I have a url, something like this localhost:8080/foo.action?param1=7¶m2=8¶m3=6
When this is the Url (as it is), request.getParmeter("param2") gives me 8 [Correct]
i) When the encoding converts this url to localhost:8080/foo.action?param1=7%26param2=8%26param3=6
In this case, request.getParameter("param1") gives me 7¶m2=8¶m3=6
ii) When the encoding converts this url to localhost:8080/foo.action?param1=7¶m2=8¶m3=6
In this case, request.getParameter("param1") gives me 7 and request.getParameter("param2") gives me null
What is the correct way of retrieving the parameters? [Assuming that using one of the two Url encoding schemes is unavoidable]
(I am using struts actions)
To prevent this do not encode parameters with delimeters, encode only parameters values. This will be the best way. If you cannot handle parameters encoding just do decoding on server side before parsing:
String queryString = request.getQueryString();
String decoded = URLDecoder.decode(queryString, "UTF-8");
String[] pares = decoded.split("&");
Map<String, String> parameters = new HashMap<String, String>();
for(String pare : pares) {
String[] nameAndValue = pare.split("=");
parameters.put(nameAndValue[0], nameAndValue[1]);
}
// Now you can get your parameter:
String valueOfParam2 = parameters.get("param2");
You can call req.getQueryString() to get the whole query parameters and then do server side decoding based on whatever encoding methods you choose.
Try using
String[] Parameters = = URLDecoder.decode(Request.getQueryString(), 'UTF-8').splitat('&') ;
Hope this helps.
I had this happen to me today. Turns out I was passing the encoded url over the wire. When the request is made it should be made as http://localhost/foo?bar=1&bat=2 not as http://localhost/foo?bar=1&bat=2.
In this case I had cut the url from an xml file and pasted it into a browser.
Related
I need to encode only parameters of the string url.
my string url is like: http://127.0.0.1:8070/app/api/fetchData?channel=abc¶m=status:new|addr:null|roomId:Default&group=iPh&reqtype=p1&serialNo=123890&codeId=A1_8uh&type=p
I want to encode value of param(key).I am working on spring boot project.
Please suggest some solution.
If you're making the request from Java, you can encode a string using base64 like this[1]:
String originalInput = "test input";
String encodedString = Base64.getEncoder().encodeToString(originalInput.getBytes());
I guess my first thought would be to encode the parameters that you want this way and then concatenate the whole thing together. What value are you trying to get out of encoding the parameters?
[1] https://www.baeldung.com/java-base64-encode-and-decode
I have problems with the character +(and maybe others) at the URIBuilder is suppose to get a decoded url but when I extract the query the + is replaced
String decodedUrl = "www.foo.com?sign=AZrhQaTRSiys5GZtlwZ+H3qUyIY=&more=boo";
URIBuilder builder = new URIBuilder(decodedUrl);
List<NameValuePair> params = builder.getQueryParams();
String sign = params.get(0).getValue();
the value of sing is AZrhQaTRSiys5GZtlwZ H3qUyIY= with a space instead +. How can I extract the correct value?
other way is:
URI uri = new URI(decodedUrl);
String query = uri.getQuery();
the value of query is sign=AZrhQaTRSiys5GZtlwZ+H3qUyIY=&more=boo in this case is correct, but I have to strip it. Is there another way to do that?
Use it differently:
String decodedUrl = "www.foo.com";
URIBuilder builder = new URIBuilder(decodedUrl);
builder.addParameter("sign", "AZrhQaTRSiys5GZtlwZ+H3qUyIY=");
builder.addParameter("more", "boo");
List<NameValuePair> params = builder.getQueryParams();
String sign = params.get(0).getValue();
addParameter method is responsible for adding parameters as to the builder. The constructor of the builder should include the base URL only.
If this URL is given to you as is, then the + is already decoded and stands for the space character. If you are the one who generates this URL, you probably skipped the URL encoding step (which can be done using the code snipped above).
Read a bit about URL encoding: http://en.wikipedia.org/wiki/Query_string#URL_encoding
That is because if you send space as parameter in url it is encoded as +. This happens because there are some rules which characters are valid in URL. See URL RFC.
It is necessary to encode any characters disallowed in a URL, including spaces and other binary data not in the allowed character set, using the standard convention of the "%" character followed by two hexadecimal digits.
If you want to have + as symbol in url you need to encode it into %2B. For example 2+2 is encoded as 2%2B2 and i am as i+am. So in your case I believe you have to correct result as AZrhQaTRSiys5GZtlwZ+H3qUyIY decodes into AZrhQaTRSiys5GZtlwZ H3qUyIY.
I am trying to call a URL from Java code in the following way:
userId = "Ankur";
template = "HelloAnkur";
value= "ParamValue";
String urlString = "https://graph.facebook.com/" + userId + "/notifications?template=" +
template + "&href=processThis.jsp?param=" + value + "&access_token=abc123";
I have the following problems in this:
When I do println(urlString), I see that the urlString only has the value upto and before the first ampersand (&). That is, it looks as: https://graph.facebook.com/Ankur/notifications?template=HelloAnkur and rest of it all (which should have been &href=processThis.jsp?param=ParamValue&access_toke=abc123) gets cut off. Why is that and how can I get and keep the full value in urlString? Does & needs to be escaped in a Java String, and if yes, how to do it?
