Reputation: 21
I get an error "Illegal character in URL
" in my code and I don't know why:
I have a token and an hash that are string type.
String currentURL = "http://platform.shopyourway.com" +
"/products/get-by-tag?tagId=220431" +
"&token=" + token +
"&hash=" + hash;
HttpURLConnection urlConnection = null;
BufferedReader reader = null;
try {
URL url = new URL(currentURL);
urlConnection = (HttpURLConnection) url.openConnection();
urlConnection.setRequestMethod("GET");
urlConnection.connect();
[...]
but when I wrote :
URL url = new URL("http://platform.shopyourway.com/products/get-by-tag?tagId=220431&token=0_11800_253402300799_1_a9c1d19702ed3a5e873fd3b3bcae6f8e3f8b845c9686418768291042ad5709f1&hash=e68e41e4ea4ed16f4dbfb32668ed02b080bf1f2cbee64c2692ef510e7f7dc26b");
it's work, but I can't write this order because I don't know the hash and the token because I generate them every time. thanks.
Upvotes: 1
Views: 5094
Reputation: 15879
From the Oracle docs on creating URLs you need to escape the "values" of your URL string.
URL addresses with Special characters
Some URL addresses contain special characters, for example the space character. Like this:
http://example.com/hello world/ To make these characters legal they need to be encoded before passing them to the URL constructor.
URL url = new URL("http://example.com/hello%20world");
Encoding the special character(s) in this example is easy as there is only one character that needs encoding, but for URL addresses that have several of these characters or if you are unsure when writing your code what URL addresses you will need to access, you can use the multi-argument constructors of the java.net.URI class to automatically take care of the encoding for you.
URI uri = new URI("http", "example.com", "/hello world/", "");
And then convert the URI to a URL.
URL url = uri.toURL();
As commented also see this other post that uses URLEncoder to replace any offending characters
Upvotes: 2