I have a string representing an URL containing spaces and want to convert it to an URI object. If I simply try to create it via
String myString = "http://myhost.com/media/File Name that has spaces inside.mp3";
URI myUri = new URI(myString);
it gives me
java.net.URISyntaxException: Illegal character in path at index X
where index X is the position of the first space in the URL string.
How can i parse myString into a URI object?
You should in fact URI-encode the "invalid" characters. Since the string actually contains the complete URL, it's hard to properly URI-encode it. You don't know which slashes / should be taken into account and which not. You cannot predict that on a raw String beforehand. The problem really needs to be solved at a higher level. Where does that String come from? Is it hardcoded? Then just change it yourself accordingly. Does it come in as user input? Validate it and show error, let the user solve itself.
At any way, if you can ensure that it are only the spaces in URLs which makes it invalid, then you can also just do a string-by-string replace with %20:
URI uri = new URI(string.replace(" ", "%20"));
Or if you can ensure that it's only the part after the last slash which needs to be URI-encoded, then you can also just do so with help of android.net.Uri utility class:
int pos = string.lastIndexOf('/') + 1;
URI uri = new URI(string.substring(0, pos) + Uri.encode(string.substring(pos)));
Do note that URLEncoder is insuitable for the task as it's designed to encode query string parameter names/values as per application/x-www-form-urlencoded rules (as used in HTML forms). See also Java URL encoding of query string parameters.
java.net.URLEncoder.encode(finalPartOfString, "utf-8");
This will URL-encode the string.
finalPartOfString is the part after the last slash - in your case, the name of the song, as it seems.
To handle spaces, #, and other unsafe characters in arbitrary locations in the url path, Use Uri.Builder in combination with a local instance of URL as I have described here:
private Uri.Builder builder;
public Uri getUriFromUrl(String thisUrl) {
URL url = new URL(thisUrl);
builder = new Uri.Builder()
.scheme(url.getProtocol())
.authority(url.getAuthority())
.appendPath(url.getPath());
return builder.build();
}
URL url = Test.class.getResource(args[0]); // reading demo file path from
// same location where class
File input=null;
try {
input = new File(url.toURI());
} catch (URISyntaxException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
I wrote this function:
public static String encode(#NonNull String uriString) {
if (TextUtils.isEmpty(uriString)) {
Assert.fail("Uri string cannot be empty!");
return uriString;
}
// getQueryParameterNames is not exist then cannot iterate on queries
if (Build.VERSION.SDK_INT < 11) {
return uriString;
}
// Check if uri has valid characters
// See https://tools.ietf.org/html/rfc3986
Pattern allowedUrlCharacters = Pattern.compile("([A-Za-z0-9_.~:/?\\#\\[\\]#!$&'()*+,;" +
"=-]|%[0-9a-fA-F]{2})+");
Matcher matcher = allowedUrlCharacters.matcher(uriString);
String validUri = null;
if (matcher.find()) {
validUri = matcher.group();
}
if (TextUtils.isEmpty(validUri) || uriString.length() == validUri.length()) {
return uriString;
}
// The uriString is not encoded. Then recreate the uri and encode it this time
Uri uri = Uri.parse(uriString);
Uri.Builder uriBuilder = new Uri.Builder()
.scheme(uri.getScheme())
.authority(uri.getAuthority());
for (String path : uri.getPathSegments()) {
uriBuilder.appendPath(path);
}
for (String key : uri.getQueryParameterNames()) {
uriBuilder.appendQueryParameter(key, uri.getQueryParameter(key));
}
String correctUrl = uriBuilder.build().toString();
return correctUrl;
}
Related
Say I have a URL
http://example.com/query?q=
and I have a query entered by the user such as:
random word £500 bank $
I want the result to be a properly encoded URL:
http://example.com/query?q=random%20word%20%A3500%20bank%20%24
What's the best way to achieve this? I tried URLEncoder and creating URI/URL objects but none of them come out quite right.
