How to download a website and put into another folder? java - java

I have this java code that download the .xml from my website and save it as an .xmlfile.
My problem is i want to save it into another folder.
When i run the code it downloads the file and saves it to the same folder where the java code is located.
I have searched about this and i cant find anything. Here is the code:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.FileWriter;
import java.io.InputStreamReader;
import java.net.URL;
public class dlxml {
public static void main(String[] args)
throws Exception {
URL url = new URL("http://localhost:8080/lab/lab.xml");
BufferedReader reader = new BufferedReader
(new InputStreamReader(url.openStream()));
BufferedWriter writer = new BufferedWriter
(new FileWriter("data.xml"));
String line;
while ((line = reader.readLine()) != null) {
System.out.println(line);
writer.write(line);
writer.newLine();
}
reader.close();
writer.close();
}
}
So basically i want to download the file and save it to another folder. Is it possible?
and what if when i save it to a folder and there is the same file and name but i want to save it the same. For example i have data.xml save it to another folder but there is another same file with data.xml but i dont want it to be data(1).xml i want it to be data.xml
Thanks


Just define the folder path while writing to the file. As shown below:
BufferedWriter writer = new BufferedWriter(new FileWriter("\path\to\folder\data.xml"));


BufferedWriter writer = new BufferedWriter(new FileWriter(FULL_PATH));


In this code segment, you will want to change the file to use a path.
BufferedWriter writer = new BufferedWriter(new FileWriter("data.xml"));
By just using the name of the file, it will create the file in the relative directory (the same folder your code is running from) You'll want to change it to an absolute path so you can specify where you're storing it.

Related

java.lang.NullPointerException: Location is required. javafx [duplicate]

My project has the following structure:
/src/main/java/
/src/main/resources/
/src/test/java/
/src/test/resources/
I have a file in /src/test/resources/test.csv and I want to load the file from a unit test in /src/test/java/MyTest.java
I have this code which didn't work. It complains "No such file or directory".
BufferedReader br = new BufferedReader (new FileReader(test.csv))
I also tried this
InputStream is = (InputStream) MyTest.class.getResourcesAsStream(test.csv))
This also doesn't work. It returns null. I am using Maven to build my project.

Try the next:
ClassLoader classloader = Thread.currentThread().getContextClassLoader();
InputStream is = classloader.getResourceAsStream("test.csv");
If the above doesn't work, various projects have been added the following class: ClassLoaderUtil1 (code here).2
Here are some examples of how that class is used:
src\main\java\com\company\test\YourCallingClass.java
src\main\java\com\opensymphony\xwork2\util\ClassLoaderUtil.java
src\main\resources\test.csv
// java.net.URL
URL url = ClassLoaderUtil.getResource("test.csv", YourCallingClass.class);
Path path = Paths.get(url.toURI());
List<String> lines = Files.readAllLines(path, StandardCharsets.UTF_8);
// java.io.InputStream
InputStream inputStream = ClassLoaderUtil.getResourceAsStream("test.csv", YourCallingClass.class);
InputStreamReader streamReader = new InputStreamReader(inputStream, StandardCharsets.UTF_8);
BufferedReader reader = new BufferedReader(streamReader);
for (String line; (line = reader.readLine()) != null;) {
// Process line
}
Notes
See it in The Wayback Machine.
Also in GitHub.

Try:
InputStream is = MyTest.class.getResourceAsStream("/test.csv");
IIRC getResourceAsStream() by default is relative to the class's package.
As #Terran noted, don't forget to add the / at the starting of the filename

Try following codes on Spring project
ClassPathResource resource = new ClassPathResource("fileName");
InputStream inputStream = resource.getInputStream();
Or on non spring project
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("fileName").getFile());
InputStream inputStream = new FileInputStream(file);

Here is one quick solution with the use of Guava:
import com.google.common.base.Charsets;
import com.google.common.io.Resources;
public String readResource(final String fileName, Charset charset) throws IOException {
return Resources.toString(Resources.getResource(fileName), charset);
}
Usage:
String fixture = this.readResource("filename.txt", Charsets.UTF_8)

Non spring project:
String filePath = Objects.requireNonNull(getClass().getClassLoader().getResource("any.json")).getPath();
Stream<String> lines = Files.lines(Paths.get(filePath));
Or
String filePath = Objects.requireNonNull(getClass().getClassLoader().getResource("any.json")).getPath();
InputStream in = new FileInputStream(filePath);
For spring projects, you can also use one line code to get any file under resources folder:
File file = ResourceUtils.getFile(ResourceUtils.CLASSPATH_URL_PREFIX + "any.json");
String content = new String(Files.readAllBytes(file.toPath()));

