I have looked online with mixed results, but is there a way to programmatically extract a zip file on the BB? Very basic my app will display different encrypted file types, and those files are delivered in a zip file. My idea was to have the user browse to the file on their SDCard, select it, and I extract what i need as a stream from the file. is this possible?
Use GZIPInputStream
Example:
try
{
InputStream inputStream = httpConnection.openInputStream();
GZIPInputStream gzis = new GZIPInputStream(inputStream);
StringBuffer sb = new StringBuffer();
char c;
while ((c = (char)gzis.read()) != -1)
{
sb.append(c);
}
String data = sb.toString();
gzis.close();
}
catch(IOException ioe)
{
}
Just two things:
In BB API there are only GZip and ZLib support, and no multiple files compression support, so it's not possible to compress several files and extract only one of them.
Up to my experience, such functionality will fly on simulator, but may be really performance-killing on real device
See How to retrieve data from a attached zip file in Blackberry application?
PS Actually you can implement custom multi-entries stream and parse it after decompress, but that seems to be useless, if you want this archive format to be supported in other applications.
Related
I'm using the Upload component in a Vaadin8 project to get a file up on the server, as shown in the source code on this page:
https://demo.vaadin.com/sampler/#ui/data-input/other/upload
After I choose the file on my pc and click upload, the window opens up just like in the sampler, and the progress bar goes all the way to the end, but the file is nowhere to be found in the project file system. Is there another step I'm supposed to be doing? How to I configure the destination folder for the uploaded files?
Taken from official documentation here Receiving upload:
The uploaded files are typically stored as files in a file system, in a database, or as temporary objects in memory. The upload component writes the received data to an java.io.OutputStream so you have plenty of freedom in how you can process the upload content.
So in your case, an uploaded file is stored as a temporary object. In V8 documentation example is cut-off, but it's presented in V7: Receiving Upload Data
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
}
public void uploadSucceeded(SucceededEvent event) {
// Show the uploaded file in the image viewer
image.setVisible(true);
image.setSource(new FileResource(file));
}
So the idea is that you could create a file yourself, once upload has succeed.
I'm working with VS2015 and ASP.Net on a webservice application which is installed in the AWS cloud.
In one of my methods i got two files, a PDF and a XML.
These files just exist as instances of type MemoryStream.
Now i have to compress these two "files" in a ZIP file before adding the zip as attachment to an E-mail (class MailMessage).
It seems that i have to save the memorystreams to files before adding them as entries to the zip.
Is ist true or do i have another possibility to add the streams as entries to the zip?
Thanks in advance!
The answer is no.
It is not necessary to save the files before adding them to the stream for the ZIP file.
I have found a solution with the Nuget package DotNetZip.
Here is a code example how to use it.
In that example there two files which only exist in MemoryStream objects, not on a local disc.
It is important to reset the Position property of the streams to zero before adding them to the ZIP stream.
At last i save the ZIP stream as a file in my local folder to control the results.
//DotNetZip from Nuget
//http://shahvaibhav.com/create-zip-file-in-memory-using-dotnetzip/
string zipFileName = System.IO.Path.GetFileNameWithoutExtension(xmlFileName) + ".zip";
var zipMemStream = new MemoryStream();
zipMemStream.Position = 0;
using (Ionic.Zip.ZipFile zip = new Ionic.Zip.ZipFile())
{
textFileStream.Position = 0;
zip.AddEntry(System.IO.Path.GetFileNameWithoutExtension(xmlFileName) + ".txt", textFileStream);
xmlFileStream.Position = 0;
zip.AddEntry(xmlFileName, xmlFileStream);
zip.Save(zipMemStream);
// Try to save the ZIP-Stream as a ZIP file. And suddenly: It works!
var zipFs = new FileStream(zipFileName, FileMode.Create);
zipMemStream.Position = 0;
zipMemStream.CopyTo(zipFs);
zipMemStream.WriteTo(zipFs);
}
I am facing an issue while exporting japanese text in CSV format. Junk characters are being exported instead of original japanese text. I am using .NET MVC FileStreamResult to export records in Csv file and used encoding format as UTF8 (I have also used some other encoding format, but no luck). I debugged my code and able to convert string from memory stream and vice versa and able to see original japanese text being exported. Once exporting completed, I opened the CSV file, but only able to see junk character instead of expected text. If I open the CSV file in NotePad ( Opening the csv file in Notepad is NOT my requirement. I am referring Notepad only to verify whether i am able to see Japanese translated language ), then i can see the expected japanese text. It would be really helpful if someone please help me find root cause of this issue and provide a resolution.