URLEncoder is the way to go. You only need to keep in mind to encode only the individual query string parameter name and/or value, not the entire URL, for sure not the query string parameter separator character & nor the parameter name-value separator character =.
String q = "random word £500 bank $";
String url = "https://example.com?q=" + URLEncoder.encode(q, StandardCharsets.UTF_8);
When you're still not on Java 10 or newer, then use StandardCharsets.UTF_8.toString() as charset argument, or when you're still not on Java 7 or newer, then use "UTF-8".
Note that spaces in query parameters are represented by +, not %20, which is legitimately valid. The %20 is usually to be used to represent spaces in URI itself (the part before the URI-query string separator character ?), not in query string (the part after ?).
Also note that there are three encode() methods. One without Charset as second argument and another with String as second argument which throws a checked exception. The one without Charset argument is deprecated. Never use it and always specify the Charset argument. The javadoc even explicitly recommends to use the UTF-8 encoding, as mandated by RFC3986 and W3C.
All other characters are unsafe and are first converted into one or more bytes using some encoding scheme. Then each byte is represented by the 3-character string "%xy", where xy is the two-digit hexadecimal representation of the byte. The recommended encoding scheme to use is UTF-8. However, for compatibility reasons, if an encoding is not specified, then the default encoding of the platform is used.
See also:
What every web developer must know about URL encoding
I would not use URLEncoder. Besides being incorrectly named (URLEncoder has nothing to do with URLs), inefficient (it uses a StringBuffer instead of Builder and does a couple of other things that are slow) Its also way too easy to screw it up.
Instead I would use URIBuilder or Spring's org.springframework.web.util.UriUtils.encodeQuery or Commons Apache HttpClient.
The reason being you have to escape the query parameters name (ie BalusC's answer q) differently than the parameter value.
The only downside to the above (that I found out painfully) is that URL's are not a true subset of URI's.
Sample code:
import org.apache.http.client.utils.URIBuilder;
URIBuilder ub = new URIBuilder("http://example.com/query");
ub.addParameter("q", "random word £500 bank \$");
String url = ub.toString();
// Result: http://example.com/query?q=random+word+%C2%A3500+bank+%24
You need to first create a URI like:
String urlStr = "http://www.example.com/CEREC® Materials & Accessories/IPS Empress® CAD.pdf"
URL url = new URL(urlStr);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
Then convert that URI to an ASCII string:
urlStr = uri.toASCIIString();
Now your URL string is completely encoded. First we did simple URL encoding and then we converted it to an ASCII string to make sure no character outside US-ASCII remained in the string. This is exactly how browsers do it.
Guava 15 has now added a set of straightforward URL escapers.
The code
URL url = new URL("http://example.com/query?q=random word £500 bank $");
URI uri = new URI(url.getProtocol(), url.getUserInfo(), IDN.toASCII(url.getHost()), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
String correctEncodedURL = uri.toASCIIString();
System.out.println(correctEncodedURL);
Prints
http://example.com/query?q=random%20word%20%C2%A3500%20bank%20$
What is happening here?
1. Split URL into structural parts. Use java.net.URL for it.
2. Encode each structural part properly!
3. Use IDN.toASCII(putDomainNameHere) to Punycode encode the hostname!
4. Use java.net.URI.toASCIIString() to percent-encode, NFC encoded Unicode - (better would be NFKC!). For more information, see: How to encode properly this URL
In some cases it is advisable to check if the URL is already encoded. Also replace '+' encoded spaces with '%20' encoded spaces.
Here are some examples that will also work properly
{
"in" : "http://نامهای.com/",
"out" : "http://xn--mgba3gch31f.com/"
},{
"in" : "http://www.example.com/‥/foo",
"out" : "http://www.example.com/%E2%80%A5/foo"
},{
"in" : "http://search.barnesandnoble.com/booksearch/first book.pdf",
"out" : "http://search.barnesandnoble.com/booksearch/first%20book.pdf"
}, {
"in" : "http://example.com/query?q=random word £500 bank $",
"out" : "http://example.com/query?q=random%20word%20%C2%A3500%20bank%20$"
}
The solution passes around 100 of the test cases provided by Web Platform Tests.