For java after 1.7
List<String> lines = Files.readAllLines(Paths.get(getClass().getResource("test.csv").toURI()));
Alternatively you can use Spring utils if you are in Spring echosystem
final val file = ResourceUtils.getFile("classpath:json/abcd.json");
To get to more behind the scene, check out following blog
https://todzhang.com/blogs/tech/en/save_resources_to_files

I faced the same issue.
The file was not found by a class loader, which means it was not packed into the artifact (jar). You need to build the project. For example, with maven:
mvn clean package
So the files you added to resources folder will get into maven build and become available to the application.
I would like to keep my answer: it does not explain how to read a file (other answers do explain that), it answers why InputStream or resource was null. Similar answer is here.

ClassLoader loader = Thread.currentThread().getContextClassLoader();
InputStream is = loader.getResourceAsStream("test.csv");
If you use context ClassLoader to find a resource then definitely it will cost application performance.

Now I am illustrating the source code for reading a font from maven created resources directory,
scr/main/resources/calibril.ttf
Font getCalibriLightFont(int fontSize){
Font font = null;
try{
URL fontURL = OneMethod.class.getResource("/calibril.ttf");
InputStream fontStream = fontURL.openStream();
font = new Font(Font.createFont(Font.TRUETYPE_FONT, fontStream).getFamily(), Font.PLAIN, fontSize);
fontStream.close();
}catch(IOException | FontFormatException ief){
font = new Font("Arial", Font.PLAIN, fontSize);
ief.printStackTrace();
}
return font;
}
It worked for me and hope that the entire source code will also help you, Enjoy!

Import the following:
import java.io.IOException;
import java.io.FileNotFoundException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
import java.util.ArrayList;
The following method returns a file in an ArrayList of Strings:
public ArrayList<String> loadFile(String filename){
ArrayList<String> lines = new ArrayList<String>();
try{
ClassLoader classloader = Thread.currentThread().getContextClassLoader();
InputStream inputStream = classloader.getResourceAsStream(filename);
InputStreamReader streamReader = new InputStreamReader(inputStream, StandardCharsets.UTF_8);
BufferedReader reader = new BufferedReader(streamReader);
for (String line; (line = reader.readLine()) != null;) {
lines.add(line);
}
}catch(FileNotFoundException fnfe){
// process errors
}catch(IOException ioe){
// process errors
}
return lines;
}

getResource() was working fine with the resources files placed in src/main/resources only. To get a file which is at the path other than src/main/resources say src/test/java you need to create it exlicitly.
the following example may help you
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.net.URISyntaxException;
import java.net.URL;
public class Main {
public static void main(String[] args) throws URISyntaxException, IOException {
URL location = Main.class.getProtectionDomain().getCodeSource().getLocation();
BufferedReader br = new BufferedReader(new FileReader(location.getPath().toString().replace("/target/classes/", "/src/test/java/youfilename.txt")));
}
}

You can use the com.google.common.io.Resources.getResource to read the url of file and then get the file content using java.nio.file.Files to read the content of file.
URL urlPath = Resources.getResource("src/main/resource");
List<String> multilineContent= Files.readAllLines(Paths.get(urlPath.toURI()));

if you are loading file in static method then
ClassLoader classLoader = getClass().getClassLoader();
this might give you an error.
You can try this
e.g. file you want to load from resources is resources >> Images >> Test.gif
import org.springframework.core.io.ClassPathResource;
import org.springframework.core.io.Resource;
Resource resource = new ClassPathResource("Images/Test.gif");
File file = resource.getFile();

To read the files from src/resources folder then try this :
DataSource fds = new FileDataSource(getFileHandle("images/sample.jpeg"));
public static File getFileHandle(String fileName){
return new File(YourClassName.class.getClassLoader().getResource(fileName).getFile());
}
in case of non static reference:
return new File(getClass().getClassLoader().getResource(fileName).getFile());

Does the code work when not running the Maven-build jar, for example when running from your IDE? If so, make sure the file is actually included in the jar. The resources folder should be included in the pom file, in <build><resources>.

The following class can be used to load a resource from the classpath and also receive a fitting error message in case there's a problem with the given filePath.
import java.io.InputStream;
import java.nio.file.NoSuchFileException;
public class ResourceLoader
{
private String filePath;
public ResourceLoader(String filePath)
{
this.filePath = filePath;
if(filePath.startsWith("/"))
{
throw new IllegalArgumentException("Relative paths may not have a leading slash!");
}
}
public InputStream getResource() throws NoSuchFileException
{
ClassLoader classLoader = this.getClass().getClassLoader();
InputStream inputStream = classLoader.getResourceAsStream(filePath);
if(inputStream == null)
{
throw new NoSuchFileException("Resource file not found. Note that the current directory is the source folder!");
}
return inputStream;
}
}

this.getClass().getClassLoader().getResource("filename").getPath()

My file in the test folder could not be found even though I followed the answers. It got resolved by rebuilding the project. It seems IntelliJ did not recognize the new file automatically. Pretty nasty to find out.

