I guess this is duplicate. But could not found what I am loking for.
I have one Java Spring MVC web application runniong as server. I have another angularJS application running as client. In AngularJS I am setting some cookies, But the string values are encoded with special characters. It looks like this
id=20; name=%22myname%22
In my Controller I am getting the cookie value id properly, but not name.
Cookie[] cookies=request.getCookies();
String name=cookies[1].getValue();
System.out.println(name);
And it prints
%22myname%22
How to get rid of this encoded characters?
You can decode your string in your controller as follows:
URLDecoder.decode(name, "utf8")
String value = URLDecode.decode(cookie.getValue(), "UTF-8");
Related
I am trying to understand what is the difference and importance of different charsets available while encoding and decoding text.
I have a scenario, where I want to call a RestAPI. The RestAPI has a base URL, for ex: https://myrestapiurl.com. Now to perform a GET request, the URL is formed by appending the id of the entity that I want to fetch, like: https://myrestapiurl.com('id')
id : It has no limitations on valid characters!
I have encountered an id: باقی ریسورس , So before calling the RestAPI, I need to encode it. Using Java's URLEncoder, I tried the following:
String s ="باقی ریسورس";
String encodedID = URLEncoder.encode(s, StandardCharsets.UTF_8.name() )
Using the encodedID, I try to make a request using PostMan. The request fails with 404 or 400 when I use different charset. It only succeeds when I encode using ISO_8859_1 as follows:
String encodedID = URLEncoder.encode(s, StandardCharsets.ISO_8859_1.name());
String URL = "https://myrestapiurl.com('" + encodedID + "')";
This works fine, through code as well as PostMan. My question is:
How can I decide which charset to use before encoding? Or should I have fallbacks? That is if it fails with UTF_8 then try with UTF_16 etc etc...but this is very in-efficient. In case if the entity actually doesn't exist, then, these tries would be overhead
Also, when I visit https://www.w3schools.com/tags/ref_urlencode.ASP and enter the text to be encoded, it provides the valid encoded string with ISO_8859_1 , how does it manage to do so?
How can this be done in Java without using any other extra libraries like apache? We don't have choice to add extra dependencies!
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.
I have a request,In Browser address bar enter:
http://localhost:8888/cmens-tops-outwear/t-b-f-a-c-s-fLoose-p-g-e-i-o.htm?'"--></style></script><script>netsparker(0x0000E1)</script>=
Tomcat6.0.35 i have set URIEncoding="UTF-8"
Use request.getQueryString() in servlet:
if chrome,i get
'%22--%3E%3C/style%3E%3C/script%3E%3Cscript%3Enetsparker(0x0000E1)%3C/script%3E=
if ie,I get
'"--></style></script><script>netsparker(0x0000E1)</script>=
Why?
Additional
I want to get request.getQueryString() to create a uri
URI uri = URI.create(url)
if ie:
java.net.URISyntaxException: Illegal character in query at index 36: /cmens/t-b-f-a-c-s-f-p-g-e-i-o.htm?'"--></style></script><script>netsparker(0x0000E1)</script>
at java.net.URI$Parser.fail(URI.java:2809)
at java.net.URI$Parser.checkChars(URI.java:2982)
at java.net.URI$Parser.parseHierarchical(URI.java:3072)
at java.net.URI$Parser.parse(URI.java:3024)
at java.net.URI.<init>(URI.java:578)
at java.net.URI.create(URI.java:840)
How to determine the queryString whether has be encoded?
The HttpServletRequest#getQueryString() is per definition undecoded. See also the javadoc (emphasis mine):
Returns:
a String containing the query string or null if the URL contains no query string. The value is not decoded by the container.
Basically, you need to URL-decode it yourself if you'd like to parse it manually instead of using getParameterXxx() methods for some reason (which implicitly decodes the parameters!).
