具有URL编码参数的Java URL编码

Question

How to encode in Java this String

http://demo.pl/sample?id=tests%trg=https%3A%2F%2Fwww.google.com%sample.html%3Fwmc%3DAFF48+_LS.%23%7NUMBER_ID%7D_%23%7NUMBER_ID%7D..

java.net.URLEncoder encode this String like this:

http%3A%2F%2Fdemo.pl%2Fsample%3Fid%3Dtests%25trg%3Dhttps%253A%252F%252Fwww.google.com%25sample.html%253Fwmc%253DAFF48%2B_LS.%2523%257NUMBER_ID%257D_%2523%257NUMBER_ID%257D..

I expect this result:

http%3A%2F%2Fdemo.pl%2Fsample%3Fid%3Dtests%25trg%3Dhttps%3A%2F%2Fwww.google.com%sample.html%3Fwmc%3DAFF48+_LS.%23%7NUMBER_ID%7D_%23%7NUMBER_ID%7D..

Answer 1

I think following code can help you:

String s = "http://demo.pl/sample?id=tests%trg=https%3A%2F%2Fwww.google.com%sample.html%3Fwmc%3DAFF48+_LS.%23%7NUMBER_ID%7D_%23%7NUMBER_ID%7D";
int i = s.indexOf("%");
String result1 = URLEncoder.encode(s.substring(0, i)) + "%25" + s.substring(i + 1);
System.out.println(result1); // print http%3A%2F%2Fdemo.pl%2Fsample%3Fid%3Dtests%25trg=https%3A%2F%2Fwww.google.com%sample.html%3Fwmc%3DAFF48+_LS.%23%7NUMBER_ID%7D_%23%7NUMBER_ID%7D

I do not want to encode encoded part of String. Needs a universal algorithm. String is not always encoded fragmentarily

I think universal algorithm is impossible in that case, what you can do that find encoded part manually and not encoded it again (see code above).