I am having a textbox and submit button in my jsp page. When submitting this button with some url in textbox, I am getting the response of that url using URLConnection
String strUrl = request.getParameter("url");
URL url = new URL(strUrl);
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
byte[] encodedBytes = Base64.encodeBase64("root:pass".getBytes());
String encoding = new String(encodedBytes);
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
connection.setRequestMethod("GET");
connection.connect();
InputStream content = (InputStream) connection.getInputStream();
BufferedReader in = new BufferedReader(new InputStreamReader(content));
try {
fWriter = new FileWriter(new File("f:\\new.html"));
writer = new BufferedWriter(fWriter);
while ((line = in.readLine()) != null) {
String s = line.toString();
writer.write(s);
}
writer.close();
} catch (Exception e) {
e.printStackTrace();
}
In the resulting html page, every css and js and images were missing as they are pointed to get from local.
for example, js is placed as followed in my generated html page.
<script src="/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
But this actual src is as follows,
<script src="https://www.url.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
I know that there are many solution to replace all src, href with url host. Found many answers related to that.
I used a solution as follows,
if (s.contains(("href="))) {
if (s.contains("\"../") || s.contains("\"/")) {
s = s.replace("\"../", "\"http://" + url.getHost() + "/");
s = s.replace("\"/", "\"http://" + url.getHost() + "/");
writer.write(s);
out.println(s);
}
}
Now I am able to get link,but its not useful in all the web sites. which means that it will helpful for only sites having that kind of host only prefix with src and hrefs.
In some websites, links are defined as href="frmArticles.aspx". In this case its not enough to add host with href url, because href and src are different even though I prefix with host. For example, folowing URL having href links as different than its URL.
http://www.nakkheeran.in/Users/frmMagazine.aspx?M=2
தை தை தை
If, I am adding host to this href it becomes as follows,
தை தை தை
And this is not available. Because, the actual url is
தை தை தை
There are essentially two ways to get the absolute URL:
Using Jsoup's abs:href attribute getter. It works like this:
Element a = myDoc.select("a").first(); //selects tue first link on the page, replace with whatever selector you need to get your link (a element)
String url = a.attr("abs:href"); //gets the absolute url of the link (href attribute)
Note that you need to provide Jsoup with the URL of the HTML document you are using, so it can resolve the URL correctly, this is done automatically if you use Jsoup.connect(myHtmlUrl).get(), if you are parsing HTML from a String or from a file, you need to provide it, use the appropriate Jsoup.parse() method which allows you to provide a base URL
The other way is with Java's built in URL class, which is probably what you should use in your case. You can use it like this:
String absoluteUrl = new URL(new URL("http://example.com/example.html"), "script.js")
Which would print:
http://example.com/script.js
To clarify a bit, the first parameter (in this case example.com) is the url your HTML document is from, and the second parameter ("script.js") is the URL found in your HTML.
In your case, you could use it like:
String absoluteUrl = new URL(new URL("https://www.url.com/"), "/ajax/libs/jquery/2.1.1/jquery.min.js")
Which will print:
https://www.url.com/ajax/libs/jquery/2.1.1/jquery.min.js
The URL class has a constructor URL(URL context, String url) that does what you tried doing with regexps.
Edit: In your case the context URL is the source URL of the parsed resource. Let's say you parse something from URL context = new URL("http://example.com/path/to/some.html#where?is+carmen+sandiego"). Then you just take the reference of any link and create a URL ref = new URL(context, src).
Related
How to normalise a URL in Java to remove the fragment. I.e. from https://www.website.com#something to https://www.website.com
This is possible with the URL.Normalize code, although in this specific use case I've only got a full absolute URL which needs to remain intact.
I'd like to be able to modify this code slightly to remove the fragment from the URL;
//The website below is just an example. In reality, this URL is unknown and could be anything. Both with and without a fragment depending on the use case
URL absUrl = new URL("https://www.website.com#something");
My thoughts so far is that this is only going to be possible by breaking down the URL into the Protocol + Domain + Path then joining it all back together which does appear to work, but there must be a more elegant way of doing this.
