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How to download monthly Treasury Files

Up till early this year the US Treasury web site posted monthly US Receipts and Outlays data in txt format. It was easy to write a program to read and store the info. All I use were:
URL url = new URL("https://www.fiscal.treasury.gov/fsreports/rpt/mthTreasStmt/mts1214.txt")
URLConnection connection.openConnection();
InputStream is = connection.getInputStream();
Then I just read the InputStream into a local file.
Now when I try same code, for May, I get an InputStream with nothing in it.
Just clicking on "https://www.fiscal.treasury.gov/fsreports/rpt/mthTreasStmt/mts0415.xlsx" opens an excel worksheet (the download path has since changed).
Which is great if you don't mind clicking on each link separately ... saving the file somewhere ... opening it manually to enable editing ... then saving it again as a real .xlsx file (because they really hand you an .xls file.)
But when I create a URL from that link, and use it to get an InputStream, the is empty. I also tried url.openStream() directly. No different.
Can anyone see a way I can resume using a program to read the new format?
In case its of interest I now use this code to write the stream to the file bit by bit... but there are no bits, so I don't know if it works.
static void copyInputStreamToFile( InputStream in, File file ) {
try {
OutputStream out = new FileOutputStream(file);
byte[] buf = new byte[1024];
System.out.println("reading: " + in.read(buf));
//This is what tells me it is empty, i.e. the loop below is ignored.
int len;
while((len=in.read(buf))>0){
out.write(buf,0,len);
}
out.close();
in.close();
} catch (Exception e) {
e.printStackTrace();
}
}
Any help is appreciated.

Creating Zip file while client is downloading

I try to develop something like dropbox(very basic one). For one file to download, it's really easy. Just use servletoutputstream. what i want is: when client asks me multiple file, i zip files in server side then send to user. But if file is big it takes too many times to zip them and send to user.
is there any way to send files while they are compressing?
thanks for your help.

Part of the Java API for ZIP files is actually desgined to provide "on the fly" compression. It all fits nicely both with the java.io API and the servlet API, which means this is even... kind of easy (no multithreading required - even for performance reason, because usually your CPU will probably be faster at ZIPping than your network will be at sending contents).
The part you'll be interacting with is ZipOutputStream. It is a FilterOutputStream (which means it is designed to wrap an outputstream that already exists - in your case, that would be the respone's OutputStream), and will compress every byte you send it, using ZIP compression.
So, say you have a get request
protected void doGet(HttpServletRequest req, HttpServletResponse resp)
throws ServletException, IOException {
// Your code to handle the request
List<YourFileObject> responseFiles = ... // Whatever you need to do
// We declare that the response will contain raw bytes
response.setContentType("application/octet-stream");
// We open a ZIP output stream
try (ZipOutputStream zipStream = new ZipOutputStream(response.getOutputStream()) {// This is Java 7, but not that different from java 6
// We need to loop over each files you want to send
for(YourFileObject fileToSend : responseFiles) {
// We give a name to the file
zipStream.putNextEntry(new ZipEntry(fileToSend.getName()));
// and we copy its content
copy(fileToSend, zipStream);
}
}
}
Of course, you should do proper exception handling. A couple quick notes though :
The ZIP file format mandates that each file has a name, so you must create a new ZipEntry each time you start a new file (you'll probably get an IllegalStateException if you do not, anyway)
Proper use of the API would be that you close each entry once you are done writing to it (at the end of the file). BUT : the Java implementation does that for you : each time you call putNextEntry it closes the previous one (if need be) all by itself
Likewise, you must not forget to close the ZIP stream, beacuse, this will properly close the last entry AND flush everything that is needed to create a proper ZIP file. Failure to do so will result in a corrupt file. Here, the try with resources statement does this : it closes the ZipOutputStream once everything is written to it.
The copy method here is just what you would use to transfert all the bytes from the original file to the outputstream, there is nothing ZIP specific about it. Just call outputStream.write(byte[] bytes).
**EDIT : ** to clarify...
For example, given a YourFileType that has the following methods :
public interface YourFileType {
public byte[] getContent();
public InputStream getContentAsStream();
}
Then the copy method could look like (this is all very basic Java IO, you could maybe use a library such as commons io to not reinvent the wheel...)
public void copy(YourFileType file, OutputStream os) throws IOException {
os.write(file.getContent());
}
Or, for a full streaming implementation :
public void copy(YourFileType file, OutputStream os) throws IOException {
try (InputStream fileContent = file.getContentAsStream()) {
byte[] buffer = new byte[4096]; // 4096 is kind of a magic number
int readBytesCount = 0;
while((readBytesCount = fileContent.read(buffer)) >= 0) {
os.write(buffer, 0, readBytesCount);
}
}
}
Using this kind of implementation, your client will start receiveing a response almost as soon as you start writing to the ZIPOutputStream (the only delay would be that of internal buffers), meaning it should not timeout (unless you spent too long buliding the content to send - but that would not be the ZIPping part fault's).

