz);

if (url.indexOf(":/") < 0) {
// Do some port magic...
url = "http://" + url;
}

Accepted Solution

04.22.2008 at 09:20PM PDT, ID: 21417761

ddipro:

thxs for reply

i try your code but page is not open.Error is not show on out windows but page is not not open.one thing i want to know you write "do some port magic " what it mean sir...

Regard's
Ddipro

04.22.2008 at 11:20PM PDT, ID: 21418097

Bart_Cr:

The port magic means you should check if there's a port defined in the URL and map the protocol accordingly.

If you receive a url as follows: www.an-address.com:443 it means the url should connect to port 443 on the server. And from the list of default port number you can state that the protocol should be https instead of http, so you're url becomes "https://www.an-address.com".

04.22.2008 at 11:27PM PDT, ID: 21418131

ddipro:

You are right but how can i set the protocol https insted to http.Plz give me tips..

04.22.2008 at 11:33PM PDT, ID: 21418161

Bart_Cr:

You can simply create a new URL objects using the above string: URL url = new URL("https://www.an-address.com");
If Java knows the protocol it will handle it correctly.

04.22.2008 at 11:50PM PDT, ID: 21418238

ddipro:

hello sir

i am useing URL url = new URL("https://www.an-address.com"); but it is not working...

04.23.2008 at 12:01AM PDT, ID: 21418284

ddipro:

hello sir

i am useing Url url in proxy.java class alraedy.Now you have told me to use Url url = new URL("").So if i use this in httprequest.java then nothing to show and when i use it in proxy.java then it show error.

so i am not grting result...

Plz help me

04.23.2008 at 02:06AM PDT, ID: 21418702

ddipro:

hello sir
i am waiting for your responce....

05.07.2008 at 02:16AM PDT, ID: 21514459

melchkishore:

please check your firewall settings otherwise try to access with the IP Address of the URL which u r trying.

05.10.2008 at 01:12AM PDT, ID: 21538198

ddipro:

hello sir

sorry for late responce ..

i stop firewall and use Ip address of url but error is same ..

20080236-EE-VQP-29 / EE_QW_2_20070628