Fragment removal is fairly simple using the conversion methods toURI and toURL. So to convert a URL to a URI:
URL url = /*what have you*/ …
URI u = url.toURI();
To remove any fragment from the URI:
if( u.getFragment() != null ) { // Remake with same parts, less the fragment:
u = new URI( u.getScheme(), u.getSchemeSpecificPart(), /*fragment*/null ); }
In reconstructing a URI from its parts like that, it’s important to use the decoded getters (as shown), not the corresponding raw ones. For authority on this usage, see e.g. the Identity section of the API.
To convert the result back to a URL:
url = u.toURL();
Fragments do not exist as a separate entity in Java URLs. But you can convert a URL into a URI and back to remove a fragment. I did it like this:
URL url;
...
if (url.toString().contains("#")) {
URI uri = null;
try {
uri = new URI(url.getProtocol(), url.getHost(), url.getPath(), null);
String file = "";
if (uri.getPath() != null) {
file += uri.getPath();
}
if (uri.getQuery() != null) {
file += uri.getQuery();
}
url = new URL(uri.getScheme(), uri.getHost(), uri.getPort(), file);
} catch (URISyntaxException e) {
...
} catch (MalformedURLException e) {
...
}
}
Here's my problem. I have a txt file called "sites.txt" . In these i type random internet sites. My Goal is to save the first image of each site. I tried to filter the Server response by the img tag and it actually works for some sites, but for some not.
The sites where it works the img src starts with http:// ... the sites it doesnt work start with anything else.
I also tried to add the http:// to the img src images which didnt have it, but i still get the same error:
Exception in thread "main" java.net.MalformedURLException: no protocol:
at java.net.URL.<init>(Unknown Source)
My current code is:
public static void main(String[] args) throws IOException{
try {
File file = new File ("sites.txt");
Scanner scanner = new Scanner (file);
String url;
int counter = 0;
while(scanner.hasNext())
{
url=scanner.nextLine();
URL page = new URL(url);
URLConnection yc = page.openConnection();
BufferedReader in = new BufferedReader(new InputStreamReader(yc.getInputStream()));
String inputLine = in.readLine();
while (!inputLine.toLowerCase().contains("img"))inputLine = in.readLine();
in.close();
String[] parts = inputLine.split(" ");
int i=0;
while(!parts[i].contains("src"))i++;
String destinationFile = "image"+(counter++)+".jpg";
saveImage(parts[i].substring(5,parts[i].length()-1), destinationFile);
String tmp=scanner.nextLine();
System.out.println(url);
}
scanner.close();
}
catch (FileNotFoundException e)
{
System.out.println ("File not found!");
System.exit (0);
}
}
public static void saveImage(String imageUrl, String destinationFile) throws IOException {
// TODO Auto-generated method stub
URL url = new URL(imageUrl);
String fileName = url.getFile();
String destName = fileName.substring(fileName.lastIndexOf("/"));
System.out.println(destName);
InputStream is = url.openStream();
OutputStream os = new FileOutputStream(destinationFile);
byte[] b = new byte[2048];
int length;
while ((length = is.read(b)) != -1) {
os.write(b, 0, length);
}
is.close();
os.close();
}
I also got a tip to use the apache jakarte http client libraries but i got absolutely no idea how i could use those i would appreciate any help.
A URL (a type of URI) requires a scheme in order to be valid. In this case, http.
When you type www.google.com into your browser, the browser is inferring you mean http:// and automatically prepends it for you. Java doesn't do this, hence your exception.
Make sure you always have http://. You can easily fix this using regex:
String fixedUrl = stringUrl.replaceAll("^((?!http://).{7})", "http://$1");
or
if(!stringUrl.startsWith("http://"))
stringUrl = "http://" + stringUrl;
An alternative solution
Simply try with ImageIO that contains static convenience methods for locating ImageReaders and ImageWriters, and performing simple encoding and decoding.
