I should use link. And it means that method GET and all parameters I can see in url. How I can hide parametrs from url like that
http://localhost:8080/MyApp/action.do?method=delete&id=0
And how after that I can get parametrs in java class.
You can't hide GET parameters from a URL. GET name value pairs have to be part of a request. However one thing you can certainly try is to encrypt your querystring into a string which only your code can decode on your host, for example
http://localhost:8080/MyApp/action.do?method=delete&id=0
could be
http://localhost:8080/MyApp/action.do?param=[ENCRYPTED/OBFUSCATED STRING]
then your application at the above url can decode that and convert back into name value pairs
That's the idea of GET method - to path parameters through URL, probably you should use POST.
Related
I am trying to achieve same thing as this: How to use query parameter represented as JSON with Spring RestTemplate?, sending JSON string as a URL parameter in restTemplate.exchange().
The accepted answer mentions that sending JSON object as request parameter is generally not a good idea since you will probably face problem with curly brackets inside JSON. That is exactly what is happening when I am trying to make a GET call to an API. Since this is an API from another system, I cannot ask them to change the format and will have to call the GET endpoint, passing JSON as parameter. How can I achieve this in restTemplate.exchange() call?
Note: The mentioned related question does not guide on how to overcome this problem and I do not have enough reputation to comment on it to ask the author of the answer.
Answering my own question. While it is a bad idea to pass JSON like this in a query/url parameter, there is a workaround as suggested here: https://jira.spring.io/browse/SPR-9220?focusedCommentId=76760&page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel#comment-76760.
Replicating the code here in case this link goes dead:
String url = "http://localhost:8983/solr/select?wt=json&indent=true&fl=*&q=*:*&fq={!geofilt}&sfield=venue_location&pt=28.0674,-80.5595&d=25";
URI uri = UriComponentsBuilder.fromUriString(url).build().encode().toUri();
System.out.println(uri);
// http://localhost:8983/solr/select?wt=json&indent=true&fl=*&q=*:*&fq=%7B!geofilt%7D&sfield=venue_location&pt=28.0674,-80.5595&d=25
Basically, instead of passing url having JSON query/url parameters as a string, pass it as a URI. Then call exchange method as before, but with URI instead of String:
restTemplate.exchange(uri, HttpMethod.GET, requestEntity, String.class)
If this is 3rd party API and you cannot control or change JSON processing on backend side - there is no solution. Even if you will encode with URLEncoder - there is no guarantee that API backend would process such request correctly.
You can use URLEncoder class to encode the URL in exchange method, e.g.:
String url = "http://www.yoursite.com/api?param={\"some_key\":\"some_value\"}";
System.out.println(URLEncoder.encode(url, StandardCharsets.UTF_8.name()));
This will encode the characters (like braces and double quotes) and server then will decode it back to json.
Is it possible to get everything after the root domain name from an HTTP request using Java?
So if the url being requested is http://example.com/my-path then I'd like to get the value of my-path
I'm using the PlayFramework so if there's a header called "Path" or something like that it should be easier to get it with:
String path = request.headers.get("path");
But this page suggests that such a thing doesn't exist:
https://en.wikipedia.org/wiki/List_of_HTTP_header_fields
If there's a different solution to get the same result, that would also be appreciated. If it is possible it will provide a solution to this question as well:
http://stackoverflow.com/questions/28503129/redirect-example-com-to-www-example-com-in-playframework-1-2-x
To get everything after the domain, use uri:
String path = request.uri();
Or if you want to cut out query string parameters, use path:
String path = request.path();
How would I get the full URL of a JSP page.
For example the URL might be http://www.example.com/news.do/?language=nl&country=NL
If I do things like the following I always get news.jsp and not .do
out.print(request.getServletPath());
out.print(request.getRequestURI());
out.print(request.getRequest());
out.print(request.getContextPath());
You need to call request.getRequestURL():
Reconstructs the URL the client used to make the request. The returned URL contains a protocol, server name, port number, and server path, but it does not include query string parameters.
Given URL = http:/localhost:8080/sample/url.jsp?id1=something&id2=something&id3=something
request.getQueryString();
it returns id1=something&id2=something&id3=something
See This
I found a solution. It may not be perfect solution. But it working solution.
String qs = request.getQueryString();
String foo = request.getRequestURL().toString()+"?"+qs;
How to read access token from this url? When I try to read this by request.getParameter() method it returns null value
http://localhost:8080/FirstPick/views/common/home.faces#access_token=xxxxxxxxxx&expires_in=4015
Besides this,
Content after the hash (#) is only be used on the client side. If you require that information on the server, you can use a different separator with query-string using '?', or you can submit it via Ajax after the page has loaded by reading it on the client with JavaScript.
The part of the URI after the hash(#) is never sent to the server, reason is that the hash identifier was originally designed to point at references within the given web page and not to new resources on the server.
Thanks
You cant read any parameter from querystring like this..
It must contain '?'.
Only the string that appears after '?' is called 'QueryString'.
and from 'QueryString' you can get the value.
And you menstion the url here, is not contain '?' so it doesnt have 'QueryString'.
and You cant use request.getParameter() method.
On client side (i.e. from JavaScript) you can check window.location.hash to get hash. On server side, general answer is 'it is impossible' since hash is not sent in request to server.
Is there any way I can get url address of an application from java code, I mean complete address not only value from getContextPath(). Something like http://localhost:8080/etc
Try with getRequestUrl().
I hope it helps you
In a servlet or JSP, you can call javax.servlet.http.HttpUtils.getRequestURL(request)
It returns a StringBuffer containing the entire URL up to the servlet
From the javadoc
Reconstructs the URL the client used to make the request, using information in the HttpServletRequest object. The returned URL contains a protocol, server name, port number, and server path, but it does not include query string parameters.
If you only want up to the context path, you'll have to remove your servlet path
There is no method that returns the entire URL including the query parameters. You need to use something like:
req.getRequestURL()+"?"+req.getQueryString();
Or if you don't have query parameters you can use getRequestURL
It is pretty confusing, but here is a graphic that helps sort it out: