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How to copy lines from a text file to another text file? [duplicate]

I want to copy a file from one location to another location in Java. What is the best way to do this?
Here is what I have so far:
import java.io.File;
import java.io.FilenameFilter;
import java.util.ArrayList;
import java.util.List;
public class TestArrayList {
public static void main(String[] args) {
File f = new File(
"D:\\CBSE_Demo\\Demo_original\\fscommand\\contentplayer\\config");
List<String>temp=new ArrayList<String>();
temp.add(0, "N33");
temp.add(1, "N1417");
temp.add(2, "N331");
File[] matchingFiles = null;
for(final String temp1: temp){
matchingFiles = f.listFiles(new FilenameFilter() {
public boolean accept(File dir, String name) {
return name.startsWith(temp1);
}
});
System.out.println("size>>--"+matchingFiles.length);
}
}
}
This does not copy the file, what is the best way to do this?

You can use this (or any variant):
Files.copy(src, dst, StandardCopyOption.REPLACE_EXISTING);
Also, I'd recommend using File.separator or / instead of \\ to make it compliant across multiple OS, question/answer on this available here.
Since you're not sure how to temporarily store files, take a look at ArrayList:
List<File> files = new ArrayList();
files.add(foundFile);
To move a List of files into a single directory:
List<File> files = ...;
String path = "C:/destination/";
for(File file : files) {
Files.copy(file.toPath(),
(new File(path + file.getName())).toPath(),
StandardCopyOption.REPLACE_EXISTING);
}

Update:
see also
https://stackoverflow.com/a/67179064/1847899
Using Stream
private static void copyFileUsingStream(File source, File dest) throws IOException {
InputStream is = null;
OutputStream os = null;
try {
is = new FileInputStream(source);
os = new FileOutputStream(dest);
byte[] buffer = new byte[1024];
int length;
while ((length = is.read(buffer)) > 0) {
os.write(buffer, 0, length);
}
} finally {
is.close();
os.close();
}
}
Using Channel
private static void copyFileUsingChannel(File source, File dest) throws IOException {
FileChannel sourceChannel = null;
FileChannel destChannel = null;
try {
sourceChannel = new FileInputStream(source).getChannel();
destChannel = new FileOutputStream(dest).getChannel();
destChannel.transferFrom(sourceChannel, 0, sourceChannel.size());
}finally{
sourceChannel.close();
destChannel.close();
}
}
Using Apache Commons IO lib:
private static void copyFileUsingApacheCommonsIO(File source, File dest) throws IOException {
FileUtils.copyFile(source, dest);
}
Using Java SE 7 Files class:
private static void copyFileUsingJava7Files(File source, File dest) throws IOException {
Files.copy(source.toPath(), dest.toPath());
}
Or try Googles Guava :
https://github.com/google/guava
docs:
https://guava.dev/releases/snapshot-jre/api/docs/com/google/common/io/Files.html

Use the New Java File classes in Java >=7.
Create the below method and import the necessary libs.
public static void copyFile( File from, File to ) throws IOException {
Files.copy( from.toPath(), to.toPath() );
}
Use the created method as below within main:
File dirFrom = new File(fileFrom);
File dirTo = new File(fileTo);
try {
copyFile(dirFrom, dirTo);
} catch (IOException ex) {
Logger.getLogger(TestJava8.class.getName()).log(Level.SEVERE, null, ex);
}
NB:- fileFrom is the file that you want to copy to a new file fileTo in a different folder.
Credits - #Scott: Standard concise way to copy a file in Java?