Sample code:
// read a image from the URL
// I used the URL that is your profile pic on StackOverflow
BufferedImage image = ImageIO
.read(new URL(
"https://www.gravatar.com/avatar/3935223a285ab35a1b21f31248f1e721?s=32&d=identicon&r=PG&f=1"));
// save the image
ImageIO.write(image, "jpg", new File("resources/avatar.jpg"));
When you're scraping the site's HTML for image elements and their src attributes, you'll run into several different representations of URLs.
Some examples are:
resource = https://google.com/images/srpr/logo9w.png
resource = google.com/images/srpr/logo9w.png
resource = //google.com/images/srpr/logo9w.png
resource = /images/srpr/logo9w.png
resource = images/srpr/logo9w.png
For the second through fifth ones, you'll need to build the rest of the URL.
The second one may be more difficult to differentiate from the fourth and fifth ones, but I'm sure there are workarounds. The URL Standard leads me to believe you won't see it as often, because I don't think it's technically valid.
The third case is pretty simple. If the resource variable starts with //, then you just need to prepend the protocol/scheme to it. You can do this with the site object you have:
url = site.getProtocol() + ":" + resource
For the fourth and fifth cases, you'll need to prepend the resource with the entire site's URL.
Here's a sample application that uses jsoup to parse the HTML, and a simple utility method to build the resource URL. You're interested in the buildResourceUrl method. Also, it doesn't handle the second case; I'll leave that to you.
import java.io.*;
import java.net.*;
import org.jsoup.*;
import org.jsoup.nodes.*;
import org.jsoup.select.*;
public class SiteScraper {
public static void main(String[] args) throws IOException {
URL site = new URL("https://google.com/");
Document doc = Jsoup.connect(site.toString()).get();
Elements images = doc.select("img");
for (Element image : images) {
String src = image.attr("src");
System.out.println(buildResourceUrl(site, src));
}
}
static URL buildResourceUrl(URL site, String resource)
throws MalformedURLException {
if (!resource.matches("^(http|https|ftp)://.*$")) {
if (resource.startsWith("//")) {
return new URL(site.getProtocol() + ":" + resource);
} else {
return new URL(site.getProtocol() + "://" + site.getHost() + "/"
+ resource.replaceAll("^/", ""));
}
}
return new URL(resource);
}
}
This obviously won't cover everything, but it's a start. You may run into problems when the URL you're trying to access is in a subdirectory of the root of the site (i.e., http://some.place/under/the/rainbow.html). You may even encounter base64 encoded data URI's in the src attribute... It really depends on the individual case and how far you're willing to go.
So I was attempting to use this String in a URL :-
http://site-test.com/Meetings/IC/DownloadDocument?meetingId=c21c905c-8359-4bd6-b864-844709e05754&itemId=a4b724d1-282e-4b36-9d16-d619a807ba67&file=\\s604132shvw140\Test-Documents\c21c905c-8359-4bd6-b864-844709e05754_attachments\7e89c3cb-ce53-4a04-a9ee-1a584e157987\myDoc.pdf
In this code: -
String fileToDownloadLocation = //The above string
URL fileToDownload = new URL(fileToDownloadLocation);
HttpGet httpget = new HttpGet(fileToDownload.toURI());
But at this point I get the error: -
java.net.URISyntaxException: Illegal character in query at index 169:Blahblahblah
I realised with a bit of googling this was due to the characters in the URL (guessing the &), so I then added in some code so it now looks like so: -
String fileToDownloadLocation = //The above string
fileToDownloadLocation = URLEncoder.encode(fileToDownloadLocation, "UTF-8");
URL fileToDownload = new URL(fileToDownloadLocation);
HttpGet httpget = new HttpGet(fileToDownload.toURI());
However, when I try and run this I get an error when I try and create the URL, the error then reads: -
java.net.MalformedURLException: no protocol: http%3A%2F%2Fsite-test.testsite.com%2FMeetings%2FIC%2FDownloadDocument%3FmeetingId%3Dc21c905c-8359-4bd6-b864-844709e05754%26itemId%3Da4b724d1-282e-4b36-9d16-d619a807ba67%26file%3D%5C%5Cs604132shvw140%5CTest-Documents%5Cc21c905c-8359-4bd6-b864-844709e05754_attachments%5C7e89c3cb-ce53-4a04-a9ee-1a584e157987%myDoc.pdf
It looks like I can't do the encoding until after I've created the URL else it replaces slashes and things which it shouldn't, but I can't see how I can create the URL with the string and then format it so its suitable for use. I'm not particularly familiar with all this and was hoping someone might be able to point out to me what I'm missing to get string A into a suitably formatted URL to then use with the correct characters replaced?
