I'm trying to get an url parameter in jee.
So I have this kind of url :
http://MySite/MySite.jsp?page=recherche&msg=toto
First i tried with : request.getParameter("msg").toString();
it works well but if I try to search "c++" , the method "getParameter()" returns "c" and not "c++" and i understand.
So I tried another thing. I get the current URL and parse it to get the value of the message :
String msg[]= request.getQueryString().split("msg=");
message=msg[1].toString();
It works now for the research "c++" but now I can't search accent. What can I do ?
EDIT 1
I encode the message in the url
String urlString=Utils.encodeUrl(request.getParameter("msg"));
so for the URL : http://MySite/MySite.jsp?page=recherche&msg=c++
i have this encoded URL : http://MySite/MySite.jsp?page=recherche&msg=c%2B%2B
And when i need it, i decode the message of the URL
String decodedUrl = URLDecoder.decode(url, "ISO-8859-1");
Thanks everybody
Anything you send via "get" method goes as part of the url, which needs to be urlencoded to be valid in case it contains at least one of the reserved characters. So, any character will need to be encoded before sending.
In order to send c++, you would have to send c%2B%2B. That would be interpreted properly at the server side.
Here some reference you can check:
http://www.blooberry.com/indexdot/html/topics/urlencoding.htm
Now the question is, how and where do you generate your URL? According to the language, you will need to use the proper method to encode your strings.
if I try to search "c++" , the method "getParameter()" returns "c" and not "c++"
Query parameters are treated as application/x-www-form-urlencoded, so a + character in the URL means a space character in the parameter value. If you want to send a + character then it needs to be encoded in the URL as %2B:
http://MySite/MySite.jsp?page=recherche&msg=c%2B%2B
The same applies to accented characters, they need to be escaped as the bytes of their UTF-8 representation, so été would need to be:
msg=%C3%A9t%C3%A9
(é being Unicode character U+00E9, which is C3 A9 in UTF-8).
In short, it's not the fault of this code, it's the fault of whatever component is responsible for constructing the URL on the client side.
Call your URL with
msg=c%2B%2B
+ in a URL mean 'space'. It needs to be escaped.
You need to escape special characters when passing them as URL parameters. Since + means space and & means and another parameter, these cannot be used as parameter values.
See this other S.O. question.
You may want to use the Apache HTTP client library to help you with the URL encoding/decoding. The URIUtil class has what you need.
Something like this should work:
String rawParam = request.getParameter("msg");
String msgParam = URIUtil.decode(rawParam);
Your example indicates that the data is not being properly encoded on the client side. See this JavaScript question.
Related
I have setup a URI as below:
router.attach("/pmap/campaign/{campaign}/staffCat/{staffCat}/isp/{isp}", PMAPResource.class);
In the PMAPResource.class I have the following code:
public Representation represent()
{
String campaignID = (String) this.getRequestAttributes().get("campaign");
String staffCat = Reference.decode((String) this.getRequestAttributes().get("staffCat"));
String ispID = (String) this.getRequestAttributes().get("isp");
}
The staffCat field is manual input from user, it can be anything. Some examples are:
BAA2(A)
BAB1#(A)
BA B1
It works for most cases until it hits the # sign where it returns 404 Not Found error. Console dump shows the following /fwd-PMAP/pmap/campaign/1/staffCat/BAB1.
What should I do in order to read the "#" so I can get BAB1#(A) as it is?
# has special meaning in a URL. It must be escaped.
There are actually many special characters in URLs, so you should always escape arbitrary text when building a URL in the client.
In this particular case, the correct URL sent by the client would be something like:
/fwd-PMAP/pmap/campaign/1/staffCat/BAB1%23(A)/isp/XXX
Again, this is a problem that needs to be fixed on the client. The server will decode the %23 for you.
In my Rest API it should be possible to retrieve data which is inside a bounding box. Because the bounding box has four coordinates I want to design the GET requests in such way, that they accept the bounding box as JSON. Therefore I need to be able to send and document JSON strings as URL parameter.