I got it work on both running jar and in IDE by writing as
InputStream schemaStream =
ProductUtil.class.getClassLoader().getResourceAsStream(jsonSchemaPath);
byte[] buffer = new byte[schemaStream.available()];
schemaStream.read(buffer);
File tempFile = File.createTempFile("com/package/schema/testSchema", "json");
tempFile.deleteOnExit();
FileOutputStream out = new FileOutputStream(tempFile);
out.write(buffer);

This worked pretty fine for me :
InputStream in = getClass().getResourceAsStream("/main/resources/xxx.xxx");
InputStreamReader streamReader = new InputStreamReader(in, StandardCharsets.UTF_8);
BufferedReader reader = new BufferedReader(streamReader);
String content = "";
for (String line; (line = reader.readLine()) != null;) {
content += line;
}

I get it to work without any reference to "class" or "ClassLoader".
Let's say we have three scenarios with the location of the file 'example.file' and your working directory (where your app executes) is home/mydocuments/program/projects/myapp:
a)A sub folder descendant to the working directory:
myapp/res/files/example.file
b)A sub folder not descendant to the working directory:
projects/files/example.file
b2)Another sub folder not descendant to the working directory:
program/files/example.file
c)A root folder:
home/mydocuments/files/example.file (Linux; in Windows replace home/ with C:)
1) Get the right path:
a)String path = "res/files/example.file";
b)String path = "../projects/files/example.file"
b2)String path = "../../program/files/example.file"
c)String path = "/home/mydocuments/files/example.file"
Basically, if it is a root folder, start the path name with a leading slash.
If it is a sub folder, no slash must be before the path name. If the sub folder is not descendant to the working directory you have to cd to it using "../". This tells the system to go up one folder.
2) Create a File object by passing the right path:
File file = new File(path);
3) You are now good to go:
BufferedReader br = new BufferedReader(new FileReader(file));

How to add data to CSV file instead of removing what was previosly thre

I am using this method to write to a CSV file. Every time I run my code it removes all previous data from the file instead of adding to it.
import java.io.File;
import java.io.FileWriter;
import java.io.IOException;
public class Main {
public static void main(String[] args) throws IOException {
File dir = new File(".");
String loc = dir.getCanonicalPath() + File.separator + "Code.txt";
FileWriter fstream = new FileWriter(loc, true);
BufferedWriter out = new BufferedWriter(fstream);
out.write("something");
out.newLine();
//close buffer writer
out.close();
}
}```

You're better off using java.nio.file.Path, which opens up a lot of utility methods on java.nio.file.Files:
Path path = Path.of("Code.txt"); // Relative to current dir
Files.writeString(path, text, StandardCharsets.UTF_8, StandardOpenOption.APPEND);
Where text is the content to append. Files.writeString is from Java 11+. For older versions you can use:
Files.write(path, text.getBytes(StandardCharsets.UTF_8), StandardOpenOption.APPEND);

Output to a file in java

How do I specify the path where I want to save my file when creating an output to a file in java?
//Set up Printer Output file
PrintWriter pw = new PrintWriter (new BufferedWriter (new FileWriter("project61.dat")));
For some reason after running my program, I don't see that my project61.dat file is created. I can't find anywhere in my C drive.

Simple google search yields helpful examples.
Here is one taken from here:
import java.io.IOException;
import java.io.PrintWriter;
public class MainClass {
public static void main(String[] args) {
try {
PrintWriter pw = new PrintWriter("c:\\temp\\printWriterOutput.txt");
pw.println("PrintWriter is easy to use.");
pw.println(1234);
pw.close();
} catch (IOException e) {}
}
}

If you use the following
//Set up Printer Output file
PrintWriter pw = new PrintWriter (new BufferedWriter (new FileWriter("project61.dat")));
Then it will create the file under your project. Please find the file where you project available.
If you want to write this file in a particular directory then mention the absolute path.
PrintWriter pw = new PrintWriter (new BufferedWriter (new FileWriter("c:\\project61.dat")));
pw.write("Test");
pw.close();
It will create the file under "C:" directory.
If you use the absolute directory "C://temp//project61.dat" then the temp folder must be available in c drive. The folder will not be created by default.

