how to check protocol is present in URL , if not present need to append it.
is there any class to achieve this in java?
eg: String URL = www.google.com
need to get http://www.google.com
Just use String.startsWith("http://") to check this.
public String ensure_has_protocol(final String a_url)
{
if (!a_url.startsWith("http://"))
{
return "http://" + a_url;
}
return a_url;
}
EDIT:
An alternative would use a java.net.URL instance, whose constructor would throw an java.net.MalformedURLException if the URL did not contain a (legal) protocol (or was invalid for any other reason):
public URL make_url(final String a_url) throws MalformedURLException
{
try
{
return new URL(a_url);
}
catch (final MalformedURLException e)
{
}
return new URL("http://" + a_url);
}
You can use URL.toString() to obtain string representation of the URL. This is an improvement on the startsWith() approach as it guarantees that return URL is valid.
Let's say you have String url = www.google.com. String class methods would be enough for the goal of checking protocol identifiers. For example, url.startsWith("https://") would check whether a specific string is starting with the given protocol name.
However, are these controls enough for validation?
I think they aren't enough. First of all, you should define a list of valid protocol identifiers, e.g. a String array like {"http", "ftp", "https", ...}. Then you can parse your input String with regex ("://") and test your URL header whether it belongs to the list of valid protocol identifiers. And domain name validation methods are beyond this question, you can/should handle it with different techniques as well.
Just for completeness, I would do something like the following:
import com.google.common.base.Strings;
private static boolean isUrlHttps(String url){
if(Strings.isNullOrEmpty(url))
return false;
return url.toLowerCase().startsWith("https://");
}
Related
I need to convert a list of URLS to their host name. SO tried the below mentioned code:
URL netUrl = new URL(url);
String host = netUrl.getHost();
The above mentioned code is producing output as shown below:
a95-101-128-242.deploy.akamaitechnologies.com
a23-1-242-192.deploy.static.akamaitechnologies.com
edge-video-shv-01-lht6.fbcdn.net
I want only the website name from the above output like as shown below:
akamaitechnologies
akamaitechnologies
fbcdn
Please someone help.
Thanks
If you want to parse a URL, use java.net.URI. java.net.URL has a bunch of problems -- its equals method does a DNS lookup which means code using it can be vulnerable to denial of service attacks when used with untrusted inputs.
public static String getDomainName(String url) throws URISyntaxException {
URI uri = new URI(url);
String domain = uri.getHost();
return domain.startsWith("www.") ? domain.substring(4) : domain;
}
This should work.
I have a text field to acquire location information (String type) from User. It could be file directory based (e.g. C:\directory) or Web url (e.g. http://localhost:8008/resouces). The system will read some predetermined metadata file from the location.
Given the input string, how can I detect the nature of the path location whether it is a file based or Web URL effectively.
So far I have tried.
URL url = new URL(location); // will get MalformedURLException if it is a file based.
url.getProtocol().equalsIgnoreCase("http");
File file = new File(location); // will not hit exception if it is a url.
file.exist(); // return false if it is a url.
I am still struggling to find a best way to tackle both scenarios. :-(
Basically I would not prefer to explicitly check the path using the prefix such as http:// or https://
Is there an elegant and proper way of doing this?
You can check if the location starts with http:// or https://:
String s = location.trim().toLowerCase();
boolean isWeb = s.startsWith("http://") || s.startsWith("https://");
Or you can use the URI class instead of URL, URI does not throw MalformedURLException like the URL class:
URI u = new URI(location);
boolean isWeb = "http".equalsIgnoreCase(u.getScheme())
|| "https".equalsIgnoreCase(u.getScheme())
Although new URI() may also throw URISyntaxException if you use backslash in location for example. Best way would be to either use prefix check (my first suggestion) or create a URL and catch MalformedURLException which if thrown you'll know it cannot be a valid web url.
