How to resolve java.net.MalformedURLException: Protocol not found: 9 in android

Question

I am trying to load images in my android application from a url (http://www.elifeshopping.com/images/stories/virtuemart/product/thumbnail (2).jpg) using BitmapFactory the code is below :

try {
            // ImageView i = (ImageView)findViewById(R.id.image);
            bitmap = BitmapFactory.decodeStream((InputStream) new URL(url)
                    .getContent());
            i.setImageBitmap(bitmap);
        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

here i get

05-03 15:57:13.156: W/System.err(1086): java.net.MalformedURLException: Protocol not found: 9
05-03 15:57:13.167: W/System.err(1086):     at java.net.URL.<init>(URL.java:273)
05-03 15:57:13.167: W/System.err(1086):

at java.net.URL.<init>(URL.java:157).

Please help by telling what I am doing wrong.

Where do you get the url from. Check that your variable url input string starts with either http:// or https:// print the url string in the exception — user1369689
– user1369689, Commented May 3, 2012 at 13:27

Shruti · Accepted Answer · 2012-05-11 10:58:15Z

6

I used

productImgUrl = productImgUrl.replaceAll(" ", "%20");

i replaced all the spaces by %20

and its working for me ..

Thanks everybody for their responses

answered May 11, 2012 at 10:58

Shruti

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1 Comment

Hanuman Over a year ago

works for me, I checked in browser it is showing in Image but when I checked in app it shows that Malformed exception, So I added this small code It works fine, It is not showing error. Thanks @Shruti

Stephen C · Accepted Answer · 2012-05-04 07:09:13Z

2

Please help by telling what I am doing wrong.

I think that the problem is that you are calling the URL constructor with an invalid URL string. Indeed, the exception message implies that the URL string starts with "9:". (The 'protocol' component is the sequence of characters before the first colon character of the URL.)

This doesn't make a lot of sense if the URL string really is:

"http://www.elifeshopping.com/images/stories/virtuemart/product/thumbnail (2).jpg"

so I'd infer that it is ... in fact ... something else. Print it out before you call the URL constructor to find out what it really is.

(You should also %-escape the space characters in the URL's path ... but I doubt that will fix this particular exception incarnation.)

edited May 4, 2012 at 7:09

answered May 3, 2012 at 13:30

Stephen C

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2 Comments

Shruti Over a year ago

i didn't got this (exception message implies that the URL string starts with "9:". (The 'protocol' component is the sequence of characters before the first colon character.)) ... do i need to edit http ?

Stephen C Over a year ago

@Shruti - I'm saying that that string is not the one that you are passing to the URL constructor.

Changwei Yao · Accepted Answer · 2012-05-03 13:28:13Z

0

Change your url to http://www.elifeshopping.com/images/stories/virtuemart/product/thumbnail%20%282%29.jpg

answered May 3, 2012 at 13:28

Changwei Yao

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