So I'm making a program that extracts the lyrics from a user given song off of AZ Lyrics.
The problem I'm having is that after converting the string to a URL, it says Jsoup is not able to parse it because it doesn't accept strings despite the variable being a URL that we are passing in.
String strURL = "http://www.azlyrics.com/lyrics/" + artist + "/" + song + ".html";
URL url = new URL(strURL);
Document doc = Jsoup.parse(url);
What should I do?
I dont know which version of jsoup your are using, but as per latest version the parse method with url alone is not available. You need to pass a timeOut. So try
Document doc = Jsoup.parse(url, 30000);
There is a connect method which would be the best option (IMO). You could pass the stringURL variable directly. Try
Document doc = Jsoup.connect(strURL).get();
If these didn't help check the value of artist and song variables.
You don't need to convert the string strURL to URL, this should work:
Document doc = Jsoup.connect("http://www.azlyrics.com/lyrics/" + artist + "/" + song + ".html").timeout(10000).get();
String html = doc.text();
I've set a timeout of 10 seconds, adjust to fit your needs.
You can take a look at the available methods here
Related
I'm automating my test scenario for validation of a pdf document. This document opens in a new browser tab once clicked on the document link(anchor tag). I want to validate a few important contents in a document for which I'm using Apache PDFBox. But, the document URL has a prefix 'blob' because of which, java.net.URL class is throwing MalformedException for unknown protocol: blob. how should I define/add that protocol in java?
Please let me know how to get rid of this error so that I can successfully use PDFBox to parse my pdf file.
Java version - 1.8
This is the screenshot of pdf document after it opens in a browser.
This is HTML source of document. But, as it's a pdf view, cannot perform any operations such as fetching text/windowTitle etc.
following is a sample code snippet -
public void readPdfContents() throws IOException {
String url = "blob:https://cpswebqa.testcbidata.com/f9ad63bc-700e-4f49-a4fb-807ad1a44b01";
URL pdfUrl = new URL(url);
InputStream ips = pdfUrl.openStream();
BufferedInputStream bis = new BufferedInputStream(ips);
PDFParser pdfParser = new PDFParser(bis);
pdfParser.parse();
String pdfData = new PDFTextStripper().getText(pdfParser.getPDDocument());
System.out.println("PDF Data is - " + pdfData);
}
Error stack trace -
Exception in thread "main" java.net.MalformedURLException: unknown protocol: blob
at java.net.URL.<init>(URL.java:600)
at java.net.URL.<init>(URL.java:490)
at java.net.URL.<init>(URL.java:439)
at com.cbsh.automation.file.testrunner.WEB.Sample.main(Sample.java:11)
I got the same problem and found a solution injecting Javascript like in here:
How to download an image with Python 3/Selenium if the URL begins with “blob:”?
I wrote in Java and it worked very well, here is the code:
private String getBytesBase64FromBlobURI(ChromeDriver driver, String uri) {
String script = " "
+ "var uri = arguments[0];"
+ "var callback = arguments[1];"
+ "var toBase64 = function(buffer){for(var r,n=new Uint8Array(buffer),t=n.length,a=new Uint8Array(4*Math.ceil(t/3)),i=new Uint8Array(64),o=0,c=0;64>c;++c)i[c]='ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'.charCodeAt(c);for(c=0;t-t%3>c;c+=3,o+=4)r=n[c]<<16|n[c+1]<<8|n[c+2],a[o]=i[r>>18],a[o+1]=i[r>>12&63],a[o+2]=i[r>>6&63],a[o+3]=i[63&r];return t%3===1?(r=n[t-1],a[o]=i[r>>2],a[o+1]=i[r<<4&63],a[o+2]=61,a[o+3]=61):t%3===2&&(r=(n[t-2]<<8)+n[t-1],a[o]=i[r>>10],a[o+1]=i[r>>4&63],a[o+2]=i[r<<2&63],a[o+3]=61),new TextDecoder('ascii').decode(a)};"
+ "var xhr = new XMLHttpRequest();"
+ "xhr.responseType = 'arraybuffer';"
+ "xhr.onload = function(){ callback(toBase64(xhr.response)) };"
+ "xhr.onerror = function(){ callback(xhr.status) };"
+ "xhr.open('GET','"+ uri +"');"
+ "xhr.send();";
String result = (String) driver.executeAsyncScript(script, uri);
return result;
}
I hope it help someone.
Cheers!
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).
I tried to make a image links downloader with jsoup. I have made a downloader HTML code part, and when I have done a parse part, I recognized, that sometimes links to images appeared without main part. So I found absUrl solution, but by some reasons it did not work (it gave me null). So I tried use uri.resolve(), but it gave me unchanged result. So now I do not know how to solve it. I attached part of my code, that responsible for parsing ant writing url to string:
public static String finalcode(String textin) throws Exception {
String text = source(textin);
Document doc = Jsoup.parse(text);
Elements images = doc.getElementsByTag("img");
String Simages = images.toString();
int Limages = countLines(Simages);
StringBuilder src = new StringBuilder();
while (Limages > 0) {
Limages--;
Element image = images.get(Limages);
String href = image.attr("src");
src.append(href);
src.append("\n");
}
String result = src.toString();
return result;
}
It looks like you are parsing HTML from String, not from URL. Because of that jsoup can't know from which URL this HTML codes comes from, so it can't create absolute path.
To set this URL for Document you should parse it using Jsoup.parse(String html, String baseUri) version, like
String url = "http://server/pages/document.htlm";
String text = "<img src = '../images/image_name1.jpg'/><img src = '../images/image_name2.jpg'/>'";
Document doc = Jsoup.parse(text, url);
Elements images = doc.getElementsByTag("img");
for (Element image : images){
System.out.println(image.attr("src")+" -> "+image.attr("abs:src"));
}
Output:
../images/image_name1.jpg -> http://server/images/image_name1.jpg
../images/image_name2.jpg -> http://server/images/image_name2.jpg
Other option would be letting Jsoup parse page directly by supplying URL instead of String with HTML
Document doc = Jsoup.connect("http://example.com").get();
This way Document will know from which URL it came, so it will be able to create absolute paths.
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 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()