public static void copyFile(File oldLocation, File newLocation) throws IOException {
if ( oldLocation.exists( )) {
BufferedInputStream reader = new BufferedInputStream( new FileInputStream(oldLocation) );
BufferedOutputStream writer = new BufferedOutputStream( new FileOutputStream(newLocation, false));
try {
byte[] buff = new byte[8192];
int numChars;
while ( (numChars = reader.read( buff, 0, buff.length ) ) != -1) {
writer.write( buff, 0, numChars );
}
} catch( IOException ex ) {
throw new IOException("IOException when transferring " + oldLocation.getPath() + " to " + newLocation.getPath());
} finally {
try {
if ( reader != null ){
writer.close();
reader.close();
}
} catch( IOException ex ){
Log.e(TAG, "Error closing files when transferring " + oldLocation.getPath() + " to " + newLocation.getPath() );
}
}
} else {
throw new IOException("Old location does not exist when transferring " + oldLocation.getPath() + " to " + newLocation.getPath() );
}
}

Copy a file from one location to another location means,need to copy the whole content to another location.Files.copy(Path source, Path target, CopyOption... options) throws IOException this method expects source location which is original file location and target location which is a new folder location with destination same type file(as original).
Either Target location needs to exist in our system otherwise we need to create a folder location and then in that folder location we need to create a file with the same name as original filename.Then using copy function we can easily copy a file from one location to other.
public static void main(String[] args) throws IOException {
String destFolderPath = "D:/TestFile/abc";
String fileName = "pqr.xlsx";
String sourceFilePath= "D:/TestFile/xyz.xlsx";
File f = new File(destFolderPath);
if(f.mkdir()){
System.out.println("Directory created!!!!");
}
else {
System.out.println("Directory Exists!!!!");
}
f= new File(destFolderPath,fileName);
if(f.createNewFile()) {
System.out.println("File Created!!!!");
} else {
System.out.println("File exists!!!!");
}
Files.copy(Paths.get(sourceFilePath), Paths.get(destFolderPath, fileName),REPLACE_EXISTING);
System.out.println("Copy done!!!!!!!!!!!!!!");
}

You can do it with the Java 8 Streaming API, PrintWriter and the Files API
try (PrintWriter pw = new PrintWriter(new File("destination-path"), StandardCharsets.UTF_8)) {
Files.readAllLines(Path.of("src/test/resources/source-file.something"), StandardCharsets.UTF_8)
.forEach(pw::println);
}
If you want to modify the content on-the-fly while copying, check out this link for the extended example https://overflowed.dev/blog/copy-file-and-modify-with-java-streams/

I modified one of the answers to make it a bit more efficient.
public void copy(){
InputStream in = null;
try {
in = new FileInputStream(Files);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
try {
OutputStream out = new FileOutputStream();
try {
// Transfer bytes from in to out
byte[] buf = new byte[1024];
while (true) {
int len = 0;
try {
if (!((len = in.read(buf)) > 0)) break;
} catch (IOException e) {
e.printStackTrace();
}
try {
out.write(buf, 0, len);
} catch (IOException e) {
e.printStackTrace();
}
}
} finally {
try {
out.close();
} catch (IOException e) {
e.printStackTrace();
}
}
} finally {
try {
in.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
private void moveFile() {
copy();
File dir = getFilesDir();
File file = new File(dir, "my_filename");
boolean deleted = file.delete();
}

Files.exists()
Files.createDirectory()
Files.copy()
Overwriting Existing Files:
Files.move()
Files.delete()
Files.walkFileTree()
enter link description here

You can use
FileUtils.copy(sourceFile, destinationFile);
https://commons.apache.org/proper/commons-io/apidocs/org/apache/commons/io/FileUtils.html

while downloading the object from amazon s3 i cant able to download it to diffrent folder rather then the file uploaded path