If you're open to the use of a try/catch scenario being "elegant", here is a way that is more specific:
try {
processURL(new URL(location));
}
catch (MalformedURLException ex){
File file = new File(location);
if (file.exists()) {
processFile(file);
}
else {
throw new PersonalException("Can't find the file");
}
}
This way, you're getting the automatic URL syntax checking and, that failing, the check for file existence.
you can try:
static public boolean isValidURL(String urlStr) {
try {
URI uri = new URI(urlStr);
return uri.getScheme().equals("http") || uri.getScheme().equals("https");
}
catch (Exception e) {
return false;
}
}
note that this will return false for any other reason that invalidates the url, ofor a non http/https url: a malformed url is not necessarily an actual file name, and a good file name can be referring to a non exisiting one, so use it in conjunction with you file existence check.
public boolean urlIsFile(String input) {
if (input.startsWith("file:")) return true;
try { return new File(input).exists(); } catch (Exception e) {return false;}
}
This is the best method because it is hassle free, and will always return true if you have a file reference. For instance, other solutions don't and cannot cover the plethora of protocol schemes available such as ftp, sftp, scp, or any future protocol implementations. So this one is the one for all uses and purposes; with the caveat of the file must exist, if it doesn't begin with the file protocol.
if you look at the logic of the function by it's name, you should understand that, returning false for a non existent direct path lookup is not a bug, that is the fact.
I am writing something like an image proxy where I receive URLs from my site front-end, and I download images , re-size them, and return smaller images for the front end and client to download from the "proxy".
This means I need to take care of all-sorts of url patterns, this is why I chose to decode the given url and than encode it using URIUtils.decode:
private String fixUrl(String fromUrl) throws URIException {
fromUrl = URIUtil.decode(fromUrl);
fromUrl = URIUtil.encodeQuery(fromUrl);
return fromUrl;
}
This should help me take care of urls that are already encoded.
My problem is that some of the urls are double encoded, and from what I saw, URIUtils.decode, performs recursive decode and this means that in cases of double encoded urls I will get a bad url that does not work.
Is there a simple way to decode only once?
I'd try to check that URL still contains character %. If it does not contain any % it is not encoded and you can stop your decoding procedure.
Easiest option I know is to use the built-in
java.net.URLDecoder.decode
In order to decode automatically only once.
If like me you have the case that sometimes URL's are double / triple encoded - you can use this recursive function in order to decode again and again until there are no "%" or "+" :
private static String completeDecode(String url) {
if(url.contains("%") || url.contains("+"))
{
try
{
return(completeDecode(java.net.URLDecoder.decode(url, "UTF-8")));
}
catch (UnsupportedEncodingException e)
{
e.printStackTrace();
}
}
return url;
}
Cheers
I wanted to know if there is any standard APIs in Java to validate a given URL?
I want to check both if the URL string is right i.e. the given protocol is valid and then to check if a connection can be established.
I tried using HttpURLConnection, providing the URL and connecting to it. The first part of my requirement seems to be fulfilled but when I try to perform HttpURLConnection.connect(), 'java.net.ConnectException: Connection refused' exception is thrown.
Can this be because of proxy settings? I tried setting the System properties for proxy but no success.
Let me know what I am doing wrong.
For the benefit of the community, since this thread is top on Google when searching for
"url validator java"
Catching exceptions is expensive, and should be avoided when possible. If you just want to verify your String is a valid URL, you can use the UrlValidator class from the Apache Commons Validator project.
For example:
String[] schemes = {"http","https"}; // DEFAULT schemes = "http", "https", "ftp"
UrlValidator urlValidator = new UrlValidator(schemes);
if (urlValidator.isValid("ftp://foo.bar.com/")) {
System.out.println("URL is valid");
} else {
System.out.println("URL is invalid");
}
The java.net.URL class is in fact not at all a good way of validating URLs. MalformedURLException is not thrown on all malformed URLs during construction. Catching IOException on java.net.URL#openConnection().connect() does not validate URL either, only tell wether or not the connection can be established.