Receiving files over socket

I am implementing a Direct Connect client. I am using the NMDC protocol. I can connect to a hub and other connected clients. I am trying to retrieve the file list from each client, I understand that in order to do that one must download the file files.xml.bz2 from the other client. The protocol to download a file is as follows:
-> $ADCGET file <filename> <params>|
<- $ADCSND file <fileName> <params>|
<- (*** binary data is now transfered from client B to client A ***)
I am trying to create a file named files.xml.bz2 using the binary data received. Here's my code:
//filesize is provided through the $ADCSND response from other client
byte[] data = new byte[filesize];
/*
Reading binary data from socket inputstream
*/
int read = 0;
for (int i=0; read<filesize;){
int available = in2.available();
int leftspace = filesize-read;
if (available>0){
in2.read(data, read, available>leftspace? leftspace:available);
++i;
}
read += (available>leftspace? leftspace:available)+1;
}
/*
writing the bytes to an actual file
*/
ByteArrayInputStream f = new ByteArrayInputStream(data);
FileOutputStream file = new FileOutputStream("files.xml.bz2");
file.write(data);
file.close();
The file is created, however, the contents (files.xml) are not readable. Opening it in firefox gives:
XML Parsing Error: not well-formed
Viewing the contents in the terminal only reads binary data. What am i doing wrong?
EDIT
I also tried Decompressing the file using the bz2 libray from Apache Ant.
ByteArrayInputStream f = new ByteArrayInputStream(data);
BZip2CompressorInputStream bzstream = new BZip2CompressorInputStream(f);
FileOutputStream xmlFile = new FileOutputStream("files.xml");
byte[] bytes = new byte[1024];
while((bzstream.read(bytes))!=-1){
xmlFile.write(bytes);
}
xmlFile.close();
bzstream.close();
I get an error, here's the stacktrace:
java.io.IOException: Stream is not in the BZip2 format
at org.apache.commons.compress.compressors.bzip2.BZip2CompressorInputStream.init(BZip2CompressorInputStream.java:240)
at org.apache.commons.compress.compressors.bzip2.BZip2CompressorInputStream.<init>(BZip2CompressorInputStream.java:132)
at org.apache.commons.compress.compressors.bzip2.BZip2CompressorInputStream.<init>(BZip2CompressorInputStream.java:109)
at control.Controller$1.run(Controller.java:196)

Usual, typical misuse of available(). All you need to copy a stream in Java is as follows:
while ((count = in.read(buffer)) >= 0)
{
out.write(buffer, 0, count);
}
Use this with any size buffer greater than zero, but preferably several kilobytes. You don't need a new buffer per iteration, and you don't need to know how much data is available to read without blocking, as you have to block, otherwise you're just smoking the CPU. But you do need to know how much data was actually read per iteration, and this is the first place where your code falls down.

The error java.io.IOException: Stream is not in the BZip2 format is generated by the constructor of class BZip2CompressorInputStream. I decided to scan the bytes, looking for the magic number to make sure that the file was bz2 format, it turns out that Java was right -- it wasnt in bz2 format.
Upon examining the source code of Jucy, I saw that the reason for this was a slight error in the command I sent to the other client, in essence, this error was caused a mistake in my protocol implementation. The solution was:
Replace:
$ADCGET file files.xml.bz2 0 -1 ZL1|
With:
$ADCGET file files.xml.bz2 0 -1|
ZL1 specifies compression of the files being sent (Not necessary).

Problem sending XML via HTTP

I want to have an application which parses various RSS feeds and send the information to a remote server. The information is sent in xml format via http. At first I tried to deploy this application on my own server, so I send the xml using the method shown in this tutorial by Java Tips. Here is my code which is replicated from the example:
First Method
String strURL = "http://localhost/readme/readme_xml";
String strXMLFilename = "output.xml";
File input = new File(strXMLFilename);
PostMethod post = new PostMethod(strURL);
post.setRequestEntity(new InputStreamRequestEntity(
new FileInputStream(input), input.length()));
post.setRequestHeader(
"Content-type", "text/xml; charset=ISO-8859-1");
HttpClient httpclient = new HttpClient();
try {
int result = httpclient.executeMethod(post);
System.out.println("Response status code: " + result);
System.out.println("Response body: ");
System.out.println(post.getResponseBodyAsString());
} finally {
post.releaseConnection();
}
This works perfectly (I even tested using a remote server outside the localhost). Then, somehow I cant use my own server to deploy this application, so I decided to migrate to Google Apps Engine. One thing about it, as we know it, is that not all libraries are allowed in the environment. So I try another method shown in ExampleDepot.com (I can't find where the exact url though) as below:
Second Method
try {
/* fill up this url with the remote server url */
URL url = new URL("http://localhost/readme/readme_xml");
FileReader fr = new FileReader("output.xml");
char[] buffer = new char[1024*10];
int len = 0;
if ((len = fr.read(buffer)) != -1){
/* send http request to remote server */
URLConnection conn = url.openConnection();
conn.setRequestProperty("Content-Type","text/xml;charset=ISO-8859-1"); /* need to specify the content type */
conn.setDoOutput(true);
conn.setDoOutput(true);
PrintWriter pw = new PrintWriter(conn.getOutputStream());
pw.write(buffer, 0, len);
pw.flush();
/* receive response from remote server*/
BufferedReader bf = new BufferedReader(new InputStreamReader(conn.getInputStream()));
String input = null;
while ((input = bf.readLine()) != null){
System.out.println(input);
}
}
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
The second method though, doesn't work and gives the following error (I use SimpleXMLElement (php) object to parse xml in the remote hosting):
Error message from remote server
Here's the php code from the remote server (In here, I just want the SimpleXMLElement to parse the xml without doing anything else fancy for now)
$xml = new SimpleXMLElement('php://input', NULL, TRUE);
foreach ($xml -> attributes() as $name => $val){
echo "[".$name."] = ".$val."\n";
}
I thought the cause of this problem is the malfunction xml file (because the eclipse IDE indicates there's error of "invalid byte 1 of 1-byte utf-8 sequence"). Then I use the same exact input xml file to the first method, but it still works perfectly.
So is there any adjustment that I need to make to the second method? Or is there any other method that I can use to send xml file to remote server? Let me know if I need to add some other details. Thanks for your help.
NOTE: I actually solved this problem by using the solution given in the comments. I didn't use approaches suggested in the answers, even though those answers are pretty useful. So, I didn't select the best answer out of those answers given. Nonetheless, I still appreciate all of your helps, thus deserve my upvote. Cheers!

