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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

Use of properties in java

I'm new here and i would like to ask on how properties can work with the java codes i mean the values inside of the properties will be use as variables. For example i have file1.txt and file2.txt inside config.properties and store it in an Arraylist then scan the folder and if the files are found copy it. My work only shows the names of the data from the properties that is stored in the arraylist but my another problem is how these data will be copied to another folder.
so far this is my code
public class MainClass {
static Properties prop = new Properties();
static InputStream input = null;
static String filename = "";
public static void main(String[] args) throws IOException {
File source = new File("D:/ojt");
File dest = new File("D:/ojt/New folder");
Properties prop = new Properties();
InputStream input = null;
try {
// getFiles(filename);
} catch (IOException ex) {
ex.printStackTrace();
} finally {
if (input != null) {
try {
input.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
}
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();
}
}
private static void copy(String fromPath, String outputPath)
{
// filter = new FileTypeOrFolderFilter(fileType);
File currentFolder = new File(fromPath);
File outputFolder = new File(outputPath);
scanFolder(currentFolder, outputFolder);
}
private static void getFiles(String path) throws IOException{
//Put filenames in arraylist<string>
String filename = "bydatefilesdir.props";
input = MainClass.class.getClassLoader().getResourceAsStream(filename);
Scanner s = new Scanner(input);
File dir = new File(path);
final ArrayList<String> list = new ArrayList<String>();
while (s.hasNextLine()){
list.add(s.nextLine());
}
// ArrayList<String> filenames = new ArrayList<String>();
// for(File file : dir.listFiles()){
// filenames.add(file.getName());
// }
prop.load(input);
//Check if the files are in the arraylist
for (int i = 0; i < list.size(); i++){
String s1 = list.get(i);
System.out.println("File "+i+" : "+s1);
}
System.out.println("\n");
}
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();
}
}
private static void copy(String source, String dest)
{
// filter = new FileTypeOrFolderFilter(fileType);
File currentFolder = new File(source);
File outputFolder = new File(dest);
scanFolder(currentFolder, outputFolder);
}
private static void scanFolder(File source, File dest)
{
System.out.println("Scanning folder [" + source.toString() + "]...\n");
File[] files = source.listFiles();
for (File file : files) {
if (file.isDirectory()) {
scanFolder(source, dest);
} else {
try {
copyFileUsingStream(source, dest);
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
}
}
PS: sorry for the poor programming just new in java.. still learning
Edited: I've included the updated codes above..

I suppose java.util.Properties has build in methods to achieve what you are trying.
please refer this example and you will find a better solution.
public class App {
public static void main( String[] args ){
Properties prop = new Properties();
InputStream input = null;
try {
String filename = "config.properties";
input = App.class.getClassLoader().getResourceAsStream(filename);
if(input==null){
System.out.println("Sorry, unable to find " + filename);
return;
}
//load a properties file from class path, inside static method
prop.load(input);
//get the property value and print it out
System.out.println(prop.getProperty("file1.txt"));
System.out.println(prop.getProperty("file2.txt"));
} catch (IOException ex) {
ex.printStackTrace();
} finally{
if(input!=null){
try {
input.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
}
}

How to make a copy of a file in android?

In my app I want to save a copy of a certain file with a different name (which I get from user)
Do I really need to open the contents of the file and write it to another file?
What is the best way to do so?

To copy a file and save it to your destination path you can use the method below.
public static void copy(File src, File dst) throws IOException {
InputStream in = new FileInputStream(src);
try {
OutputStream out = new FileOutputStream(dst);
try {
// Transfer bytes from in to out
byte[] buf = new byte[1024];
int len;
while ((len = in.read(buf)) > 0) {
out.write(buf, 0, len);
}
} finally {
out.close();
}
} finally {
in.close();
}
}
On API 19+ you can use Java Automatic Resource Management:
public static void copy(File src, File dst) throws IOException {
try (InputStream in = new FileInputStream(src)) {
try (OutputStream out = new FileOutputStream(dst)) {
// Transfer bytes from in to out
byte[] buf = new byte[1024];
int len;
while ((len = in.read(buf)) > 0) {
out.write(buf, 0, len);
}
}
}
}

Alternatively, you can use FileChannel to copy a file. It might be faster than the byte copy method when copying a large file. You can't use it if your file is bigger than 2GB though.
public void copy(File src, File dst) throws IOException {
FileInputStream inStream = new FileInputStream(src);
FileOutputStream outStream = new FileOutputStream(dst);
FileChannel inChannel = inStream.getChannel();
FileChannel outChannel = outStream.getChannel();
inChannel.transferTo(0, inChannel.size(), outChannel);
inStream.close();
outStream.close();
}

Kotlin extension for it
fun File.copyTo(file: File) {
inputStream().use { input ->
file.outputStream().use { output ->
input.copyTo(output)
}
}
}

