I want to download a .txt file from website and my code works, so I don't get an error and it loads the document, but the document is full of hmtl code, instead of my content.
public static void main(String[] args) {
try {
URL website = new URL("http://www.file-upload.net/download-11700212/document.txt.html");
String filepath = "C://Users//" + System.getProperty("user.name") + "//Desktop//document.txt";
ReadableByteChannel channel = Channels.newChannel(website.openStream());
FileOutputStream stream = new FileOutputStream(filepath);
stream.getChannel().transferFrom(channel, 0, Long.MAX_VALUE);
System.out.println("Download successfull.");
} catch (Exception e) {
System.out.println("Download was not successfull.");
}
}
The download itself works, I got the txt file on my desktop, but the content is wrong and full of html code.
Please help.
Thanks.
The URL you are trying to download from is an HTML page, rather than the document itself. The link on that page you should be trying to download from is...
http://www.file-upload.net/download5.php?valid=451.69031370715&id=11700212&name=document.txt
However, if you wish to guarantee that you're downloading a text file, then you should choose a text file to download directly e.g.
http://humanstxt.org/humans.txt
I have a Python project called Python Webscraper which can read a URL and copy its textual contents to a text file without the HTML.
You'll need to install a package called Beautiful Soup then run the code from the GitHub repo.
Related
I'm interested in having a button which takes a pic from the user's computer and uploads it on my server.
I managed to solve the server uploading part, but I'm having difficulties in handling the path from the user computer. Vaadin Upload does not provide me full path, but I want it to be dynamic. Looking at the documentation, they use some temp location, but I don't know how to implement that.
public OutputStream receiveUpload(String filename,
String mimeType) {
// Create upload stream
FileOutputStream fos = null; // Stream to write to
try {
// Open the file for writing.
file = new File("/tmp/uploads/" + filename);
fos = new FileOutputStream(file);
} catch (final java.io.FileNotFoundException e) {
new Notification("Could not open file<br/>",
e.getMessage(),
Notification.Type.ERROR_MESSAGE)
.show(Page.getCurrent());
return null;
}
return fos; // Return the output stream to write to
}
I'm expecting that when the File Chooser closes, I get some kind of file path or handler so I can put it on my server.
In the filename argument you will have the name of uploaded file.
The path of the file however is not sent to the server, this is one of the restrictions of web applications / web browsers.
With the code you use, you will have a copy of the uploaded file on your server in the tmp folder.
There is no way to directly access files on client computers through the web browser.
I am trying to display the chm file containing the help which is loaded from resources:
try
{
URL url = this.getClass().getResource("/resources/help.chm");
File file = new File(url.toURI());
Desktop.getDesktop().open(file); //Exception
}
catch (Exception e)
{
e.printStackTrace();
}
When the project is run from NetBeans, the help file is displayed correctly.
Unfortunately, it does not work, when the program is run from the jar file; it leads to an exception.
In my opinion, the internal structure of jar described by URI has not been recognized... Is there any better way? For example, using the BufferReader class?
BufferedReader in = new BufferedReader( new InputStreamReader(url.openStream()));
An analogous problem with the jpg file has been fixed with the BufferedImage class
BufferedImage img = null;
URL url = this.getClass().getResource("/resources/test.jpg");
if (url!= null)
{
img = ImageIO.read(url);
}
without any conversion to URI...
Thanks for your help...
A .jar file is a zip file with a different extension. An entry in a .jar file is not itself a file, and trying to create a File object from a .jar resource URL will never work. Use getResourceAsStream and copy the stream to a temporary file:
Path chmPath = Files.createTempFile(null, ".chm");
try (InputStream chmResource =
getClass().getResourceAsStream("/resources/help.chm")) {
Files.copy(chmResource, chmPath,
StandardCopyOption.REPLACE_EXISTING);
}
Desktop.getDesktop().open(chmPath.toFile());
As an alternative, depending on how simple your help content is, you could just store it as a single HTML file, and pass the resource URL to a non-editable JEditorPane. If you want to have a table of contents, an index, and searching, you might want to consider learning how to use JavaHelp.
I am using org.apache.poi.xwpf.converter.xhtml.XHTMLConverter class to convert docx to html. Below is my groovy code
public Map convert(String wordDocPath, String htmlPath,
Map conversionParams)
{
log.info("Converting word file "+wordDocPath)
try
{
...
