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HTTP URL Address Encoding in Java
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i need java code to encode URL to avoid special characters such as spaces and % and & ...etc
URL construction is tricky because different parts of the URL have different rules for what characters are allowed: for example, the plus sign is reserved in the query component of a URL because it represents a space, but in the path component of the URL, a plus sign has no special meaning and spaces are encoded as "%20".
RFC 2396 explains (in section 2.4.2) that a complete URL is always in its encoded form: you take the strings for the individual components (scheme, authority, path, etc.), encode each according to its own rules, and then combine them into the complete URL string. Trying to build a complete unencoded URL string and then encode it separately leads to subtle bugs, like spaces in the path being incorrectly changed to plus signs (which an RFC-compliant server will interpret as real plus signs, not encoded spaces).
In Java, the correct way to build a URL is with the URI class. Use one of the multi-argument constructors that takes the URL components as separate strings, and it'll escape each component correctly according to that component's rules. The toASCIIString() method gives you a properly-escaped and encoded string that you can send to a server. To decode a URL, construct a URI object using the single-string constructor and then use the accessor methods (such as getPath()) to retrieve the decoded components.
Don't use the URLEncoder class! Despite the name, that class actually does HTML form encoding, not URL encoding. It's not correct to concatenate unencoded strings to make an "unencoded" URL and then pass it through a URLEncoder. Doing so will result in problems (particularly the aforementioned one regarding spaces and plus signs in the path).
I also spent quite some time with this issue, so that's my solution:
String urlString2Decode = "http://www.test.com/äüö/path with blanks/";
String decodedURL = URLDecoder.decode(urlString2Decode, "UTF-8");
URL url = new URL(decodedURL);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
String decodedURLAsString = uri.toASCIIString();
If you don't want to do it manually use Apache Commons - Codec library. The class you are looking at is: org.apache.commons.codec.net.URLCodec
String final url = "http://www.google.com?...."
String final urlSafe = org.apache.commons.codec.net.URLCodec.encode(url);
Here is my solution which is pretty easy:
Instead of encoding the url itself i encoded the parameters that I was passing because the parameter was user input and the user could input any unexpected string of special characters so this worked for me fine :)
String review="User input"; /*USER INPUT AS STRING THAT WILL BE PASSED AS PARAMTER TO URL*/
try {
review = URLEncoder.encode(review,"utf-8");
review = review.replace(" " , "+");
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
}
String URL = "www.test.com/test.php"+"?user_review="+review;
I would echo what Wyzard wrote but add that:
for query parameters, HTML encoding is often exactly what the server is expecting; outside these, it is correct that URLEncoder should not be used
the most recent URI spec is RFC 3986, so you should refer to that as a primary source
I wrote a blog post a while back about this subject: Java: safe character handling and URL building
Related
Pages with spaces in the URL don't get correctly translated:
i.e.
http://www.streetinsider.com/Press Releases/National Trends Reflected in Plano Housing Market/9778767.html
or
http://www.streetinsider.com/Press%20Releases/National+Trends+Reflected+in+Plano+Housing+Market/9778767.html
Gives 404. Please note "Press Releases" is encoded as "Press%20Releases".
However following two versions work fine where "Press Releases" is encoded as "Press+Releases".
http://www.streetinsider.com/Press+Releases/National+Trends+Reflected+in+Plano+Housing+Market/9778767.html
The article parses fine with plus signs or HEX spaces %20.
http://www.streetinsider.com/Press+Releases/National%20Trends%20Reflected%20in%20Plano%20Housing%20Market/9778767.html
Both + and %20 represent spaces. Then why this behavior.
And also, in java what could I use to get the correct encoded URL
Both + and %20 represent spaces
Only in query strings. Elsewhere in a URL a plus is a plus, not a space. In this case the web server gives you the same content for the two different URLs
http://www.streetinsider.com/Press+Releases/National+Trends+Reflected+in+Plano+Housing+Market/9778767.html
and
http://www.streetinsider.com/Press+Releases/National%20Trends%20Reflected%20in%20Plano%20Housing%20Market/9778767.html
but the two URLs are distinct, they're not alternative representations of the same URL.
Officially + might only be used in the query string (after ?).
This is what URLEncoder is for:
"?x=" + URLEncoder.encode("Hello World", "UTF-8");
"?x=" + URLEncoder.encode("ŝi estas ĉarma", "UTF-8");
?x=Hello+World
?x=%C5%9Di+estas+%C4%89arma
The more universal class URI, obeys the specification for spaces to be replaced, using %.
URI uri = new URI("http", "www.streetinsider.com",
"/Press Releases/National Trends Reflected in Plano Housing Market/9778767.html",
"?x=ŝi estas ĉarma");
String u = uri.toString();
http://www.streetinsider.com/Press%20Releases/National%20Trends%20
Reflected%20in%20Plano%20Housing%20Market/9778767.html#?x=ŝi%20estas%20ĉarma
One sometime encounters URI as generalisation for File and others, and then has to be careful not introducing %20 in file names.
So probably there is a partial remapping on streetinsider of + or even %20 as it seems; in order to reach the same code.
Your statement
Both + and %20 represent spaces.
is not exactly true in all cases.
Space characters may only be encoded as "+" in one context: application/x-www-form-urlencoded key-value pairs.