This is simple on Android O (API 26), As you see:
#RequiresApi(api = Build.VERSION_CODES.O)
public static void copy(File origin, File dest) throws IOException {
Files.copy(origin.toPath(), dest.toPath());
}

These worked nice for me
public static void copyFileOrDirectory(String srcDir, String dstDir) {
try {
File src = new File(srcDir);
File dst = new File(dstDir, src.getName());
if (src.isDirectory()) {
String files[] = src.list();
int filesLength = files.length;
for (int i = 0; i < filesLength; i++) {
String src1 = (new File(src, files[i]).getPath());
String dst1 = dst.getPath();
copyFileOrDirectory(src1, dst1);
}
} else {
copyFile(src, dst);
}
} catch (Exception e) {
e.printStackTrace();
}
}
public static void copyFile(File sourceFile, File destFile) throws IOException {
if (!destFile.getParentFile().exists())
destFile.getParentFile().mkdirs();
if (!destFile.exists()) {
destFile.createNewFile();
}
FileChannel source = null;
FileChannel destination = null;
try {
source = new FileInputStream(sourceFile).getChannel();
destination = new FileOutputStream(destFile).getChannel();
destination.transferFrom(source, 0, source.size());
} finally {
if (source != null) {
source.close();
}
if (destination != null) {
destination.close();
}
}
}

Much simpler now with Kotlin:
File("originalFileDir", "originalFile.name")
.copyTo(File("newFileDir", "newFile.name"), true)
trueorfalse is for overwriting the destination file
https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.io/java.io.-file/copy-to.html

It might be too late for an answer but the most convenient way is using
FileUtils's
static void copyFile(File srcFile, File destFile)
e.g. this is what I did
`
private String copy(String original, int copyNumber){
String copy_path = path + "_copy" + copyNumber;
try {
FileUtils.copyFile(new File(path), new File(copy_path));
return copy_path;
} catch (IOException e) {
e.printStackTrace();
}
return null;
}
`

in kotlin , just :
val fileSrc : File = File("srcPath")
val fileDest : File = File("destPath")
fileSrc.copyTo(fileDest)

Here is a solution that actually closes the input/output streams if an error occurs while copying. This solution utilizes apache Commons IO IOUtils methods for both copying and handling the closing of streams.
public void copyFile(File src, File dst) {
InputStream in = null;
OutputStream out = null;
try {
in = new FileInputStream(src);
out = new FileOutputStream(dst);
IOUtils.copy(in, out);
} catch (IOException ioe) {
Log.e(LOGTAG, "IOException occurred.", ioe);
} finally {
IOUtils.closeQuietly(out);
IOUtils.closeQuietly(in);
}
}

in Kotlin: a short way
// fromPath : Path the file you want to copy
// toPath : The path where you want to save the file
// fileName : name of the file that you want to copy
// newFileName: New name for the copied file (you can put the fileName too instead of put a new name)
val toPathF = File(toPath)
if (!toPathF.exists()) {
path.mkdir()
}
File(fromPath, fileName).copyTo(File(toPath, fileName), replace)
this is work for any file like images and videos

now in kotlin you could just use
file1.copyTo(file2)
where file1 is an object of the original file and file2 is an object of the new file you want to copy to

Simple and easy way...!
import android.os.FileUtils;
try (InputStream in = new FileInputStream(sourceFile);
OutputStream out = new FileOutputStream(destinationFile) ){
FileUtils.copy(in, out);
}catch(Exception e){
Log.d("ReactNative","Error copying file: "+e.getMessage());
}

What is the "Jodatime" / "Apache commons" of Zip/Unzip Java utilities?

I'm sure that there is a mature, widely used ZIP file utility out there, I just can't seem to find out. Something with the same maturity as Apache Commons, Google Collections, Joda Time
I'm trying to do the simplest task of getting a zip file as a byte array (ZipInputStream) and extract it to a folder. this seems like a very tedious task.
I would hope for a syntactic sugar API that does somethnig like this:
public class MyDreamZIPUtils
public static void extractToFolder(ZipInputStream zin, File outputFolderRoot){
...
}
public static void extractToFolder(ZipFile zf, File outputFolderRoot){
...
}
public static zipFolder(File folderToZip, File zippedFileLocation){
...
}
public static zipFolder(File folderToZip, ByteArrayOutputStream zipResult){
...
}
Anything like this?
Am I missing something?

http://commons.apache.org/compress/
I am sure you can write the "syntactic sugar" on top of that.
Javadoc: http://commons.apache.org/compress/apidocs/index.html