Consider this piece of code:
try {
new URL("http://.com");
new URL("http://com.");
new URL("http:// ");
new URL("ftp://::::#example.com");
} catch (MalformedURLException malformedURLException) {
malformedURLException.printStackTrace();
}
..which does not throw any exceptions.
I recommend using some validation API implemented using a context free grammar, or in very simplified validation just use regular expressions. However I need someone to suggest a superior or standard API for this, I only recently started searching for it myself.
Note
It has been suggested that URL#toURI() in combination with handling of the exception java.net. URISyntaxException can facilitate validation of URLs. However, this method only catches one of the very simple cases above.
The conclusion is that there is no standard java URL parser to validate URLs.
You need to create both a URL object and a URLConnection object. The following code will test both the format of the URL and whether a connection can be established:
try {
URL url = new URL("http://www.yoursite.com/");
URLConnection conn = url.openConnection();
conn.connect();
} catch (MalformedURLException e) {
// the URL is not in a valid form
} catch (IOException e) {
// the connection couldn't be established
}
Using only standard API, pass the string to a URL object then convert it to a URI object. This will accurately determine the validity of the URL according to the RFC2396 standard.
Example:
public boolean isValidURL(String url) {
try {
new URL(url).toURI();
} catch (MalformedURLException | URISyntaxException e) {
return false;
}
return true;
}
Use the android.webkit.URLUtil on android:
URLUtil.isValidUrl(URL_STRING);
Note: It is just checking the initial scheme of URL, not that the entire URL is valid.
There is a way to perform URL validation in strict accordance to standards in Java without resorting to third-party libraries:
boolean isValidURL(String url) {
try {
new URI(url).parseServerAuthority();
return true;
} catch (URISyntaxException e) {
return false;
}
}
The constructor of URI checks that url is a valid URI, and the call to parseServerAuthority ensures that it is a URL (absolute or relative) and not a URN.
Just important to point that the URL object handle both validation and connection. Then, only protocols for which a handler has been provided in sun.net.www.protocol are authorized (file,
ftp, gopher, http, https, jar, mailto, netdoc) are valid ones. For instance, try to make a new URL with the ldap protocol:
new URL("ldap://myhost:389")
You will get a java.net.MalformedURLException: unknown protocol: ldap.
You need to implement your own handler and register it through URL.setURLStreamHandlerFactory(). Quite overkill if you just want to validate the URL syntax, a regexp seems to be a simpler solution.
Are you sure you're using the correct proxy as system properties?
Also if you are using 1.5 or 1.6 you could pass a java.net.Proxy instance to the openConnection() method. This is more elegant imo:
//Proxy instance, proxy ip = 10.0.0.1 with port 8080
Proxy proxy = new Proxy(Proxy.Type.HTTP, new InetSocketAddress("10.0.0.1", 8080));
conn = new URL(urlString).openConnection(proxy);
I think the best response is from the user #b1nary.atr0phy. Somehow, I recommend combine the method from the b1nay.atr0phy response with a regex to cover all the possible cases.