String notificationWorkingFolder = "C:\tomcats\Notification\store\Notification1234"
FileInputStream fis = new FileInputStream(wordDocPath);
XWPFDocument document = new XWPFDocument(fis);
XHTMLOptions options = XHTMLOptions.create().URIResolver(new FileURIResolver(new File(notificationWorkingFolder)));
File htmlFile = new File(htmlPath);
OutputStream out = new FileOutputStream(htmlFile)
XHTMLConverter.getInstance().convert(document, out, options);
log.info("Converted to HTML file "+htmlPath)
return [success:true,htmlFileName:getFileName(htmlPath)]
}
catch(Exception e)
{
log.error("Exception :"+e.getMessage(),e)
return [success:false]
}
}
The above code is converting docx to html successfully, but if docx contains any images it puts <img src="C:\tomcats\Notification\store\Notification1234\word\media\image1.png"> but do not copy the image to that folder. As a result, when I open html tag, all images appears empty. Am I missing something in code? Is there a way to generate an image srouce link instead of absolute path, like <img src="http://localhost:8080/webapp/image1.png">
I got answer for first question from this link lychaox.com/java/poi/Word07toHtml.html. I had to add one line of code options.setExtractor(new FileImageExtractor(imageFolderFile)); to generate images.
Second question I resolved by pattern search and replacement.
Even with proper usage, it's worth noting that XHTMLConverter uses XHTMLMapper, which does not process headers, footers, or VML Images. Any images falling into those categories will be lost.
The PDFConverter is more fully featured, but also uses the GPL licensed library, iText.
I'm coding a Spring MVC project on Eclipse. I'm stuck when coding the upload picture function. The client side using HTML5 API to read and send multipart file to server. The following code was used to save the image to server.
#RequestMapping(method = RequestMethod.POST)
public void processUpload(#RequestParam("pic") MultipartFile file) throws IOException {
// if (!result.hasErrors()) {
FileOutputStream outputStream = null;
String filePath = System.getProperty("java.io.tmpdir") + "/" + file.getOriginalFilename();
try {
outputStream = new FileOutputStream(new File(filePath));
outputStream.write(file.getInputStream().read());
outputStream.close();
} catch (Exception e) {
System.out.println("Error while saving file");
The file sent to server and get proceed but the file name is not original file name but some random string that java generated. I found that file in apache-tomcat-6.0.26\work\Catalina\localhost\ with name like this: upload__f20d9c4_1357767c999__7ffe_00000001. The file then disappear.
My question is where the file gone and how to correctly write uploaded file to some folder such as /uploads rather than save to temp folder?
I'm new hear so please correct me if I posted wrong :D
This might help :-
Change the file type to CommonsMultipartFile when you are fetching the file MultipartFile.
file.transferTo(new File("D:/"));
Let me know if this helps.
This may help as well :- http://www.codejava.net/frameworks/spring/spring-mvc-file-upload-tutorial-with-eclipse-ide
I am currently working on an application, where users are given an option to browse and upload excel file, I am badly stuck to get the absolute path of the file being browsed. As location could be anything (Windows/Linux).
import org.apache.myfaces.custom.fileupload.UploadedFile;
-----
-----
private UploadedFile inpFile;
-----
getters and setters
public UploadedFile getInpFile() {
return inpFile;
}
#Override
public void setInpFile(final UploadedFile inpFile) {
this.inpFile = inpFile;
}
we are using jsf 2.0 for UI development and Tomahawk library for browse button.
Sample code for browse button
t:inputFileUpload id="file" value="#{sampleInterface.inpFile}"
valueChangeListener="#{sampleInterface.inpFile}" />
Sample code for upload button
<t:commandButton action="#{sampleInterface.readExcelFile}" id="upload" value="upload"></t:commandButton>
Logic here
Browse button -> user will select the file by browsing the location
Upload button -> on Clicking upload button, it will trigger a method readExcelFile in SampleInterface.
SampleInterface Implementation File
public void readExcelFile() throws IOException {
System.out.println("File name: " + inpFile.getName());
String prefix = FilenameUtils.getBaseName(inpFile.getName());
String suffix = FilenameUtils.getExtension(inpFile.getName());
...rest of the code
......
}
File name : abc.xls
prefix : abc
suffix: xls
Please help me in getting the full path ( as in c:.....) of the file being browsed, this absolute path would then be passed to excelapachepoi class where it will get parsed and contents would be displayed/stored in ArrayList.
Why do you need the absolute file path? What can you do with this information? Creating a File? Sorry no, that is absolutely not possible if the webserver runs at a physically different machine than the webbrowser. Think once again about it. Even more, a proper webbrowser doesn't send information about the absolute file path back.
You just need to create the File based on the uploaded file's content which the client has already sent.
String prefix = FilenameUtils.getBaseName(inpFile.getName());
String suffix = FilenameUtils.getExtension(inpFile.getName());
File file = File.createTempFile(prefix + "-", "." + suffix, "/path/to/uploads");
InputStream input = inpFile.getInputStream();
OutputStream output = new FileOutputStream(file);
try {
IOUtils.copy(input, output);
} finally {
IOUtils.closeQuietly(output);
IOUtils.closeQuietly(input);
}
// Now you can use File.
See also:
How to get the file path from HTML input form in Firefox 3
I remember to have some problem with this in the past too. If I am not mistaken, I think you cannot get the full file path when uploading a file. I think the browser won't tell you it for security purposes.