Using Spring's UriComponentsBuilder:
UriComponentsBuilder
.fromUriString(url)
.build()
.encode()
.toUri()
The Apache HttpComponents library provides a neat option for building and encoding query parameters.
With HttpComponents 4.x use:
URLEncodedUtils
For HttpClient 3.x use:
EncodingUtil
Here's a method you can use in your code to convert a URL string and map of parameters to a valid encoded URL string containing the query parameters.
String addQueryStringToUrlString(String url, final Map<Object, Object> parameters) throws UnsupportedEncodingException {
if (parameters == null) {
return url;
}
for (Map.Entry<Object, Object> parameter : parameters.entrySet()) {
final String encodedKey = URLEncoder.encode(parameter.getKey().toString(), "UTF-8");
final String encodedValue = URLEncoder.encode(parameter.getValue().toString(), "UTF-8");
if (!url.contains("?")) {
url += "?" + encodedKey + "=" + encodedValue;
} else {
url += "&" + encodedKey + "=" + encodedValue;
}
}
return url;
}
In Android, I would use this code:
Uri myUI = Uri.parse("http://example.com/query").buildUpon().appendQueryParameter("q", "random word A3500 bank 24").build();
Where Uri is a android.net.Uri
In my case I just needed to pass the whole URL and encode only the value of each parameters.
I didn't find common code to do that, so (!!) so I created this small method to do the job:
public static String encodeUrl(String url) throws Exception {
if (url == null || !url.contains("?")) {
return url;
}
List<String> list = new ArrayList<>();
String rootUrl = url.split("\\?")[0] + "?";
String paramsUrl = url.replace(rootUrl, "");
List<String> paramsUrlList = Arrays.asList(paramsUrl.split("&"));
for (String param : paramsUrlList) {
if (param.contains("=")) {
String key = param.split("=")[0];
String value = param.replace(key + "=", "");
list.add(key + "=" + URLEncoder.encode(value, "UTF-8"));
}
else {
list.add(param);
}
}
return rootUrl + StringUtils.join(list, "&");
}
public static String decodeUrl(String url) throws Exception {
return URLDecoder.decode(url, "UTF-8");
}
It uses Apache Commons' org.apache.commons.lang3.StringUtils.
Use this：
URLEncoder.encode(query, StandardCharsets.UTF_8.displayName());
or this:
URLEncoder.encode(query, "UTF-8");
You can use the following code.
String encodedUrl1 = UriUtils.encodeQuery(query, "UTF-8"); // No change
String encodedUrl2 = URLEncoder.encode(query, "UTF-8"); // Changed
String encodedUrl3 = URLEncoder.encode(query, StandardCharsets.UTF_8.displayName()); // Changed
System.out.println("url1 " + encodedUrl1 + "\n" + "url2=" + encodedUrl2 + "\n" + "url3=" + encodedUrl3);
Say I have a URL
http://example.com/query?q=
and I have a query entered by the user such as:
random word £500 bank $
I want the result to be a properly encoded URL:
http://example.com/query?q=random%20word%20%A3500%20bank%20%24
What's the best way to achieve this? I tried URLEncoder and creating URI/URL objects but none of them come out quite right.
URLEncoder is the way to go. You only need to keep in mind to encode only the individual query string parameter name and/or value, not the entire URL, for sure not the query string parameter separator character & nor the parameter name-value separator character =.
String q = "random word £500 bank $";
String url = "https://example.com?q=" + URLEncoder.encode(q, StandardCharsets.UTF_8);
When you're still not on Java 10 or newer, then use StandardCharsets.UTF_8.toString() as charset argument, or when you're still not on Java 7 or newer, then use "UTF-8".