The RFC-1866 (HTML 2.0 specification), paragraph 8.2.1. subparagraph 1. says: "The form field names and values are escaped: space characters are replaced by `+', and then reserved characters are escaped").
Here is an example of such a string in URL where RFC-1866 allows encoding spaces as pluses: "http://example.com/over/there?name=foo+bar". So, only after "?", spaces can be replaced by pluses (in other cases, spaces should be encoded to %20). This way of encoding form data is also given in later HTML specifications, for example, look for relevant paragraphs about application/x-www-form-urlencoded in HTML 4.01 Specification, and so on.
The URL that you have provided is not a form data containing key/value pairs, it's just a path to a 9778767.html file:
http://www.streetinsider.com/Press%20Releases/National+Trends+Reflected+in+Plano+Housing+Market/9778767.html
So, it is illegal to use pluses here. The correct URL in this case should have been the following:
http://www.streetinsider.com/Press%20Releases/National%20Trends%20Reflected%20in%20Plano%20Housing%20Market/9778767.html
I have a problem calling WS.url() in play framework 2.3.3 with url containing spaces. All other characters all url encoded automatically but not spaces. When i try to change all spaces to "%20", WS convert it to "%2520" because of "%" character. With spaces i've got java.net.URISyntaxException: Illegal character in query. How can i handle this ?
part of the URL's query String:
&input=/mnt/mp3/music/folder/01 - 23.mp3
The code looks like this:
Promise<JsonNode> jsonPromise = WS.url(url).setAuth("", "cube", WSAuthScheme.BASIC).get().map(
new Function<WSResponse, JsonNode>() {
public JsonNode apply(WSResponse response) {
System.out.println(response.getBody());
JsonNode json = response.asJson();
return json;
}
}
);
You should "build" your URL based on the way java.net.URL(which Play! uses for it's WS) does it. WS.url() follows the same logic.
The use of URLEncoder/Decoder is recommended only for form data.
From JavaDoc:
"Note, the java.net.URI class does perform escaping of its component
fields in certain circumstances. The recommended way to manage the
encoding and decoding of URLs is to use java.net.URI, and to convert
between these two classes using toURI() and URI.toURL(). The
URLEncoder and URLDecoder classes can also be used, but only for HTML
form encoding, which is not the same as the encoding scheme defined
in RFC2396."
So, the solution is to use THIS:
WS.url(baseURL).setQueryString(yourQueryString);
Where:
baseURL is your scheme + host + path etc.
yourQueryString is... well, your query String, but WITHOUT the ?: input=/mnt/mp3/music/folder/01 - 23.mp3
Or, if you want to use a more flexible, programmatic approach, THIS:
WS.url(baseURL).setQueryParameter(param, value);
Where:
param is the parameter's name in the query String
value is the value of the parameter
If you want multiple parameters with values in your query you need to chain them by adding another .setQueryParameter(...). This implies that this approach is not very accomodating for complex, multi-parameter query Strings.
Cheers!
If you check the console you will find that the exception is : java.net.URISyntaxException: Illegal character in path at index ...
That's because play Java api uses java.net.URL (as you can see here in line 47).
You can use java.net.URLEncoder to encode your URL
WS.url("http://" + java.net.URLEncoder.encode("google.com/test me", "UTF-8"))
UPDATE
If you want an RFC 2396 compliant method you can do this :
java.net.URI u = new java.net.URI(null, null, "http://google.com/test me",null);
System.out.println("encoded url " + u.toASCIIString());
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.
Say i have space in a url, what is the right way to convert it to %20? no 'replace' suggestion please.
For example, if you put "http://test.com/test and test/a" into the browser window, it converts to http://test.com/test%20and%20test/a
If I use URLEncoder, I get even the / converted. which is not what i want.
Thanks,
this is the right way, seems like. to add to the question, what if there is also some non ascii code in the path that I want convert to valid url with utf8 encode? e.g.： test.com:8080/test and test/pierlag2_carré/a?query=世界 I'd want it to be converted to test.com:8080/test%20and%20test/pierlag2_carr%C3%A9/a?query=%E4%B8%96%E7%95%8C
Try splitting into a URI with the aid of the URL class:
String sUrl = "http://test.com:8080/test and test/a?query=world";
URL url = new URL(sUrl);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
String canonical = uri.toString();
System.out.println(canonical);
Output:
http://test.com:8080/test%20and%20test/a?query=world
The correct way to build URLs in Java is to create a URI object and fill out each part of the URL. The URI class handles the encoding rules for the distinct parts of the URL as they differ from one to the next.
URLEncoder is not what you want, despite its name, as that actually does HTML form encoding.
EDIT:
Based on your comments, you are receiving the URL as input to your application and do not control the initial generation of the URL. The real problem you are currently experiencing is that the input you are receiving, the URL, is not a valid URL. URLs / URIs cannot contain spaces per the spec (hence the %20 in the browser).
Since you have no control over the invalid input you are going to be forced to split the incoming URL into its parts:
scheme
host
path
Then you are going to have to split the path and separately encode each part to ensure that you do not inadvertently encode the / that delimits your path fragments.
Finally, you can put all of them back together in a URI object and then pass that around your application.
You may find useful this code to replace blank spaces in your URL:
String myUrl = "http://test.com/test and test/a";
myUrl = myUrl.replaceAll(" ", "%20");
URI url = new URI(myUrl);
System.out.print(url.toString());
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.