What is wrong with this URI URL? IllegalArgumentException, Illegal character

From what I've seen, every attempt either misses something, encodes something it shouldn't, such as the '?', or double-encodes something, thereby url-encoding the '%' in the url encoding.

How about just encoding the bit you care about escaping, and doing it exactly once?

String apiURI =    "https://api.mongolab.com/api/1/databases/activity_recognition/collections/entropy_data?f="    + URLEncoder.encode("{\"mean0\": 1}", "UTF-8")    + "&apiKey=myApiKey";

If you wanted to use java.net.URI, you'd have to include the query string separately, e.g.:

new URI(    "https",    "api.mongolab.com",    "/api/1/databases/activity_recognition/collections/entropy_data",    "f={\"mean0\": 1}&apiKey=myApiKey",    null  ).toURL()

Another way to do this would be:

uri = new URI("https", "api.mongolab.com", "/api/1/databases/activity_recognition/collections/entropy_data?f={\"mean\": 1}&apiKey=myApiKey", null);URL url = uri.toURL();

Note that I changed the %22 (urlencoded quotes) to \" (escaped quotes) otherwise you will wind up with your % sign being urlencoded.

To clarify, the point of this is that if you do this:

String query = "https://api.mongolab.com...";query = URLEncoder.encode(query, "utf-8");

you will wind up with https%3A%2F%2Fapi.mongolab.com.

I guess you are looking for something like this

String flag1 = URLEncoder.encode("This string has spaces", "UTF-8");

You can refer to the documentation from Oracle URL Encoder or refer to SOF