Note that spaces in query parameters are represented by +, not %20, which is legitimately valid. The %20 is usually to be used to represent spaces in URI itself (the part before the URI-query string separator character ?), not in query string (the part after ?).
Also note that there are three encode() methods. One without Charset as second argument and another with String as second argument which throws a checked exception. The one without Charset argument is deprecated. Never use it and always specify the Charset argument. The javadoc even explicitly recommends to use the UTF-8 encoding, as mandated by RFC3986 and W3C.
All other characters are unsafe and are first converted into one or more bytes using some encoding scheme. Then each byte is represented by the 3-character string "%xy", where xy is the two-digit hexadecimal representation of the byte. The recommended encoding scheme to use is UTF-8. However, for compatibility reasons, if an encoding is not specified, then the default encoding of the platform is used.
See also:
What every web developer must know about URL encoding
I would not use URLEncoder. Besides being incorrectly named (URLEncoder has nothing to do with URLs), inefficient (it uses a StringBuffer instead of Builder and does a couple of other things that are slow) Its also way too easy to screw it up.
Instead I would use URIBuilder or Spring's org.springframework.web.util.UriUtils.encodeQuery or Commons Apache HttpClient.
The reason being you have to escape the query parameters name (ie BalusC's answer q) differently than the parameter value.
The only downside to the above (that I found out painfully) is that URL's are not a true subset of URI's.
Sample code:
import org.apache.http.client.utils.URIBuilder;
URIBuilder ub = new URIBuilder("http://example.com/query");
ub.addParameter("q", "random word £500 bank \$");
String url = ub.toString();
// Result: http://example.com/query?q=random+word+%C2%A3500+bank+%24
You need to first create a URI like:
String urlStr = "http://www.example.com/CEREC® Materials & Accessories/IPS Empress® CAD.pdf"
URL url = new URL(urlStr);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
Then convert that URI to an ASCII string:
urlStr = uri.toASCIIString();
Now your URL string is completely encoded. First we did simple URL encoding and then we converted it to an ASCII string to make sure no character outside US-ASCII remained in the string. This is exactly how browsers do it.
Guava 15 has now added a set of straightforward URL escapers.
The code
URL url = new URL("http://example.com/query?q=random word £500 bank $");
URI uri = new URI(url.getProtocol(), url.getUserInfo(), IDN.toASCII(url.getHost()), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
String correctEncodedURL = uri.toASCIIString();
System.out.println(correctEncodedURL);
Prints
http://example.com/query?q=random%20word%20%C2%A3500%20bank%20$
What is happening here?
1. Split URL into structural parts. Use java.net.URL for it.
2. Encode each structural part properly!
3. Use IDN.toASCII(putDomainNameHere) to Punycode encode the hostname!
4. Use java.net.URI.toASCIIString() to percent-encode, NFC encoded Unicode - (better would be NFKC!). For more information, see: How to encode properly this URL
In some cases it is advisable to check if the URL is already encoded. Also replace '+' encoded spaces with '%20' encoded spaces.
Here are some examples that will also work properly
{
"in" : "http://نامهای.com/",
"out" : "http://xn--mgba3gch31f.com/"
},{
"in" : "http://www.example.com/‥/foo",
"out" : "http://www.example.com/%E2%80%A5/foo"
},{
"in" : "http://search.barnesandnoble.com/booksearch/first book.pdf",
"out" : "http://search.barnesandnoble.com/booksearch/first%20book.pdf"
}, {
"in" : "http://example.com/query?q=random word £500 bank $",
"out" : "http://example.com/query?q=random%20word%20%C2%A3500%20bank%20$"
}
The solution passes around 100 of the test cases provided by Web Platform Tests.
Using Spring's UriComponentsBuilder:
UriComponentsBuilder
.fromUriString(url)
.build()
.encode()
.toUri()
The Apache HttpComponents library provides a neat option for building and encoding query parameters.