while downloading the object from amazon s3 i cant able to download it to diffrent folder rather then the file uploaded path...why it is happening like this may be it is metaData problem....please post your valuable comments Thanks in advance..below am posting the code for upload and download
public void AmazonUpload(String fileObj) throws IOException {
try {
this.key = fileObj;
try {
if (this.key == null) {
} else {
if (readFile(this.key) != null) {
// this.key="1";
this.putObjResult = this.amzObj.putObject(new PutObjectRequest(this.bucketName, this.key, readFile(this.key)));
}
}
} catch (AmazonServiceException ae) {
System.out.println(ae.getMessage());
}
} catch (AmazonServiceException ex) {
Logger.getLogger(AthinioCloudMigration.class.getName()).log(Level.SEVERE, null, ex);
}
}
public void AmazonDownload(String dirName, String xmlFilename, String amazonid) throws ParserConfigurationException, SAXException, TransformerException, IOException {
String cloudid;
cloudid = amazonid;
this.comm = new CommonResources(xmlFilename);
this.RequestFiles=new ArrayList();
try {
this.RequestFiles = this.comm.getXML(cloudid);
if (this.RequestFiles != null) {
int len = this.RequestFiles.size();
System.out.println(len);
for (int index = 0; index < len; index++) {
this.CRobj = (CommonResources) this.RequestFiles.get(index);
if (cloudid.equals(this.CRobj.getCloudID())) {
this.newFile = new File(dirName + this.CRobj.getFileName().concat(".rec"));
System.out.println(newFile);
newFile.createNewFile();
this.metaData = this.amzObj.getObject(new GetObjectRequest(this.bucketName, (dirName + this.CRobj.getFileName())), this.newFile);
System.out.println(metaData);
java.io.File tmp = new java.io.File(dirName + this.CRobj.getFileName());
System.out.println(tmp);
tmp.delete();
76,23 87%

Since in Amazon S3 there is no folder structure you receive everything as a object.
For Example: In your bucket you store file in a folder like structure but when you request fro the objects from S3 you will receve your file like folder1/folder2/demo.txt.
So try this one, Get the InputStream for the S3 for your file like amazonS3.getObject(bucket, "folder1/folder2/demo.txt").getObjectContent();. After getting your InputStream pass your File Name, Download location and InputStream to the below method. If you use Java 7 use FileSystems.getDefault().getPath(fullPathWithFileName).getFileName().toString() to get file name from your object name.
public void saveFile(String uploadFileName, String path, InputStream inputStream) throws Exception {
DataOutputStream dos = null;
OutputStream out = null;
try {
File newDirectory = new File(path);
if (!newDirectory.exists()) {
newDirectory.mkdirs();
}
File uploadedFile = new File(path, uploadFileName);
out = new FileOutputStream(uploadedFile);
byte[] fileAsBytes = new byte[inputStream.available()];
inputStream.read(fileAsBytes);
dos = new DataOutputStream(out);
dos.write(fileAsBytes);
} catch (IOException io) {
io.printStackTrace();
} catch (Exception e) {
e.printStackTrace();
} finally {
try {
if (out != null) {
out.close();
}
if (dos != null) {
dos.close();
}
} catch (IOException e) {
e.printStackTrace();
}
}
}

