Reputation: 497
I have following
username [email protected]
how "+" is encoded as %2B
encoded url is :http://test.in/api/voi/login?password=anshu&username=amit24%[email protected]
secondly if username is this amit24*[email protected] then how we encode this url ?
I had tried like this :
String url=ConstantNet.URL_LOGIN+ "?password=" + password.getText().toString() + "&username=" + email_mobile.getText().toString();
String encodedurl = null;
try {
encodedurl = URLEncoder.encode(url,"UTF-8");
Log.e("urlEncoded",""+encodedurl);
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
}
Result i am getting from encoded url is like this :
https%3A%2F%2Fgoturbo.in%2Fapi%2Fvo%2Flogin%3Fpassword%anshu%26username%3Damit%2B1%40gmail.com
but the actual result i want like this :
http://test.in/api/voi/login?password=anshu&username=amit24%[email protected]
Upvotes: 0
Views: 78
Reputation: 173
Try this.
public static String encodeUrl(String url)
{
String encoded="";
for(String i:Uri.parse(url).getQuery().split("&"))
encoded+=(i.split("=")[0]+"="+ URLEncoder.encode(i.split("=")[1])+"&");
return Uri.parse(url).getScheme()+Uri.parse(url).getSchemeSpecificPart()+encoded;
}
Upvotes: 0
Reputation: 2182
Please check below code.
String userName = "amit24*[email protected]";
String password = "anshu";
try {
String encodePassword = Uri.encode(password);
String encodeUserName = Uri.encode(userName);
android.util.Log.e(TAG, "onCreate: " + String.format("http://test.in/api/voi/login?password=%s&username=%s", encodePassword, encodeUserName));
} catch (Exception e) {
e.printStackTrace();
}
Upvotes: 2