Ex. 東京都品川区大崎 gets written as æ±äº¬éƒ½å“å·åŒºå¤§å´Ž
Note: I can see expected japanese text is exported properly if I opened the sample .CSV file using LibreOffice Calc, Linux default gEdit. But the issue is with opening this csv file using MS Office.
Please find the below attached code -
Controller/Action to execute while clicking on export to Csv button
================================================================================
[HttpPost]
[ValidateInput(false)]
public FileStreamResult SaveCustomerInfo()
{
return ExportToCsv();
}
================================================================================
private static FileStreamResult ExportToCsv()
{
var exportedData = new StringBuilder();
exportedData
.AppendLine("実行日,口座番号,支店番号,アカウント名,支店名,の/受益秩序,ステートメント日,入力日,お問い合わせ番号, ,Date Range")
.Append(
"CS0001,Demo FName,Demo LName,8/20/2015,\"Demo User Address\",City,Country,08830,0123456789,15813,Absolute from 8/20/2015 to 8/22/2015");
var stream = PrintingHelper.StringToMemoryStream(Encoding.UTF8, exportedData.ToString());
var fileStreamResult = new FileStreamResult(stream, "text/csv")
{
FileDownloadName =
new StringBuilder("TestExportedFileInCsv")
.Append(".csv").ToString()
};
return fileStreamResult;
}
It sound as though you haven't installed the language pack for MS Office on the machine that you are trying to open the csv on.
I am using the Export plugin in Grails for generating PDF/Excel report. I am able to generate single report on PDF/Excel button click. But, now I want to generate multiple reports on single button click. I tried for loop, method calls but no luck.
Reference links are ok. I don't expect entire code, needs reference only.
If you take a look at the source code for the ExportService in the plugin you will notice there are various export methods. Two of which support OutputStreams. Using either of these methods (depending on your requirements for the other parameters) will allow you to render a report to an output stream. Then using those output streams you can create a zip file which you can deliver to the HTTP client.
Here is a very rough example, which was written off the top of my head so it's really just an idea rather than working code:
// Assumes you have a list of maps.
// Each map will have two keys.
// outputStream and fileName
List files = []
// call the exportService and populate your list of files
// ByteArrayOutputStream outputStream = new ByteArrayOutputStream()
// exportService.export('pdf', outputStream, ...)
// files.add([outputStream: outputStream, fileName: 'whatever.pdf'])
// ByteArrayOutputStream outputStream2 = new ByteArrayOutputStream()
// exportService.export('pdf', outputStream2, ...)
// files.add([outputStream: outputStream2, fileName: 'another.pdf'])
// create a tempoary file for the zip file
File tempZipFile = File.createTempFile("temp", "zip")
ZipOutputStream out = new ZipOutputStream(new FileOutputStream(tempZipFile))
// set the compression ratio
out.setLevel(Deflater.BEST_SPEED);
// Iterate through the list of files adding them to the ZIP file
files.each { file ->
// Associate an input stream for the current file
ByteArrayInputStream input = new ByteArrayInputStream(file.outputStream.toByteArray())
// Add ZIP entry to output stream.
out.putNextEntry(new ZipEntry(file.fileName))
// Transfer bytes from the current file to the ZIP file
org.apache.commons.io.IOUtils.copy(input, out);
// Close the current entry
out.closeEntry()
// Close the current input stream
input.close()
}
// close the ZIP file
out.close()
// next you need to deliver the zip file to the HTTP client
response.setContentType("application/zip")
response.setHeader("Content-disposition", "attachment;filename=WhateverFilename.zip")
org.apache.commons.io.IOUtils.copy((new FileInputStream(tempZipFile), response.outputStream)
response.outputStream.flush()
response.outputStream.close()
That should give you an idea of how to approach this. Again, the above is just for demonstration purposes and isn't production ready code, nor have I even attempted to compile it.
