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I am using grid view for displaying image using xml parsing,i got some exception like
java.lang.IllegalArgumentException: Illegal character in path at index 80: http://www.theblacksheeponline.com/party_img/thumbspps/912big_361999096_Flicking Off Douchebag.jpg
How to solve this problem? I want to display all kind of url,anybody knows please give sample code for me.
Thanks All
844
Since URLs often contain characters outside the ASCII set, the URL has to be converted into a valid ASCII format. URL encoding replaces unsafe ASCII characters with a "%" followed by two hexadecimal digits. URLs cannot contain spaces. URL encoding normally replaces a space with a plus (+) sign or with %20.
Spaces are not allowed in URLs. They should be replaced by the string %20. In the query string part of the URL, %20 can be abbreviated using a plus sign (+).
A space is assigned number 32, which is 20 in hexadecimal. When you see “%20,” it represents a space in an encoded URL, for example, http://www.example.com/products%20and%20services.html.
Simply put, URL encoding translates special characters from the URL to a representation that adheres to the spec and can be correctly understood and interpreted.
URL encoding is done in the same way on android as in Java SE;
try {
String url = "http://www.example.com/?id=123&art=abc";
String encodedurl = URLEncoder.encode(url,"UTF-8");
Log.d("TEST", encodedurl);
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
}
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