The test itself works, but I can not document these requests with Spring RestDocs (1.0.0.RC1). I reproduced the problem with a simpler method. See below:
#Test public void ping_username() throws Exception
{
String query = "name={\"user\":\"Müller\"}";
String encodedQuery = URLEncoder.encode(query, "UTF-8");
mockMvc.perform(get(URI.create("/ping?" + encodedQuery)))
.andExpect(status().isOk())
.andDo(document("ping_username"));
}
When I remove .andDo(document("ping_username")) the test passes.
Stacktrace:
java.lang.IllegalArgumentException: Illegal character in query at index 32: http://localhost:8080/ping?name={"user":"Müller"}
at java.net.URI.create(URI.java:852)
at org.springframework.restdocs.mockmvc.MockMvcOperationRequestFactory.createOperationRequest(MockMvcOperationRequestFactory.java:79)
at org.springframework.restdocs.mockmvc.RestDocumentationResultHandler.handle(RestDocumentationResultHandler.java:93)
at org.springframework.test.web.servlet.MockMvc$1.andDo(MockMvc.java:158)
at application.rest.RestApiTest.ping_username(RestApiTest.java:65)
After I received the suggestion to encode the URL I tried it, but the problem remains.
The String which is used to create the URI in my test is now /ping?name%3D%7B%22user%22%3A%22M%C3%BCller%22%7D.
I checked the class MockMvcOperationRequestFactory which appears in the stacktrace, and in line 79 the following code is executed:
URI.create(getRequestUri(mockRequest)
+ (StringUtils.hasText(queryString) ? "?" + queryString : ""))
The problem here is that a not encoded String is used (in my case http://localhost:8080/ping?name={"user":"Müller"}) and the creation of the URI fails.
Remark:
Andy Wilkinson's answer is the solution for the problem. Although I think that David Sinfield is right and JSONs should be avoided in the URL to keep it simple. For my bounding box I will use a comma separated string, as it is used in WMS 1.1: BBOX=x1,y1,x2,y2
You haven't mentioned the version of Spring REST Docs that you're using, but I would guess that the problem is with URIUtil. I can't tell for certain as I can't see where URIUtil is from.
Anyway, using the JDK's URLEncoder works for me with Spring REST Docs 1.0.0.RC1:
String query = "name={\"user\":\"Müller\"}";
String encodedQuery = URLEncoder.encode(query, "UTF-8");
mockMvc.perform(get(URI.create("/baz?" + encodedQuery)))
.andExpect(status().isOk())
.andDo(document("ping_username"));
You can then use URLDecoder.decode on the server side to get the original JSON:
URLDecoder.decode(request.getQueryString(), "UTF-8")
The problem is that URIs have to be encoded as ACII. And ü is not a valid ASCII character, so it must be escaped in the url with % escaping.
If you are using Tomcat, you can use URIEncoding="UTF-8" in the Connector element of the server.xml, to configure UTF-8 escaping as default. If you do this, ü will be automatically converted to %C3%BC, which is the ASCII representation of the \uC3BC Unicode code-point (which represents ü).
Edit: It seems that I have missed the exact point of the error, but it is still the same error. Curly braces are invalid in a URI. Only the following characters are acceptable according to RFC 3986:
ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789-._~:/?#[]#!$&'()*+,;=%
So these must be escaped too.
I have a servlet running on tomcat 6 which should be called as follows:
http://<server>/Address/Details?summary="Acme & co"
However: when I iterate through the parameters in the servlet code:
//...
while (paramNames.hasMoreElements()) {
paramName = (String) paramNames.nextElement();
if (paramName.equals("summary")) {
summary = request.getParameter(paramName).toString();
}
}
//...
the value of summary is "Acme ".
I assume tomcat ignores the quotes - so it sees "& co" as a second parameter (albeit improperly formed: there's no =...).
So: is there any way to avoid this? I want the value of summary to be "Acme & co". I tried replacing '&' in the URL with & but that doesn't work (presumably because it's decoded back to a straight '&' before the params are parsed out).
Thanks.
Use http://<server>/Address/Details?summary="Acme %26 co". Because in URL special http symbol(e.g. &,/, //) does not work as parameters.