Download/Save file that is obtainable directly from the given URL

I have got links like this link, which directly ask for the filename to save with, and start downloading in the browser.
How can I download or save this file programmatically?
I tried with the following method:
static void DownloadFile(String url, String fileName) throws MalformedURLException, IOException
{
url = "http://dbpedia.org/sparql?default-graph-uri=http%3A%2F%2Fdbpedia.org&query=DESCRIBE+<"+ url +">&format=text%2Fcsv";
URL link = new URL(url); //The file that you want to download
InputStream in = new BufferedInputStream(link.openStream());
ByteArrayOutputStream out = new ByteArrayOutputStream();
byte[] buf = new byte[1024];
int n = 0;
while (-1!=(n=in.read(buf)))
{
out.write(buf, 0, n);
}
out.close();
in.close();
byte[] response = out.toByteArray();
FileOutputStream fos = new FileOutputStream(fileName);
fos.write(response);
fos.close();
System.out.println("Finished");
}
but this save the file having only the first line as ""subject","predicate","object"
" and not the complete file.
EDIT:
As suggested in an answer I tried the following, but that too gave only the first line of the file:
static void DownloadFile(String s_url, String fileName) throws MalformedURLException, IOException
{
s_url = "http://dbpedia.org/sparql?default-graph-uri=http%3A%2F%2Fdbpedia.org&query=DESCRIBE+<"+ s_url +">&format=text%2Fcsv";
//url = "http://dbpedia.org/data/Sachin_Tendulkar.rdf";
try {
URL url = new URL(s_url); //The file that you want to download
// read text returned by server
BufferedReader in = new BufferedReader(new InputStreamReader(url.openStream()));
PrintWriter out = new PrintWriter(fileName);
String line;
while ((line = in.readLine()) != null) {
out.println(line);
}
in.close();
out.close();
}
catch (MalformedURLException e) {
System.out.println("Malformed URL: " + e.getMessage());
}
catch (IOException e) {
System.out.println("I/O Error: " + e.getMessage());
}
System.out.println("Finished");
}
EDIT:
I tried with Apache FileUtils too, but that too gave only the first line of the file.
static void DownloadFile(String s_url, String fileName) throws MalformedURLException, IOException
{
s_url = "http://dbpedia.org/sparql?default-graph-uri=http%3A%2F%2Fdbpedia.org&query=DESCRIBE+<"+ s_url +">&format=text%2Fcsv";
URL url = new URL(s_url); //The file that you want to download
FileUtils.copyURLToFile(url, new File(fileName));
System.out.println("Finished");
}

if you want to download a file, Apache Commons have just what you are looking for, works great!
org.apache.commons.io.FileUtils.copyURLToFile(new URL("URL")), new File("path/to/file"));
if the url returns text, you can try something like this:
try {
URL url = new URL("http://www.google.com:80/");
// read text returned by server
BufferedReader in = new BufferedReader(new InputStreamReader(url.openStream()));
PrintWriter out = new PrintWriter("filename.txt");
String line;
while ((line = in.readLine()) != null) {
out.println(line);
}
in.close();
out.close();
}
catch (MalformedURLException e) {
System.out.println("Malformed URL: " + e.getMessage());
}
catch (IOException e) {
System.out.println("I/O Error: " + e.getMessage());
}

Downloaded jar file corrupted

I've got this code that downloads a .jar file from a specific URL and places it into a specific folder. The jar file downloaded is a mod for a game, meaning that it has to be downloaded and run correctly without being corrupted.
The problem is, each time I try downloading the file, it ends up being corrupted in some way and causing errors when it is loaded.
This is my download code:
final static int size=1024;
public static void downloadFile(String fAddress, String localFileName, String destinationDir, String modID) {
OutputStream outStream = null;
URLConnection uCon = null;
InputStream is = null;
try {
URL Url;
byte[] buf;
int ByteRead,ByteWritten=0;
Url= new URL(fAddress);
outStream = new BufferedOutputStream(new
FileOutputStream(destinationDir+"/"+localFileName));
uCon = Url.openConnection();
is = uCon.getInputStream();
buf = new byte[size];
while ((ByteRead = is.read(buf)) != -1) {
outStream.write(buf, 0, ByteRead);
ByteWritten += ByteRead;
}
System.out.println("Downloaded Successfully.");
System.out.println("File name:\""+localFileName+ "\"\nNo ofbytes :" + ByteWritten);
System.out.println("Writing info file");
WriteInfo.createInfoFile(localFileName, modID);
}catch (Exception e) {
e.printStackTrace();
}
finally {
try {
is.close();
outStream.close();
}
catch (IOException e) {
e.printStackTrace();
}
}
}
Any ideas what is wrong with this code?