Any suggestions greatly appreciated!
You need to encode your parameter's values before concatenating them to URL.
Backslash \ is special character which have to be escaped as %5C
Escaping example:
String paramValue = "param\\with\\backslash";
String yourURLStr = "http://host.com?param=" + java.net.URLEncoder.encode(paramValue, "UTF-8");
java.net.URL url = new java.net.URL(yourURLStr);
The result is http://host.com?param=param%5Cwith%5Cbackslash which is properly formatted url string.
I have the same problem, i read the url with an properties file:
String configFile = System.getenv("system.Environment");
if (configFile == null || "".equalsIgnoreCase(configFile.trim())) {
configFile = "dev.properties";
}
// Load properties
Properties properties = new Properties();
properties.load(getClass().getResourceAsStream("/" + configFile));
//read url from file
apiUrl = properties.getProperty("url").trim();
URL url = new URL(apiUrl);
//throw exception here
URLConnection conn = url.openConnection();
dev.properties
url = "https://myDevServer.com/dev/api/gate"
it should be
dev.properties
url = https://myDevServer.com/dev/api/gate
without "" and my problem is solved.
According to oracle documentation
Thrown to indicate that a malformed URL has occurred. Either no legal protocol could be found in a specification string or the string
could not be parsed.
So it means it is not parsed inside the string.
You want to use URI templates. Look carefully at the README of this project: URLEncoder.encode() does NOT work for URIs.
Let us take your original URL:
http://site-test.test.com/Meetings/IC/DownloadDocument?meetingId=c21c905c-8359-4bd6-b864-844709e05754&itemId=a4b724d1-282e-4b36-9d16-d619a807ba67&file=\s604132shvw140\Test-Documents\c21c905c-8359-4bd6-b864-844709e05754_attachments\7e89c3cb-ce53-4a04-a9ee-1a584e157987\myDoc.pdf
and convert it to a URI template with two variables (on multiple lines for clarity):
http://site-test.test.com/Meetings/IC/DownloadDocument
?meetingId={meetingID}&itemId={itemID}&file={file}
Now let us build a variable map with these three variables using the library mentioned in the link:
final VariableMap = VariableMap.newBuilder()
.addScalarValue("meetingID", "c21c905c-8359-4bd6-b864-844709e05754")
.addScalarValue("itemID", "a4b724d1-282e-4b36-9d16-d619a807ba67e")
.addScalarValue("file", "\\\\s604132shvw140\\Test-Documents"
+ "\\c21c905c-8359-4bd6-b864-844709e05754_attachments"
+ "\\7e89c3cb-ce53-4a04-a9ee-1a584e157987\\myDoc.pdf")
.build();
final URITemplate template
= new URITemplate("http://site-test.test.com/Meetings/IC/DownloadDocument"
+ "meetingId={meetingID}&itemId={itemID}&file={file}");
// Generate URL as a String
final String theURL = template.expand(vars);
This is GUARANTEED to return a fully functional URL!