public static final URL validateURL(String url, Logger logger) {
URL u = null;
try {
Pattern regex = Pattern.compile("(?i)^(?:(?:https?|ftp)://)(?:\\S+(?::\\S*)?#)?(?:(?!(?:10|127)(?:\\.\\d{1,3}){3})(?!(?:169\\.254|192\\.168)(?:\\.\\d{1,3}){2})(?!172\\.(?:1[6-9]|2\\d|3[0-1])(?:\\.\\d{1,3}){2})(?:[1-9]\\d?|1\\d\\d|2[01]\\d|22[0-3])(?:\\.(?:1?\\d{1,2}|2[0-4]\\d|25[0-5])){2}(?:\\.(?:[1-9]\\d?|1\\d\\d|2[0-4]\\d|25[0-4]))|(?:(?:[a-z\\u00a1-\\uffff0-9]-*)*[a-z\\u00a1-\\uffff0-9]+)(?:\\.(?:[a-z\\u00a1-\\uffff0-9]-*)*[a-z\\u00a1-\\uffff0-9]+)*(?:\\.(?:[a-z\\u00a1-\\uffff]{2,}))\\.?)(?::\\d{2,5})?(?:[/?#]\\S*)?$");
Matcher matcher = regex.matcher(url);
if(!matcher.find()) {
throw new URISyntaxException(url, "La url no está formada correctamente.");
}
u = new URL(url);
u.toURI();
} catch (MalformedURLException e) {
logger.error("La url no está formada correctamente.");
} catch (URISyntaxException e) {
logger.error("La url no está formada correctamente.");
}
return u;
}
This is what I use to validate CDN urls (must start with https, but that's easy to customise). This will also not allow using IP addresses.
public static final boolean validateURL(String url) {
var regex = Pattern.compile("^[https:\\/\\/(www\\.)?a-zA-Z0-9#:%._\\+~#=]{2,256}\\.[a-z]{2,6}\\b([-a-zA-Z0-9#:%_\\+.~#?&//=]*)");
var matcher = regex.matcher(url);
return matcher.find();
}
Thanks. Opening the URL connection by passing the Proxy as suggested by NickDK works fine.
//Proxy instance, proxy ip = 10.0.0.1 with port 8080
Proxy proxy = new Proxy(Proxy.Type.HTTP, new InetSocketAddress("10.0.0.1", 8080));
conn = new URL(urlString).openConnection(proxy);
System properties however doesn't work as I had mentioned earlier.
Thanks again.
Regards,
Keya
I am trying to get a java.net.URI object from a String. The string has some characters which will need to be replaced by their percentage escape sequences. But when I use URLEncoder to encode the String with UTF-8 encoding, even the / are replaced with their escape sequences.
How can I get a valid encoded URL from a String object?
http://www.google.com?q=a b gives http%3A%2F%2www.google.com... whereas I want the output to be http://www.google.com?q=a%20b
Can someone please tell me how to achieve this.
I am trying to do this in an Android app. So I have access to a limited number of libraries.
You might try: org.apache.commons.httpclient.util.URIUtil.encodeQuery in Apache commons-httpclient project
Like this (see URIUtil):
URIUtil.encodeQuery("http://www.google.com?q=a b")
will become:
http://www.google.com?q=a%20b
You can of course do it yourself, but URI parsing can get pretty messy...
Android has always had the Uri class as part of the SDK:
http://developer.android.com/reference/android/net/Uri.html
You can simply do something like:
String requestURL = String.format("http://www.example.com/?a=%s&b=%s", Uri.encode("foo bar"), Uri.encode("100% fubar'd"));
I'm going to add one suggestion here aimed at Android users. You can do this which avoids having to get any external libraries. Also, all the search/replace characters solutions suggested in some of the answers above are perilous and should be avoided.
Give this a try:
String urlStr = "http://abc.dev.domain.com/0007AC/ads/800x480 15sec h.264.mp4";
URL url = new URL(urlStr);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
url = uri.toURL();
You can see that in this particular URL, I need to have those spaces encoded so that I can use it for a request.
This takes advantage of a couple features available to you in Android classes. First, the URL class can break a url into its proper components so there is no need for you to do any string search/replace work. Secondly, this approach takes advantage of the URI class feature of properly escaping components when you construct a URI via components rather than from a single string.
The beauty of this approach is that you can take any valid url string and have it work without needing any special knowledge of it yourself.
Even if this is an old post with an already accepted answer, I post my alternative answer because it works well for the present issue and it seems nobody mentioned this method.