Not sure if this will solve your problem but you should flush your buffer at the end.
outStream.flush();

your code look like quite right; try this
public static void downloadFromUrl(String srcAddress, String userAgent, String destDir, String destFileName, boolean overwrite) throws Exception
{
InputStream is = null;
FileOutputStream fos = null;
try
{
File destFile = new File(destDir, destFileName);
if(overwrite && destFile.exists())
{
boolean deleted = destFile.delete();
if (!deleted)
{
throw new Exception(String.format("d'ho, an immortal file %s", destFile.getAbsolutePath()));
}
}
URL url = new URL(srcAddress);
URLConnection urlConnection = url.openConnection();
if(userAgent != null)
{
urlConnection.setRequestProperty("User-Agent", userAgent);
}
is = urlConnection.getInputStream();
fos = new FileOutputStream(destFile);
byte[] buffer = new byte[4096];
int len, totBytes = 0;
while((len = is.read(buffer)) > 0)
{
totBytes += len;
fos.write(buffer, 0, len);
}
System.out.println("Downloaded successfully");
System.out.println(String.format("File name: %s - No of bytes: %,d", destFile.getAbsolutePath(), totBytes));
}
finally
{
try
{
if(is != null) is.close();
}
finally
{
if(fos != null) fos.close();
}
}
}

Programmatically downloading files pushed through a PHP page

Some PHP sites use a page to act as a middle man for handling file downloads.
With a browser this works transparently. There seems to a be a slight pause while the php page processes the request.
However, attempting a download through Java using a URL or HttpURLConnection returns a plain html page. How could I get the file downloads working in the same way?
Edit: Here is an example link:
http://depot.eice.be/index.php?annee_g=jour&cours=poo
Edit: Here is some of the code I've been testing:
// This returns an HTML page
private void downloadURL(String theURL) {
URL url;
InputStream is = null;
DataInputStream dis;
String s;
StringBuffer sb = new StringBuffer();
try {
url = new URL(theURL);
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.setRequestMethod("GET");
conn.connect();
if (conn.getResponseCode()!=HttpURLConnection.HTTP_OK)
return;
InputStream in = conn.getInputStream();
ByteArrayOutputStream bos = new ByteArrayOutputStream();
int i;
while ((i = in.read()) != -1) {
bos.write(i);
}
byte[] b = bos.toByteArray();
FileOutputStream fos = new FileOutputStream( getNameFromUrl( theURL ) );
fos.write(b);
fos.close();
conn.disconnect();
} catch (MalformedURLException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
// This will throw Exceptions if the URL isn't in the expected format
public String getNameFromUrl(String url) {
int slashIndex = url.lastIndexOf('/');
int dotIndex = url.lastIndexOf('.');
System.out.println("url:" + url + "," + slashIndex + "," + dotIndex);
if (dotIndex == -1) {
return url.substring(slashIndex + 1);
} else {
try {
return url.substring(slashIndex + 1, url.length());
} catch (StringIndexOutOfBoundsException e) {
return "";
}
}
}

Considering no other constrains, you can read the redirected URL from the HTTP header and connect to that URL directly from JAVA.

There is an API setting to follow redirects automatically – but it should be true by default. How do you access the URL?
See Java API docs...

I think I've found a solution using HttpUnit. The source of the framework is available if you wish to see how this is handled.
public void downloadURL(String url) throws IOException {
WebConversation wc = new WebConversation();
WebResponse indexResp = wc.getResource(new GetMethodWebRequest(url));
WebLink[] links = new WebLink[1];
try {
links = indexResp.getLinks();
} catch (SAXException ex) {
// Log
}
for (WebLink link : links) {
try {
link.click();
} catch (SAXException ex) {
// Log
}
WebResponse resp = wc.getCurrentPage();
String fileName = resp.getURL().getFile();
fileName = fileName.substring(fileName.lastIndexOf("/") + 1);
System.out.println("filename:" + fileName);
File file = new File(fileName);
BufferedInputStream bis = new BufferedInputStream(
resp.getInputStream());
BufferedOutputStream bos = new BufferedOutputStream(
new FileOutputStream(file.getName()));
int i;
while ((i = bis.read()) != -1) {
bos.write(i);
}
bis.close();
bos.close();
}
System.out.println("Done downloading.");
}