With HttpComponents 4.x use:
URLEncodedUtils
For HttpClient 3.x use:
EncodingUtil
Here's a method you can use in your code to convert a URL string and map of parameters to a valid encoded URL string containing the query parameters.
String addQueryStringToUrlString(String url, final Map<Object, Object> parameters) throws UnsupportedEncodingException {
if (parameters == null) {
return url;
}
for (Map.Entry<Object, Object> parameter : parameters.entrySet()) {
final String encodedKey = URLEncoder.encode(parameter.getKey().toString(), "UTF-8");
final String encodedValue = URLEncoder.encode(parameter.getValue().toString(), "UTF-8");
if (!url.contains("?")) {
url += "?" + encodedKey + "=" + encodedValue;
} else {
url += "&" + encodedKey + "=" + encodedValue;
}
}
return url;
}
In Android, I would use this code:
Uri myUI = Uri.parse("http://example.com/query").buildUpon().appendQueryParameter("q", "random word A3500 bank 24").build();
Where Uri is a android.net.Uri
In my case I just needed to pass the whole URL and encode only the value of each parameters.
I didn't find common code to do that, so (!!) so I created this small method to do the job:
public static String encodeUrl(String url) throws Exception {
if (url == null || !url.contains("?")) {
return url;
}
List<String> list = new ArrayList<>();
String rootUrl = url.split("\\?")[0] + "?";
String paramsUrl = url.replace(rootUrl, "");
List<String> paramsUrlList = Arrays.asList(paramsUrl.split("&"));
for (String param : paramsUrlList) {
if (param.contains("=")) {
String key = param.split("=")[0];
String value = param.replace(key + "=", "");
list.add(key + "=" + URLEncoder.encode(value, "UTF-8"));
}
else {
list.add(param);
}
}
return rootUrl + StringUtils.join(list, "&");
}
public static String decodeUrl(String url) throws Exception {
return URLDecoder.decode(url, "UTF-8");
}
It uses Apache Commons' org.apache.commons.lang3.StringUtils.
Use this：
URLEncoder.encode(query, StandardCharsets.UTF_8.displayName());
or this:
URLEncoder.encode(query, "UTF-8");
You can use the following code.
String encodedUrl1 = UriUtils.encodeQuery(query, "UTF-8"); // No change
String encodedUrl2 = URLEncoder.encode(query, "UTF-8"); // Changed
String encodedUrl3 = URLEncoder.encode(query, StandardCharsets.UTF_8.displayName()); // Changed
System.out.println("url1 " + encodedUrl1 + "\n" + "url2=" + encodedUrl2 + "\n" + "url3=" + encodedUrl3);
public static String getMimeType(Context context, Uri uri) {
String extension; //Check uri format to avoid null
if (uri.getScheme().equals(ContentResolver.SCHEME_CONTENT)) {
//If scheme is a content
final MimeTypeMap mime = MimeTypeMap.getSingleton();
extension = mime.getExtensionFromMimeType(context.getContentResolver().getType(uri));
} else {
//If scheme is a File
//This will replace white spaces with %20 and also other special characters. This will avoid returning null values on file name with spaces and special characters.
extension = MimeTypeMap.getFileExtensionFromUrl(Uri.fromFile(new File(uri.getPath())).toString());
}
return extension;
}
I tried this not working
I have some URLs' in string form, and from these URLs' I want to generate a URI using java.net.URI.
These URLs' are actually hyperlinks in an Android Webview:
clc://C# or clc://C++
final URI u = new URI(newURL);
final String sScheme = u.getScheme();
final String sHost = u.getHost();
final String sPath = u.getPath();
But in the above code, if a URL has # or + then getHost() returns null.
I tried to encode the URL as follows, but it doesn't work:
String encodedUrl = URLEncoder.encode(url, "UTF-8");
I also tried putting %23 for #, then too it doesn`t work.