How do I load a file from resource folder?

My project has the following structure:
/src/main/java/
/src/main/resources/
/src/test/java/
/src/test/resources/
I have a file in /src/test/resources/test.csv and I want to load the file from a unit test in /src/test/java/MyTest.java
I have this code which didn't work. It complains "No such file or directory".
BufferedReader br = new BufferedReader (new FileReader(test.csv))
I also tried this
InputStream is = (InputStream) MyTest.class.getResourcesAsStream(test.csv))
This also doesn't work. It returns null. I am using Maven to build my project.

Try the next:
ClassLoader classloader = Thread.currentThread().getContextClassLoader();
InputStream is = classloader.getResourceAsStream("test.csv");
If the above doesn't work, various projects have been added the following class: ClassLoaderUtil1 (code here).2
Here are some examples of how that class is used:
src\main\java\com\company\test\YourCallingClass.java
src\main\java\com\opensymphony\xwork2\util\ClassLoaderUtil.java
src\main\resources\test.csv
// java.net.URL
URL url = ClassLoaderUtil.getResource("test.csv", YourCallingClass.class);
Path path = Paths.get(url.toURI());
List<String> lines = Files.readAllLines(path, StandardCharsets.UTF_8);
// java.io.InputStream
InputStream inputStream = ClassLoaderUtil.getResourceAsStream("test.csv", YourCallingClass.class);
InputStreamReader streamReader = new InputStreamReader(inputStream, StandardCharsets.UTF_8);
BufferedReader reader = new BufferedReader(streamReader);
for (String line; (line = reader.readLine()) != null;) {
// Process line
}
Notes
See it in The Wayback Machine.
Also in GitHub.

Try:
InputStream is = MyTest.class.getResourceAsStream("/test.csv");
IIRC getResourceAsStream() by default is relative to the class's package.
As #Terran noted, don't forget to add the / at the starting of the filename

Try following codes on Spring project
ClassPathResource resource = new ClassPathResource("fileName");
InputStream inputStream = resource.getInputStream();
Or on non spring project
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("fileName").getFile());
InputStream inputStream = new FileInputStream(file);

Here is one quick solution with the use of Guava:
import com.google.common.base.Charsets;
import com.google.common.io.Resources;
public String readResource(final String fileName, Charset charset) throws IOException {
return Resources.toString(Resources.getResource(fileName), charset);
}
Usage:
String fixture = this.readResource("filename.txt", Charsets.UTF_8)

Non spring project:
String filePath = Objects.requireNonNull(getClass().getClassLoader().getResource("any.json")).getPath();
Stream<String> lines = Files.lines(Paths.get(filePath));
Or
String filePath = Objects.requireNonNull(getClass().getClassLoader().getResource("any.json")).getPath();
InputStream in = new FileInputStream(filePath);
For spring projects, you can also use one line code to get any file under resources folder:
File file = ResourceUtils.getFile(ResourceUtils.CLASSPATH_URL_PREFIX + "any.json");
String content = new String(Files.readAllBytes(file.toPath()));

For java after 1.7
List<String> lines = Files.readAllLines(Paths.get(getClass().getResource("test.csv").toURI()));
Alternatively you can use Spring utils if you are in Spring echosystem
final val file = ResourceUtils.getFile("classpath:json/abcd.json");
To get to more behind the scene, check out following blog
https://todzhang.com/blogs/tech/en/save_resources_to_files

I faced the same issue.
The file was not found by a class loader, which means it was not packed into the artifact (jar). You need to build the project. For example, with maven:
mvn clean package
So the files you added to resources folder will get into maven build and become available to the application.
I would like to keep my answer: it does not explain how to read a file (other answers do explain that), it answers why InputStream or resource was null. Similar answer is here.

ClassLoader loader = Thread.currentThread().getContextClassLoader();
InputStream is = loader.getResourceAsStream("test.csv");
If you use context ClassLoader to find a resource then definitely it will cost application performance.

Now I am illustrating the source code for reading a font from maven created resources directory,
scr/main/resources/calibril.ttf
Font getCalibriLightFont(int fontSize){
Font font = null;
try{
URL fontURL = OneMethod.class.getResource("/calibril.ttf");
InputStream fontStream = fontURL.openStream();
font = new Font(Font.createFont(Font.TRUETYPE_FONT, fontStream).getFamily(), Font.PLAIN, fontSize);
fontStream.close();
}catch(IOException | FontFormatException ief){
font = new Font("Arial", Font.PLAIN, fontSize);
ief.printStackTrace();
}
return font;
}
It worked for me and hope that the entire source code will also help you, Enjoy!