I used only Java API calls... I did not do all your methods. you can figure them out from here... Please note i do not claim that the code is bug free... use at your own risk :)
public static void extractToFolder(ZipInputStream zin, File outputFolderRoot)
throws IOException {
FileOutputStream fos = null;
byte[] buf = new byte[1024];
ZipEntry zipentry;
for (zipentry = zin.getNextEntry(); zipentry != null; zipentry = zin.getNextEntry()) {
try {
String entryName = zipentry.getName();
System.out.println("Extracting: " + entryName);
int n;
File newFile = new File(outputFolderRoot, entryName);
if (zipentry.isDirectory()) {
newFile.mkdirs();
continue;
} else {
newFile.getParentFile().mkdirs();
newFile.createNewFile();
}
fos = new FileOutputStream(newFile);
while ((n = zin.read(buf, 0, 1024)) > -1)
fos.write(buf, 0, n);
fos.close();
zin.closeEntry();
} catch (Exception e) {
e.printStackTrace();
} finally {
if (fos != null)
try {
fos.close();
} catch (Exception ignore) {
}
}
}
zin.close();
}
public static void zipFolder(File folderToZip, File zippedFileLocation) throws IOException {
// create a ZipOutputStream to zip the data to
ZipOutputStream zos = new ZipOutputStream(new FileOutputStream(zippedFileLocation));
String path = "";
zipDir(folderToZip, zos, path);
// close the stream
zos.close();
}
private static void zipDir(File directory, ZipOutputStream zos, String path) throws IOException {
File zipDir = directory;
// get a listing of the directory content
String[] dirList = zipDir.list();
byte[] readBuffer = new byte[2156];
int bytesIn = 0;
// loop through dirList, and zip the files
for (int i = 0; i < dirList.length; i++) {
File f = new File(zipDir, dirList[i]);
if (f.isDirectory()) {
zipDir(new File(f.getPath()), zos, path + f.getName() + "/");
continue;
}
FileInputStream fis = new FileInputStream(f);
try {
ZipEntry anEntry = new ZipEntry(path + f.getName());
zos.putNextEntry(anEntry);
bytesIn = fis.read(readBuffer);
while (bytesIn != -1) {
zos.write(readBuffer, 0, bytesIn);
bytesIn = fis.read(readBuffer);
}
} finally {
fis.close();
}
}
}
Reference Java2s

Copy and validate a Zip File via Java

After some research:
How to create a Zip File
and some google research i came up with this java function:
static void copyFile(File zipFile, File newFile) throws IOException {
ZipFile zipSrc = new ZipFile(zipFile);
ZipOutputStream zos = new ZipOutputStream(new FileOutputStream(newFile));
Enumeration srcEntries = zipSrc.entries();
while (srcEntries.hasMoreElements()) {
ZipEntry entry = (ZipEntry) srcEntries.nextElement();
ZipEntry newEntry = new ZipEntry(entry.getName());
zos.putNextEntry(newEntry);
BufferedInputStream bis = new BufferedInputStream(zipSrc
.getInputStream(entry));
while (bis.available() > 0) {
zos.write(bis.read());
}
zos.closeEntry();
bis.close();
}
zos.finish();
zos.close();
zipSrc.close();
}
This code is working...but it is not nice and clean at all...anyone got a nice idea or an example?
Edit:
I want to able to add some type of validation if the zip archive got the right structure...so copying it like an normal file without regarding its content is not working for me...or would you prefer checking it afterwards...i am not sure about this one

You just want to copy the complete zip file? Than it is not needed to open and read the zip file... Just copy it like you would copy every other file.
public final static int BUF_SIZE = 1024; //can be much bigger, see comment below
public static void copyFile(File in, File out) throws Exception {
FileInputStream fis = new FileInputStream(in);
FileOutputStream fos = new FileOutputStream(out);
try {
byte[] buf = new byte[BUF_SIZE];
int i = 0;
while ((i = fis.read(buf)) != -1) {
fos.write(buf, 0, i);
}
}
catch (Exception e) {
throw e;
}
finally {
if (fis != null) fis.close();
if (fos != null) fos.close();
}
}

Try: http://commons.apache.org/io/api-release/org/apache/commons/io/FileUtils.html#copyFile
Apache Commons FileUtils#copyFile

My solution:
import java.io.*;
import javax.swing.*;
public class MovingFile
{
public static void copyStreamToFile() throws IOException
{
FileOutputStream foutOutput = null;
String oldDir = "F:/UPLOADT.zip";
System.out.println(oldDir);
String newDir = "F:/NewFolder/UPLOADT.zip"; // name as the destination file name to be done
File f = new File(oldDir);
f.renameTo(new File(newDir));
}
public static void main(String[] args) throws IOException
{
copyStreamToFile();
}
}

I have updated your code to Java 9+, FWIW
try (ZipFile srcFile = new ZipFile(inputName)) {
try (ZipOutputStream destFile = new ZipOutputStream(
Files.newOutputStream(Paths.get(new File(outputName).toURI())))) {
Enumeration<? extends ZipEntry> entries = srcFile.entries();
while (entries.hasMoreElements()) {
ZipEntry src = entries.nextElement();
ZipEntry dest = new ZipEntry(src.getName());
destFile.putNextEntry(dest);
try (InputStream content = srcFile.getInputStream(src)) {
content.transferTo(destFile);
}
destFile.closeEntry();
}
destFile.finish();
}
}