I guess you need to change the content type to multipart/form-data. See an already answered question in detailed. The file upload is discussed at the bottom of this example

I would, as the first answer suggest, read the file with an InputStream. Converting from byte to char and back again is unnecessary and a source of error. Also, verify that the input file really is using the ISO-8859-1 encoding.
UPDATE:
When using a FileReader, you accept the default encoding (i.e. how to make chars from bytes). This encoding must match the encoding used for the input file, otherwise there's a great risk that the result is corrupted. The default Java encoding is different for different platforms, so it is generally not a good idea to rely on it.
In your second example, there's no reason to read the file as characters, since it will be sent on the wire as bytes anyway. Using byte streams all the way also avoids the encoding issue (apart from the information in the content-type header).

never read a file as chars unless you are reading a text file. xml is not text, it is a binary format. copy the file using normal InputStreams and byte[]s.
also, as #beny23 suggested in his comment, make sure you always copy streams using a loop, not a single read() (even if your buffer is big enough, it is not guaranteed that the InputStream will give you all the bytes in one call, even for a FileInputStream).

Writing to CSV Files and then Zipping it up in Appengine (Java)

I'm currently working on a project that is done in Java, on google appengine.
Appengine does not allow files to be stored so any on-disk representation objects cannot be used. Some of these include the File class.
I want to write data and export it to a few csv files, and then zip it up, and allow the user to download it.
How may I do this without using any File classes? I'm not very experienced in file handling so I hope you guys can advise me.
Thanks.

You can create a zip file and add to it while the user is downloading it. If you are using a servlet, this is straigthforward:
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// ..... process request
// ..... then respond
response.setContentType("application/zip");
response.setStatus(HttpServletResponse.SC_OK);
// note : intentionally no content-length set, automatic chunked transfer if stream is larger than the internal buffer of the response
ZipOutputStream zipOut = new ZipOutputStream(response.getOutputStream());
byte[] buffer = new byte[1024 * 32];
try {
// case1: already have input stream, typically ByteArrayInputStream from a byte[] full of previoiusly prepared csv data
InputStream in = new BufferedInputStream(getMyFirstInputStream());
try {
zipOut.putNextEntry(new ZipEntry("FirstName"));
int length;
while((length = in.read(buffer)) != -1) {
zipOut.write(buffer, 0, length);
}
zipOut.closeEntry();
} finally {
in.close();
}
// case 2: write directly to output stream, i.e. you have your raw data but need to create csv representation
zipOut.putNextEntry(new ZipEntry("SecondName"));
// example setup, key is to use the below outputstream 'zipOut' write methods
Object mySerializer = new MySerializer(); // i.e. csv-writer
Object myData = getMyData(); // the data to be processed by the serializer in order to make a csv file
mySerizalier.setOutput(zipOut);
// write whatever you have to the zipOut
mySerializer.write(myData);
zipOut.closeEntry();
// repeat for the next file.. or make for-loop
}
} finally {
zipOut.close();
}
}
There is no reason to store your data in files unless you have memory constraints. Files give you InputStream and OutputStream, both which have in-memory equivalents.
Note that creating a csv writer usually means doing something like this, where the point is to take a piece of data (array list or map, whatever you have) and make it into byte[] parts. Append the byte[] parts into an OutputStream using a tool like DataOutputStream (make your own if you like) or OutputStreamWriter.

If your data is not huge, meaning can stay in memory then exporting to CSV and zipping up and streaming it for downloading can all be done on-they-fly. Caching can be done at any of these steps which greatly depends on your application's business logic.