I'm new to search engines and web crawlers. Now I want to store all the original pages in a particular web site as html files, but with Apache Nutch I can only get the binary database files. How do I get the original html files with Nutch?
Does Nutch support it? If not, what other tools can I use to achieve my goal.(The tools that support distributed crawling are better.)
Well, nutch will write the crawled data in binary form so if if you want that to be saved in html format, you will have to modify the code. (this will be painful if you are new to nutch).
If you want quick and easy solution for getting html pages:
If the list of pages/urls that you intend to have is quite low, then better get it done with a script which invokes wget for each url.
OR use HTTrack tool.
EDIT:
Writing a your own nutch plugin will be great. Your problem will get solved plus you can contribute to nutch by submitting your work !!! If you are new to nutch (in terms of code & design), then you will have to invest lot of time building a new plugin ... else its easy to do.
Few pointers for helping your initiative:
Here is a page which talks about writing own nutch plugin.
Start with Fetcher.java. See lines 647-648. That is the place where you can get the fetched content on per url basis (for those pages which got fetched successfully).
pstatus = output(fit.url, fit.datum, content, status, CrawlDatum.STATUS_FETCH_SUCCESS);
updateStatus(content.getContent().length);
You should add code right after this to invoke your plugin. Pass content object to it. By now, you would have guessed that content.getContent() is the content for url you want. Inside the plugin code, write it to some file. Filename should be based on the url name else it will be difficult to work with that. Url can be obtained by fit.url.
You must do modifications in run Nutch in Eclipse.
When you are able to run, open Fetcher.java and add the lines between "content saver" command lines.
case ProtocolStatus.SUCCESS: // got a page
pstatus = output(fit.url, fit.datum, content, status, CrawlDatum.STATUS_FETCH_SUCCESS, fit.outlinkDepth);
updateStatus(content.getContent().length);'
//------------------------------------------- content saver ---------------------------------------------\\
String filename = "savedsites//" + content.getUrl().replace('/', '-');
File file = new File(filename);
file.getParentFile().mkdirs();
boolean exist = file.createNewFile();
if (!exist) {
System.out.println("File exists.");
} else {
FileWriter fstream = new FileWriter(file);
BufferedWriter out = new BufferedWriter(fstream);
out.write(content.toString().substring(content.toString().indexOf("<!DOCTYPE html")));
out.close();
System.out.println("File created successfully.");
}
//------------------------------------------- content saver ---------------------------------------------\\
To update this answer -
It is possible to post process the data from your crawldb segment folder, and read in the html (including other data nutch has stored) directly.
Configuration conf = NutchConfiguration.create();
FileSystem fs = FileSystem.get(conf);
Path file = new Path(segment, Content.DIR_NAME + "/part-00000/data");
SequenceFile.Reader reader = new SequenceFile.Reader(fs, file, conf);
try
{
Text key = new Text();
Content content = new Content();
while (reader.next(key, content))
{
System.out.println(new String(content.GetContent()));
}
}
catch (Exception e)
{
}
The answers here are obsolete. Now, it is simply possible to get the plain HTML-files with nutch dump. Please see this answer.
In apache Nutch 2.3.1
You can save the raw HTML by edit the Nutch code firstly run the nutch in eclipse by following https://wiki.apache.org/nutch/RunNutchInEclipse
After you finish ruunning nutch in eclipse edit file FetcherReducer.java , add this code to the output method, run ant eclipse again to rebuild the class
Finally the raw html will added to reportUrl column in your database
if (content != null) {
ByteBuffer raw = fit.page.getContent();
if (raw != null) {
ByteArrayInputStream arrayInputStream = new ByteArrayInputStream(raw.array(), raw.arrayOffset() + raw.position(), raw.remaining());
Scanner scanner = new Scanner(arrayInputStream);
scanner.useDelimiter("\\Z");//To read all scanner content in one String
String data = "";
if (scanner.hasNext()) {
data = scanner.next();
}
fit.page.setReprUrl(StringUtil.cleanField(data));
scanner.close();
}