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How can I prevent the automatic decoding of RequestParam Strings in a Spring controller

I have an HTTP GET request controller endpoint where I take in a fileName as a query param and pass that on to another service. For this request the param the filename could include any sort of special characters and I would like to keep these values encoded when passing them on. 2 Characters that have been causing issues are spaces (%20) and +(%2B).
How can I keep these characters encoded in the request params.
So far I have tried using the #RequestParam annotation as well as retrieving the params via HttpServletRequest.getParameterValues(String) but both return the decoded values as spaces.
Any help is appreciated thanks!

Yes, these are automatically decoded by the servlet API. You should be able to re-encode them -
encodedValue = URLEncoder.encode(value, StandardCharsets.UTF_8);

I found out that I could get the actual value passed in by using the HttpServletRequest.getQueryString() method. Parsing this query string I was able to get the un-decoded version of the fileName being passed in. I hope this helps someone in the future.

How do I parse an HttpExchange request after an ending slash?

I have a request as follows:
localhost:8000/location/:01
My code takes as input an HttpContext request.
func(HttpExchange r) {
String area_path = r.getRequestURI(); // Equals string "/location/"
}
How do I parse an HttpExchange correctly so I can pull out the "01" from this path and store it as a variable?

That (localhost:8000/location/:01) is not a valid URL or URI
A plain colon character is not legal in the path of a URL or URI. If you want to put a colon in the path, it must be percent-encoded. Furthermore, if this was a URL, it would start with a protocol; e.g. http:.
Now ... it is unclear what the HTTP stack you are using will do with a syntactically incorrect URL / URI, but it could simply be ignoring the colon and the characters after it.
Your code looks a bit odd too. You have tagged the question as [java]. But the code looks like JavaScript rather than Java; i.e. func is a Javascript keyword. But it also looks like you are using the (deprecated) com.sun.net.httpserver.HttpExchange Java class. I don't know what to make of that ...
My advice:
Don't use a colon character in the URL path.
If you must do it, then percent-encode the colon it.
If you cannot encode it properly, then you may need to find and use a different framework for your HTTP request handling. One that will accept and handle a malformed URL / URI in the way that you want. (Good luck finding one!)
Unfortunately, the details in your question are too sketchy to give more detailed advice.

Prevent java.net.URL from stripping "?" if there are no query paramters

I am trying to use java.net.HttpURLConnection in order to make a request to a URL like this:
https://example.com/app/?#/something=else&someting2=else2
In order to do this, I need to construct a java.net.url, but the constructor strips away the question mark "?":
java.net.URL url = new URL("https://example.com/app/?#/something=else&someting2=else2");
String string = url.toString();
// String is https://example.com/app/#/something=else&someting2=else2
And when I create the connection with url.openConnection(), I just get a 404.
I get why it does this. It righfully recognizes everything after and including the hash # as a url fragment, which means that the actual url becomes https://example.com/app/?. And then, it just strips the trailing question mark. But I need the URL to be rendered as is.
This is for an Android app.
How can I force HTTPUrlConnection to make the GET request with the question mark in the URL?

In a URL, the part after ? is a query string, and the part after # is a fragment identifier, as you already noted.
While the query string is sent to the server, the fragment identifier is not, so trying to send one does not make sense.
2 possibilities:
if it is a fragment identifier, you don't need to send it and Android's behavior (normalizing the URL) is correct. Stripping the trailing ? (empty query string) should not be a problem as it makes no sense alone (for both server and client). If you want to preserve the fragment in your client code, what you want is to create a URI, use that to display wherever you want, then convert to URL with toURL() when you need to speak with server.
if it is not actually a fragment identifier, it's supposed to be part of the query string, so you need to send a hash sign as part of the URL: in that case, you need to url-encode it to %23, the part after the ? won't be stripped, and the server will know what to do with the encoded %23.

Need to replace spaces inside string with percentual symbol Java

I need to replace the spaces inside a string with the % symbol but I'm having some issues, what I tried is:
imageUrl = imageUrl.replace(' ', "%20");
But It gives me an error in the replace function.
Then:
imageUrl = imageUrl.replace(' ', "%%20");
But It still gives me an error in the replace function.
The I tried with the unicode symbol:
imageUrl = imageUrl.replace(' ', (char) U+0025 + "20");
But it still gives error.
Is there an easy way to do it?