Please help me to resolve this.....
URLEncoder doesn't always provide the correct output, especially when URIs' are involved.
Try the following approach instead:
Uri u = Uri.parse(newURL)
.buildUpon()
.appendQueryParameter("param", param)
.build();
String url = u.toString();
where param is a web service parameter (if using any). This will encode the URL in UTF-8 format correctly. Then,
final String sScheme = u.getScheme( );
final String sHost = u.getHost( );
final String sPath = u.getPath( );
It will work as expected.
I'm trying to write some JUnit tests for a Java method that takes a base URL and target URL and returns the target URL relative to the given base URL.
I'm using the category based partition to make my test set. Currently i'm testing to check the following:
check the two input URL's have the
same protocol and host;
check for
when the paths aren't the same and
that the relative URL adjusts
correctly;
check for when the base
URL is longer than the target URL;
check for when the target URL is
longer than the base URL;
check for
when the base URL and target URL are
identical;
I was wondering how other people would test this method using JUnit? Am i missing any criteria?
/**
* This method converts an absolute url to an url relative to a given base-url.
* The algorithm is somewhat chaotic, but it works (Maybe rewrite it).
* Be careful, the method is ".mm"-specific. Something like this should be included
* in the librarys, but I couldn't find it. You can create a new absolute url with
* "new URL(URL context, URL relative)".
*/
public static String toRelativeURL(URL base, URL target) {
// Precondition: If URL is a path to folder, then it must end with '/' character.
if( (base.getProtocol().equals(target.getProtocol())) &&
(base.getHost().equals(target.getHost()))) {
String baseString = base.getFile();
String targetString = target.getFile();
String result = "";
//remove filename from URL
baseString = baseString.substring(0, baseString.lastIndexOf("/")+1);
//remove filename from URL
targetString = targetString.substring(0, targetString.lastIndexOf("/")+1);
StringTokenizer baseTokens = new StringTokenizer(baseString,"/");//Maybe this causes problems under windows
StringTokenizer targetTokens = new StringTokenizer(targetString,"/");//Maybe this causes problems under windows
String nextBaseToken = "", nextTargetToken = "";
//Algorithm
while(baseTokens.hasMoreTokens() && targetTokens.hasMoreTokens()) {
nextBaseToken = baseTokens.nextToken();
nextTargetToken = targetTokens.nextToken();
System.out.println("while1");
if (!(nextBaseToken.equals(nextTargetToken))) {
System.out.println("if1");
while(true) {
result = result.concat("../");
System.out.println(result);
if (!baseTokens.hasMoreTokens()) {
System.out.println("break1");
break;
}
System.out.println("break2");
nextBaseToken = baseTokens.nextToken();
}
while(true) {
result = result.concat(nextTargetToken+"/");
System.out.println(result);
if (!targetTokens.hasMoreTokens()) {
System.out.println("break3");
break;
}
System.out.println("break4");
nextTargetToken = targetTokens.nextToken();
}
String temp = target.getFile();
result = result.concat(temp.substring(temp.lastIndexOf("/")+1,temp.length()));
System.out.println("1");
return result;
}
}
while(baseTokens.hasMoreTokens()) {
result = result.concat("../");
baseTokens.nextToken();
}
while(targetTokens.hasMoreTokens()) {
nextTargetToken = targetTokens.nextToken();
result = result.concat(nextTargetToken + "/");
}
String temp = target.getFile();
result = result.concat(temp.substring(temp.lastIndexOf("/")+1,temp.length()));
System.out.println("2");
return result;
}
System.out.println("3");
return target.toString();
}
}
Just a few thoughts...
You might want to test if either (or both) your URL input is null. :)
If the target URL has parameters (ex: http://host/app/bla?param1=value¶m2=value), does the generated relative URL contain the parameters?
If the target URL is just http://host, will it cause IndexOutOfBoundException on targetString.lastIndexOf("/")... the same applies to base URL.