Are you encoding and decoding the URL with URLEncode ? If so, can you check what the input and output of those are ? Seems like one of the special characters is not being properly encoded/decoded
Try %26 for the &
Try your parameter like
summary="Acme & co"
& is part reserved characters. Refer RFC2396 section
2.2. Reserved Characters.
how to encode URL to avoid special characters in java
Characters allowed in GET parameter
HTTP URL - allowed characters in parameter names
http://illegalargumentexception.blogspot.in/2009/12/java-safe-character-handling-and-url.html
I encrypt a text "good-bye, friend" using BasicTextEncryptor. So the encrypt value looks like below,
3qe80L1ap+cR2zRU9csFwOffw5NtWTueLRYgSXyjctI=
Then I email a URL to the user where the above parameter as a token.
Then the user copies the below URL and presses enter,
http://localhost:8080/token=3qe80L1ap+cR2zRU9csFwOffw5NtWTueLRYgSXyjctI=
But when I access the parameter in Struts 2 application through the action method it gives me the encrypt parameter as below,
3qe80L1ap cR2zRU9csFwOffw5NtWTueLRYgSXyjctI=
The + is replaced by " ". So when I decrypt it, it gives me EncryptionOperationNotPossibleException.
Does struts decode the + to " " assuming browser + is a encode character? In that case it ok before I proceed with decrypt, I replace the space with + ?
A better way would be to "URL encode" the string before appending it to actual URL.
URLEncoder.encode("3qe80L1ap+cR2zRU9csFwOffw5NtWTueLRYgSXyjctI=", "ISO-8859-1");
This would make sure the token is correctly decoded.
To, answer your question, struts does not have any role in decoding the URL parameter. Its the core functionality of the application server to decode the URL parameter. So every HTTP parameter is subjected to decoding before reaching the application code.
Whatever is decoded by the server is available by to the application (i.e. Struts in your case. )
Now to explain why the + is not reaching your struts.
java.net.URLDecoder.decode("3qe80L1ap+cR2zRU9csFwOffw5NtWTueLRYgSXyjctI="));
it returns 3qe80L1ap cR2zRU9csFwOffw5NtWTueLRYgSXyjctI=
which means that + is not getting URL Decoded.
So, reiterating, every HTTP parameter (querystring or form POST) is subjected to decoding before reaching the application code.
When you URL encode your string, + is encoded as %2B and your struts application will receive the correct decoded string.
You'll need to not put the base64 encoded string there, but encode it using the UrlEncoder, like the following:
URLEncoder.encode("3qe80L1ap+cR2zRU9csFwOffw5NtWTueLRYgSXyjctI=", "UTF-8")
That way you can put it in the link.
Consider using a so called URL safe variant of Base64. The most common variant, described in RFC 4648, uses - and _ instead of + and / respectively, and omits padding characters (=).
Most implementations of Base64 support this URL safe variant too, though if yours doesn't, it's easy enough to do manually.
URLs cannot contain spaces. URL encoding normally replaces a space with a + sign.
Thus the server decodes normally + sign to the space. See URLEncoder docs or read Java URL encoding of query string parameters.
I am using jsps and in my url I have a value for a variable like say "L & T". Now when I try to retrieve the value for it by using request.getParameter I get only "L". It recognizes "&" as a separator and thus it is not getting considered as a whole string.
How do I solve this problem?
java.net.URLEncoder.encode("L & T", "utf8")
this outputs the URL-encoded, which is fine as a GET parameter:
L+%26+T
A literal ampersand in a URL should be encoded as: %26
// Your URL
http://www.example.com?a=l&t
// Encoded
http://www.example.com?a=l%26t
You need to "URL encode" the parameters to avoid this problem. The format of the URL query string is:
...?<name>=<value>&<name>=<value>&<etc>
All <name>s and <value>s need to be URL encoded, which basically means transforming all the characters that could be interpreted wrongly (like the &) into %-escaped values. See this page for more information:
http://www.w3schools.com/TAGS/ref_urlencode.asp
If you're generating the problem URL with Java, you use this method:
String str = URLEncoder.encode(input, "UTF-8");
Generating the URL elsewhere (some templates or JS or raw markup), you need to fix the problem at the source.
You can use UriUtils#encode(String source, String encoding) from Spring Web. This utility class also provides means for encoding only some parts of the URL, like UriUtils#encodePath.