String decodedQueryString = URLDecoder.decode(request.getQueryString(), "UTF-8");
As to why Chrome sends it encoded while IE not, that's because Chrome is doing a better job of handling HTTP requests the safe/proper way. This is beyond your control. Just always URL-decode the query string yourself if you intend to parse it manually for some reason. The URIEncoding="UTF-8" configuration has only effect on getParameterXxx() methods during GET requests.
The Chrome version is URLEncoded while the IE string is decoded.
Use this tool to compare the URLEncoded and decoded versions: http://meyerweb.com/eric/tools/dencoder/
Chrome uses the URL encoding way, but IE is using strings.
For example: " is %22 in URL encoding.
< is %3E
and > is %3C
Chrome is doing it the "right way" but IE just can't do as all the others.
You can find complete list of URL characters here: http://www.w3schools.com/tags/ref_urlencode.asp
Chrome sends the url encoded. Try decoding the query string using
URLDecoder.decode(queryString, "UTF-8");
As stated by the javadoc, the query string is not decoded by the container:
returns a String containing the query string or null if the URL contains no query string. The value is not decoded by the container.
javadoc
From the following URL in OathCallBack page I want extract access_token and token_type using Java. Any idea how to do it?
http://myserver.com/OathCallBack#state=/profile&access_token=ya29.AHES6ZQLqtYrPKuw2pMzURJtWuvINspm8-Vf5x-MZ5YzqVy5&token_type=Bearer&expires_in=3600
I tried the following, but unable to extract required information.
{
String scheme = req.getScheme(); // http
String serverName = req.getServerName(); // myserver.com
int serverPort = req.getServerPort(); // 80
String contextPath = req.getContextPath();
String servletPath = req.getServletPath();
String pathInfo = req.getPathInfo(); // return null and exception
String queryString = req.getQueryString(); // return null
}
<---------------------------------------------------------->
I am going to edit my question
Thank you every one for nice reply,
google did it,
you can refer to that link by URL
http://developers.google.com/accounts/docs/OAuth2Login
inside above URL page there is following link
http://accounts.google.com/o/oauth2/auth? scope=https%3A%2F%2Fwww.googleapis.com%2Fauth%2Fuserinfo.email+https%3A%2F%2Fwww.googleapis.com%2Fauth%2Fuserinfo.profile& state=%2Fprofile& redirect_uri=https%3A%2F%2Foauth2-login-demo.appspot.com%2Foauthcallback& response_type=token& client_id=812741506391.apps.googleusercontent.com
when you click on above link, then you will get your gmail login account access_token, and that token is after # sign
Some characters cannot be part of a URL (for example, the space) and some other characters have a special meaning in a URL: for example, the character # can be used to further specify a subsection (or fragment) of a document; the character = is used to separate a name from a value.
see http://en.wikipedia.org/wiki/Query_string for more:
It looks like the '#' should be a '?'.
In a normal URL, the parameters are passed as key value pairs following a '?' and multiple parameters chained together using '&'. A URL might look as follows:
http: //someserver.com/somedir/somepage.html?param1=value1¶m2=value2¶m3=value3.
Normally the Java servlet container would return everything after the '?' when calling getQueryString() but due to the absence of the '?' it returns null.
As #Sandeep Nair has suggested getRequestURL() should return this full URL to you and you could parse it using regular expressions to get the information you want. A possible regular expression to use would be along the lines of:
(?<=access_token=)[a-zA-Z0-9.-]*
However, getRequestURL() does NOT normally return the query string, so using this method is relying on the fact that there is a '#' rather and a '?' and is therefore probably not a great solution. See here.
I would advise that you find out why you are getting a '#' instead of a '?' and try to get this changed, if you can do this then the servlet container should manage the URL parameters for you and call to request.getAttribute("access_token") and request.getAttribute("token_type") (see here) will return both values as strings.
You get query string by calling
String queryString = req.getQueryString();
It correctly returns null in your case, as there is no query string. The characters after "#" are anchor specification, which is only visible to the browser and not sent to server.
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.