How to encode the HTTP URL address in Java

Question

My Java standalone application gets a URL (which points to a file) from the user and I need to hit it and download it. The problem I am facing is that I am not able to encode the HTTP URL address properly...

Example:

URL:  http://search.barnesandnoble.com/booksearch/first book.pdf

java.net.URLEncoder.encode(url.toString(), "ISO-8859-1");

returns me:

http%3A%2F%2Fsearch.barnesandnoble.com%2Fbooksearch%2Ffirst+book.pdf

But, what I want is

http://search.barnesandnoble.com/booksearch/first%20book.pdf

(space replaced by %20)

I guess URLEncoder is not designed to encode HTTP URLs... The JavaDoc says "Utility class for HTML form encoding"... Is there any other way to do this?

Comments

Is there any of the moderators, because I can not answer a new topic?  
 Maybe I'm doing something wrong?  
 Need your help.
Thank you.

---

Is there any of the moderators, because I can not answer a new topic?  
What am I doing wrong?  
 Please tell me.  
 Yours faithfully.

---

Is there any of the moderators, because I can not answer a new topic?  
What am I doing wrong?  
 Need your help.
Thank you.

---

Can someone help how to answer a new topic ??
What am I doing wrong?  
 Need your help.
Thank you.

---

Can't answer an existing topic, what should I do ??  
 Maybe I'm not writing correctly?
Please help.  
 Yours faithfully.

Answer 1

Use one of the constructors with more than one argument, like:

URI uri = new URI(
    "http", 
    "search.barnesandnoble.com", 
    "/booksearch/first book.pdf",
    null);
URL url = uri.toURL();
//or String request = uri.toString();

(the single-argument constructor of URI does NOT escape illegal characters)

Only illegal characters get escaped by above code - it does NOT escape non-ASCII characters (see fatih's comment).
The toASCIIString method can be used to get a String only with US-ASCII characters:

URI uri = new URI(
    "http", 
    "search.barnesandnoble.com", 
    "/booksearch/é",
    null);
String request = uri.toASCIIString();

For an URL with a query like http://www.google.com/ig/api?weather=São Paulo, use the 5-parameter version of the constructor:

URI uri = new URI(
        "http", 
        "www.google.com", 
        "/ig/api",
        "weather=São Paulo",
        null);
String request = uri.toASCIIString();