Notice that I am trying to pass a (relative) URL as a parameter value in this query (the value of href as processThis.jsp?param=ParamValue. How can I pass this type of value of href without mixing it up with the query of this URL (urlString), which only has three parameters template, href and access_token? That is, how can I hide or escape ? and =? Further, what would I need to do if value was Param Value (with a space)?
Notice that the template has the value HelloAnkur (with no space). But if I wanted it to have space, as in Hello Ankur, how would I do it? Should I write it as Hello%20Ankur or Hello Ankur would be fine?
I need the solution in such a way that URL url = new URL(urlString) can be created, or url can be created via URI. Please describe your answer up to this point as creating a safe URL is not straight forward in Java.
Thanks!
(this is going to become a classic)
Use URI Templates (RFC 6570). Using this implementation (disclaimer: mine), you can avoid all encoding problems altogether:
// Immutable, can be reused as many times as you wish
final URITemplate template = new URITemplate("https://graph.facebook.com/{userId}"
+ "/notifications?template={template}"
+ "&href=processThis.jsp?param={value}"
+ "&access_token=abc123");
final Map<String, VariableValue> vars = new HashMap<String, VariableValue>();
vars.put("userId", new ScalarValue("Ankur"));
vars.put("template", new ScalarValue("HelloAnkur"));
vars.put("value", new ScalarValue("ParamValue");
// Build the expanded string
final String expanded = template.expand(vars);
// Go with the string
Note that URI templates not only allow scalar values, but also arrays (the RFC calls these "lists" -- implemented as ListValue in the above) and maps (the RFC calls these "associative arrays" -- implemented as MapValue in the above).
I am a beginner in JSP programming and I would like to know which
code ( program ) I must write to get the Unicode value from queryString
Thank you very much for your help !
For getting unicode value this code may help you.
if(request.getCharacterEncoding() == null)
request.setCharacterEncoding("UTF-8");
for more information see this :
Click Here
The HttpServletRequest#getQueryString() returns an URL-encoded query string. You should rather prefer to use getParameter(), getParameterValues() or getParameterMap() methods to retrieve respectively a single parameter, multiple parameters or the entire parameter map. Using those methods, the container will automatically URL-decode the parameter names and values for you. Inside a JSP those values are available by ${param.name}, ${paramValues.name} and ${pageContext.request.parameterMap} respectively.
If you really insist in parsing and fiddling the raw query string yourself for some unobvious reason, then you have to split them on & and = to get the individual parameters and the name=value parts respectively and finally URL-decode each part yourself. Splitting can be done using String#split() and URL-decoding can be done using URLDecoder#decode().
Here's a kickoff example:
String charset = request.getCharacterEncoding();
if (charset == null) charset = "UTF-8";
String queryString = request.getQueryString();
for (String parameter : queryString.split("&")) {
String[] parts = parameter.split("=");
String name = URLDecoder.decode(parts[0], charset);
String value = URLDecoder.decode(parts[1], charset);
// ...
}
The default charset should at least be the same as you've used to return the HTTP response. For example, when it's a JSP, it should be the same as you've specified in the pageEncoding:
<%#page pageEncoding="UTF-8" %>
How can I encode URL query parameter values? I need to replace spaces with %20, accents, non-ASCII characters etc.
I tried to use URLEncoder but it also encodes / character and if I give a string encoded with URLEncoder to the URL constructor I get a MalformedURLException (no protocol).
URLEncoder has a very misleading name. It is according to the Javadocs used encode form parameters using MIME type application/x-www-form-urlencoded.
With this said it can be used to encode e.g., query parameters. For instance if a parameter looks like &/?# its encoded equivalent can be used as:
String url = "http://host.com/?key=" + URLEncoder.encode("&/?#");
Unless you have those special needs the URL javadocs suggests using new URI(..).toURL which performs URI encoding according to RFC2396.
The recommended way to manage the encoding and decoding of URLs is to use URI
The following sample
new URI("http", "host.com", "/path/", "key=| ?/#ä", "fragment").toURL();
produces the result http://host.com/path/?key=%7C%20?/%23ä#fragment. Note how characters such as ?&/ are not encoded.
For further information, see the posts HTTP URL Address Encoding in Java or how to encode URL to avoid special characters in java.
EDIT
Since your input is a string URL, using one of the parameterized constructor of URI will not help you. Neither can you use new URI(strUrl) directly since it doesn't quote URL parameters.
So at this stage we must use a trick to get what you want:
public URL parseUrl(String s) throws Exception {
URL u = new URL(s);
return new URI(
u.getProtocol(),
u.getAuthority(),
u.getPath(),
u.getQuery(),
u.getRef()).
toURL();
}
Before you can use this routine you have to sanitize your string to ensure it represents an absolute URL. I see two approaches to this:
Guessing. Prepend http:// to the string unless it's already present.
Construct the URI from a context using new URL(URL context, String spec)
So what you're saying is that you want to encode part of your URL but not the whole thing. Sounds to me like you'll have to break it up into parts, pass the ones that you want encoded through the encoder, and re-assemble it to get your whole URL.