String.replace(String, String) is the method you want.
replace
imageUrl.replace(' ', "%");
with
imageUrl.replace(" ", "%");
System.out.println("This is working".replace(" ", "%"));
I suggest you to use a URL Encoder for Encoding Strings in java.
String searchQuery = "list of banks in the world";
String url = "http://mypage.com/pages?q=" + URLEncoder.encode(searchQuery, "UTF-8");

I've ran into issues like this in the past with certain frameworks. I don't have enough of your code to know for sure, but what might be happening is whatever http framework you are using, in my case it was spring, is encoding the URL again. I spent a few days trying to solve a similar problem where I thought that string replace and the URI.builder() was broken. What ended up being the problem was that my http framework had taken my encoded url, and encoded it again. that means that any place it saw a "%20", it would see the '%' charictor and switch it out for '%' http code, "%25", resulting in. "%2520". The request would then fail because %2520 didn't translate into the space my server was expecting. While the issue apeared to be one of my encoding not working, it was really an issue of encoding too many times. I have an example from some working code in one of my projects below
//the Url of the server
String fullUrl = "http://myapiserver.com/path/";
//The parameter to append. contains a space that will need to be encoded
String param 1 = "parameter 1"
//Use Uri.Builder to append parameter
Uri.Builder uriBuilder = Uri.parse(fullUrl).buildUpon();
uriBuilder.appendQueryParameter("parameter1",param1);
/* Below is where it is important to understand how your
http framework handles unencoded url. In my case, which is Spring
framework, the urls are encoded when performing requests.
The result is that a url that is already encoded will be
encoded twice. For instance, if you're url is
"http://myapiserver.com/path?parameter1=param 1"
and it needs to be read by the server as
"http://myapiserver.com/path?parameter1=param%201"
it makes sense to encode the url using URI.builder().append, or any valid
solutions listed in other posts. However, If the framework is already
encoding your url, then it is likely to run into the issue where you
accidently encode the url twice: Once when you are preparing the URL to be
sent, and once again when you are sending the message through the framework.
this results in sending a url that looks like
"http://myapiserver.com/path?parameter1=param%25201"
where the '%' in "%20" was replaced with "%25", http's representation of '%'
when what you wanted was
"http://myapiserver.com/path?parameter1=param%201"
this can be a difficult bug to squash because you can copy the url in the
debugger prior to it being sent and paste it into a tool like fiddler and
have the fiddler request work but the program request fail.
since my http framework was already encoding the urls, I had to unencode the
urls after appending the parameters so they would only be encoded once.
I'm not saying it's the most gracefull solution, but the code works.
*/
String finalUrl = uriBuilder.build().toString().replace("%2F","/")
.replace("%3A", ":").replace("%20", " ");
//Call the server and ask for the menu. the Menu is saved to a string
//rest.GET() uses spring framework. The url is encoded again as
part of the framework.
menuStringFromIoms = rest.GET(finalUrl);
There is likely a more graceful way to keep a url from encoding twice. I hope this example helps point you on the right direction or eliminate a possability. Good luck.

Try this:
imageUrl = imageUrl.replaceAll(" ", "%20");

Replace spaces is not enought, try this
url = java.net.URLEncoder.encode(url, "UTF-8");

java servlet: request parameter contains plus

The request parameter is like decrypt?param=5FHjiSJ6NOTmi7/+2tnnkQ==.
In the servlet, when I try to print the parameter by String param = request.getParameter("param"); I get 5FHjiSJ6NOTmi7/ 2tnnkQ==. It turns the character + into a space. How can I keep the orginal paramter or how can I properly handle the character +.
Besides, what else characters should I handle?

You have two choices
URL encode the parameter
If you have control over the generation of the URL you should choose this. If not...
Manually retrieve the parameter
If you can't change how the URL is generated (above) then you can manually retrieve the raw URL. Certain methods decode parameters for you. getParameter is one of them. On the other hand, getQueryString does not decode the String. If you have only a few parameters it shouldn't be difficult to parse the value yourself.
request.getQueryString();
//?param=5FHjiSJ6NOTmi7/+2tnnkQ==

If you want to use the '+' character in a URL you need to encode it when it is generated. For '+' the correct encoding is %2b

Use URLEncoder,URLDecoder's static methods for encoding and decoding URLs.
For example : -
Encode the URL param using
URLEncoder.encode(url,"UTF-8")
Back in the server side , decode this parameter using
URLDecoder.decode(url,"UTF-8")
decode method returns a String type of the decoded URL.

Allthough the question is some years old, I'd like to write down how I fixed the problem in my case: the download link to a file is created in a GWT page where
com.google.gwt.http.client.URL.encode(finalurl)
is used to encode the URL.
The problem was that the "+" sign a customer of us had in the filename wasn't encoded/escaped. So I had to remove the URL.encode(finalurl) and encode each parameter in the url with
URL.encodePathSegment(fileName)
I know my question is bound to GWT but it seems, URLEncoder.encode(string, encoding) should be applied to the parameter only aswell.