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.
Related
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.
I'm trying to URL percent encode my query param value while using URIBuilder to make an HTTP request to Bing API.
The url looks like
"https://api.datamarket.azure.com/Data.ashx/Bing/SearchWeb/v1/Web?$format=json&Query="
Where the Query String must be like
%27Test%20query%27
Using URLEncoder.encode(string, code), a string such as "test query", gets turned into "test+query" which is unacceptable.
URIUtil.encodeQuery()
returns "test%20query" which is almost acceptable, except it needs the %27 at the beginning and end.
When I try to just concatenate the string to make it valid as such, and then load this into URIBuilder, URIBuilder ends up with
https://api.datamarket.azure.com/Data.ashx/Bing/SearchWeb/v1/Web?%24format=json&Query=%2527test%2520query%2527
which is again unacceptable.
How can I remedy this issue? It's driving me insane.
Thanks for any help.
this is encoded URI.
$ is %24
bank is %20
if you want real URI, you need to decode .
I think decode method works well for you.
reference here:
http://hc.apache.org/httpclient-3.x/apidocs/org/apache/commons/httpclient/util/URIUtil.html
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.
I am doing an application where I have to read a URL from a webpage as a String[Its not the address of the page]. The URL that I will be reading contains query string, and I specifically need two queries from that URL. So I am using the Uri class available in Android. Now, the problem lies in the encoding/format of the URL and the query. One of the queries that I need is always an URL. Sometimes the query URL is %-encoded and sometimes not.
The URLs can be like the following :
Case 1 :
http://www.example.com/example/example.aspx?file=http%3A%2F%2FXX.XXX.XX.XXX%2FExample.file%3Ftoken%3D9dacfc85
Case 2 :
http://www.example.com/example/example.aspx?file=http://XX.XXX.XX.XXX/Example.file?token=9dacfc85
How do I get the correct Url contained in the file= query?
I am using the following [to accomplish the said work universally] :
Uri.decode(urlString.getQueryParameter("file"));
Is this the correct way to do it?
UPDATE
I have decided to first encode the whole URL regardless of its value and then get the query parameter. Theoretically, it should work.
If you are uncertain about the type of URL you would get then I would suggest you to decode every URL you get from the parameter. And when you need to use it then you can encode it.
As per my knowledge, you are doing it right.
One of parameters for action link looks like:
itemUrl=feedLink.html#xtor=RSS-3208
when I execute next code in backend in processAction():
String itemUrl = (String) request.getParameter("itemUrl");
,that I get next value: feedLink.html
e.g. request cuts itemUrl value after # symbol
escapeXml="true" in .jsp file doesn't help.
You have to URI encode the parameter names and values - your link should be itemUrl=feedLink.html%23xtor=RSS-3208.
Anything after the # in an URL specifies the location on the page the browser should display; it's not part of the URL itself. As such, if you want an actual # in your URL, it needs to be escaped (if the parser is actually compliant).
In theory, you could parse the whole URL being sent to you manually, but the better solution is to get the caller of your page to send you a correct URL in the first place (well, a URL that represents what they want, since the one in question is valid, per se).