CODEWITHSUNDEEP
javaurl
yrazlik
yrazlik

Reputation: 10777

How to convert a url that has a space in it to a valid url in java?

I have a link that has a space in it, but when i paste it to the browser, it loads the page and the link changes slightly, here is the link:

http://www.marketimyilmazlar.com/image/cache/data/DUZELTME/dvm/screenshot 7-500x500.jpg

But it returns into the following when browser loas the page:

http://www.marketimyilmazlar.com/image/cache/data/DUZELTME/dvm/screenshot%207-500x500.jpg

Here is what i do in java:

 public static Drawable LoadImageFromWeb(String link) throws URISyntaxException {
    try {
        URI uri = new URI(link);
        URL url = new URL(uri.toASCIIString());
        InputStream is = (InputStream) url.getContent();
        Drawable d = Drawable.createFromStream(is, "src name");
        return d;
    } catch (IOException e) {
        Log.e("HATA", "THERE is an error", e);
        return null;
    }
}

But the problem is, I get the following exception:

java.net.URISyntaxException: Illegal character in path at index 73:

So, can anyone help me with this? Thanks

Upvotes: 1

Views: 1881

Answers (3)

smola
smola

Reputation: 883

Doing proper URL parsing with Java standard library is far from trivial. I grew frustrated when hitting edge cases. So I developed a library for proper URL parsing in Java: galimatias. Any URL that works in a web browser should work with galimatias.

With io.mola.galimatias.URL.parse(link) you get a io.mola.galimatias.URL object. Here parsing will work with almost whatever you throw at it (unless it can't really be interpreted as a URL). Then you can transform it to a Java URL with toJavaURL().

For your use case, you would write:

 public static Drawable LoadImageFromWeb(String link) {
     try {
         URL url = io.mola.galimatias.URL.parse(link).toJavaURL();
         InputStream is = (InputStream) url.getContent();
         Drawable d = Drawable.createFromStream(is, "src name");
         return d;
     } catch (IOException e) {
         Log.e("HATA", "THERE is an error", e);
         return null;
     } catch (GalimatiasParseError e) {
         Log.e("HATA", "THERE is an error", e);
         return null;
     } catch (MalformedURLException e) {
         Log.e("HATA", "THERE is an error", e);
         return null;
     }
 }

Upvotes: 2

Ian Roberts
Ian Roberts

Reputation: 122364

The java.net.URI class may be able to help you here - its single argument constructor expects the argument to be a proper URI with illegal characters escaped, but its multiple-argument constructors can handle un-escaped parts. If your initial URL does not contain any %-escapes already, and does not have a # fragment on the end then you can do

String escaped = new URI("dummy", unescaped, null).getRawSchemeSpecificPart();

The trick here is that the three-argument constructor expects a "scheme", a "scheme-specific part" and a "fragment", all in unescaped form. So for your example in the question

new URI("dummy", unescaped, null)

will give you the properly-escaped URI

dummy:http://www.marketimyilmazlar.com/image/cache/data/DUZELTME/dvm/screenshot%207-500x500.jpg

Extracting the raw SSP from that URI will give you everything after the first colon, i.e.

http://www.marketimyilmazlar.com/image/cache/data/DUZELTME/dvm/screenshot%207-500x500.jpg

Upvotes: 7

clay
clay

Reputation: 6017

The URL must be encoded, and Java has an object for that. Use URLEncode.encode() on your string.

Upvotes: -2

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