Thanks to Erhun's answer I finally realised that my JSON mapper was returning the quotation marks around my data too! I needed to use "asText()" instead of "toString()"
It's not an uncommon issue - one's brain doesn't see anything wrong with the correct data, surrounded by quotes!
discoveryJson.path("some_endpoint").toString();
"https://what.the.com/heck"
discoveryJson.path("some_endpoint").asText();
https://what.the.com/heck
This code worked for me
public static void main(String[] args) {
try {
java.net.URL url = new java.net.URL("http://path");
System.out.println("Instantiated new URL: " + url);
}
catch (MalformedURLException e) {
e.printStackTrace();
}
}
Instantiated new URL: http://path
Very simple fix
String encodedURL = UriUtils.encodePath(request.getUrl(), "UTF-8");
Works no extra functionality needed.
I want to extract full link from a HTML file. Full link I mean absolute links. I used Tika for this purpose. Here is my code:
URL url = new URL("http://www.domainname.com/");
InputStream input = url.openStream();
LinkContentHandler linkHandler = new LinkContentHandler();
ContentHandler textHandler = new BodyContentHandler();
ToHTMLContentHandler toHTMLHandler = new ToHTMLContentHandler();
TeeContentHandler teeHandler = new TeeContentHandler(linkHandler,
textHandler, toHTMLHandler);
Metadata metadata = new Metadata();
ParseContext parseContext = new ParseContext();
HtmlParser parser = new HtmlParser();
parser.parse(input, teeHandler, metadata, parseContext);
System.out.println("title:\n" + metadata.get("title"));
for (Link link : linkHandler.getLinks()) {
System.out.println(link.getUri());
}
This give me relative url like /index.html or documents/US/economicreport.html but the absolute url in this case is http://domainname.com/index.html.
How can I get all the link correctly means the full link including domain name? How can I do that in Java?
If you have stored the base website URL in url, the following should work:
URL url = new URL("http://www.domainname.com/");
String givenUrl = ""; //This is the parsed address
if (givenUrl.charAt(0) == '/') {
String absoluteUrl = url + givenURL;
} else {
String absoluteUrl = givenUrl;
}
Slightly better than the previous, but only slightly, is
URL targetDocumentUrl = new URL("http://www.domainname.com/content.html");
String parsedUrl = link.getURI();
String absoluteLink = new URL(targetDocumentUrl, parsedURL);
However, it is still not a good solution as it has problems when the html document has the following tag
base href="/"
and the link being parsed is relative and starts with "../".
Of course you can get around this a number of ways but they involve a bit of work such as implementing a ContentHandler. I have to think for something so basic there must be a simple way to do this with the Tika LinkContentHandler.
I am trying to fetch base URL using java. I have used jtidy parser in my code to fetch the title. I am getting the title properly using jtidy, but I am not getting the base url from the given URL.
I have some URL as input:
String s1 = "http://staff.unak.is/andy/GameProgramming0910/new_page_2.htm";
String s2 = "http://www.complex.com/pop-culture/2011/04/10-hottest-women-in-fast-and-furious-movies";
From the first string, I want to fetch "http://staff.unak.is/andy/GameProgramming0910/" as a base URL and from the second string, I want "http://www.complex.com/" as a base URL.
I am using code:
URL url = new URL(s1);
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
InputStream in = conn.getInputStream();
Document doc = new Tidy().parseDOM(in, null);
String titleText = doc.getElementsByTagName("title").item(0).getFirstChild()
.getNodeValue();
I am getting titletext, but please can let me know how to get base URL from above given URL?
Try to use the java.net.URL class, it will help you:
For the second case, that it is easier, you could use new URL(s2).getHost();
For the first case, you could get the host and also use getFile() method, and remove the string after the last slash ("/"). something like: (code not tested)
URL url = new URL(s1);
String path = url.getFile().substring(0, url.getFile().lastIndexOf('/'));
String base = url.getProtocol() + "://" + url.getHost() + path;
You use the java.net.URL class to resolve relative URLs.
For the first case: removing the filename from the path:
new URL(new URL(s1), ".").toString()
For the second case: setting the root path:
new URL(new URL(s2), "/").toString()