With the java.net.URI library:
URI uri = URI.create(URLString);
And if you want a URL-formatted string corresponding to it:
String validURLString = uri.toASCIIString();
Unlike many other methods (e.g. java.net.URLEncoder) this one replaces only unsafe ASCII characters (like ç, é...).
In the above example, if URLString is the following String:
"http://www.domain.com/façon+word"
the resulting validURLString will be:
"http://www.domain.com/fa%C3%A7on+word"
which is a well-formatted URL.
If you don't like libraries, how about this?
Note that you should not use this function on the whole URL, instead you should use this on the components...e.g. just the "a b" component, as you build up the URL - otherwise the computer won't know what characters are supposed to have a special meaning and which ones are supposed to have a literal meaning.
/** Converts a string into something you can safely insert into a URL. */
public static String encodeURIcomponent(String s)
{
StringBuilder o = new StringBuilder();
for (char ch : s.toCharArray()) {
if (isUnsafe(ch)) {
o.append('%');
o.append(toHex(ch / 16));
o.append(toHex(ch % 16));
}
else o.append(ch);
}
return o.toString();
}
private static char toHex(int ch)
{
return (char)(ch < 10 ? '0' + ch : 'A' + ch - 10);
}
private static boolean isUnsafe(char ch)
{
if (ch > 128 || ch < 0)
return true;
return " %$&+,/:;=?#<>#%".indexOf(ch) >= 0;
}
You can use the multi-argument constructors of the URI class. From the URI javadoc:
The multi-argument constructors quote illegal characters as required by the components in which they appear. The percent character ('%') is always quoted by these constructors. Any other characters are preserved.
So if you use
URI uri = new URI("http", "www.google.com?q=a b");
Then you get http:www.google.com?q=a%20b which isn't quite right, but it's a little closer.
If you know that your string will not have URL fragments (e.g. http://example.com/page#anchor), then you can use the following code to get what you want:
String s = "http://www.google.com?q=a b";
String[] parts = s.split(":",2);
URI uri = new URI(parts[0], parts[1], null);
To be safe, you should scan the string for # characters, but this should get you started.
I had similar problems for one of my projects to create a URI object from a string. I couldn't find any clean solution either. Here's what I came up with :
public static URI encodeURL(String url) throws MalformedURLException, URISyntaxException
{
URI uriFormatted = null;
URL urlLink = new URL(url);
uriFormatted = new URI("http", urlLink.getHost(), urlLink.getPath(), urlLink.getQuery(), urlLink.getRef());
return uriFormatted;
}
You can use the following URI constructor instead to specify a port if needed:
URI uri = new URI(scheme, userInfo, host, port, path, query, fragment);
Well I tried using
String converted = URLDecoder.decode("toconvert","UTF-8");
I hope this is what you were actually looking for?
The java.net blog had a class the other day that might have done what you want (but it is down right now so I cannot check).
This code here could probably be modified to do what you want:
http://svn.apache.org/repos/asf/incubator/shindig/trunk/java/common/src/main/java/org/apache/shindig/common/uri/UriBuilder.java
Here is the one I was thinking of from java.net: https://urlencodedquerystring.dev.java.net/
Or perhaps you could use this class:
http://developer.android.com/reference/java/net/URLEncoder.html
Which is present in Android since API level 1.
Annoyingly however, it treats spaces specially (replacing them with + instead of %20). To get round this we simply use this fragment:
URLEncoder.encode(value, "UTF-8").replace("+", "%20");
I ended up using the httpclient-4.3.6:
import org.apache.http.client.utils.URIBuilder;
public static void main (String [] args) {
URIBuilder uri = new URIBuilder();
uri.setScheme("http")
.setHost("www.example.com")
.setPath("/somepage.php")
.setParameter("username", "Hello Günter")
.setParameter("p1", "parameter 1");
System.out.println(uri.toString());
}
Output will be:
http://www.example.com/somepage.php?username=Hello+G%C3%BCnter&p1=paramter+1