Import the following:
import java.io.IOException;
import java.io.FileNotFoundException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
import java.util.ArrayList;
The following method returns a file in an ArrayList of Strings:
public ArrayList<String> loadFile(String filename){
ArrayList<String> lines = new ArrayList<String>();
try{
ClassLoader classloader = Thread.currentThread().getContextClassLoader();
InputStream inputStream = classloader.getResourceAsStream(filename);
InputStreamReader streamReader = new InputStreamReader(inputStream, StandardCharsets.UTF_8);
BufferedReader reader = new BufferedReader(streamReader);
for (String line; (line = reader.readLine()) != null;) {
lines.add(line);
}
}catch(FileNotFoundException fnfe){
// process errors
}catch(IOException ioe){
// process errors
}
return lines;
}

getResource() was working fine with the resources files placed in src/main/resources only. To get a file which is at the path other than src/main/resources say src/test/java you need to create it exlicitly.
the following example may help you
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.net.URISyntaxException;
import java.net.URL;
public class Main {
public static void main(String[] args) throws URISyntaxException, IOException {
URL location = Main.class.getProtectionDomain().getCodeSource().getLocation();
BufferedReader br = new BufferedReader(new FileReader(location.getPath().toString().replace("/target/classes/", "/src/test/java/youfilename.txt")));
}
}

You can use the com.google.common.io.Resources.getResource to read the url of file and then get the file content using java.nio.file.Files to read the content of file.
URL urlPath = Resources.getResource("src/main/resource");
List<String> multilineContent= Files.readAllLines(Paths.get(urlPath.toURI()));

if you are loading file in static method then
ClassLoader classLoader = getClass().getClassLoader();
this might give you an error.
You can try this
e.g. file you want to load from resources is resources >> Images >> Test.gif
import org.springframework.core.io.ClassPathResource;
import org.springframework.core.io.Resource;
Resource resource = new ClassPathResource("Images/Test.gif");
File file = resource.getFile();

To read the files from src/resources folder then try this :
DataSource fds = new FileDataSource(getFileHandle("images/sample.jpeg"));
public static File getFileHandle(String fileName){
return new File(YourClassName.class.getClassLoader().getResource(fileName).getFile());
}
in case of non static reference:
return new File(getClass().getClassLoader().getResource(fileName).getFile());

Does the code work when not running the Maven-build jar, for example when running from your IDE? If so, make sure the file is actually included in the jar. The resources folder should be included in the pom file, in <build><resources>.

The following class can be used to load a resource from the classpath and also receive a fitting error message in case there's a problem with the given filePath.
import java.io.InputStream;
import java.nio.file.NoSuchFileException;
public class ResourceLoader
{
private String filePath;
public ResourceLoader(String filePath)
{
this.filePath = filePath;
if(filePath.startsWith("/"))
{
throw new IllegalArgumentException("Relative paths may not have a leading slash!");
}
}
public InputStream getResource() throws NoSuchFileException
{
ClassLoader classLoader = this.getClass().getClassLoader();
InputStream inputStream = classLoader.getResourceAsStream(filePath);
if(inputStream == null)
{
throw new NoSuchFileException("Resource file not found. Note that the current directory is the source folder!");
}
return inputStream;
}
}

this.getClass().getClassLoader().getResource("filename").getPath()

My file in the test folder could not be found even though I followed the answers. It got resolved by rebuilding the project. It seems IntelliJ did not recognize the new file automatically. Pretty nasty to find out.

I got it work on both running jar and in IDE by writing as
InputStream schemaStream =
ProductUtil.class.getClassLoader().getResourceAsStream(jsonSchemaPath);
byte[] buffer = new byte[schemaStream.available()];
schemaStream.read(buffer);
File tempFile = File.createTempFile("com/package/schema/testSchema", "json");
tempFile.deleteOnExit();
FileOutputStream out = new FileOutputStream(tempFile);
out.write(buffer);

This worked pretty fine for me :
InputStream in = getClass().getResourceAsStream("/main/resources/xxx.xxx");
InputStreamReader streamReader = new InputStreamReader(in, StandardCharsets.UTF_8);
BufferedReader reader = new BufferedReader(streamReader);
String content = "";
for (String line; (line = reader.readLine()) != null;) {
content += line;
}

I get it to work without any reference to "class" or "ClassLoader".
Let's say we have three scenarios with the location of the file 'example.file' and your working directory (where your app executes) is home/mydocuments/program/projects/myapp:
a)A sub folder descendant to the working directory:
myapp/res/files/example.file
b)A sub folder not descendant to the working directory:
projects/files/example.file
b2)Another sub folder not descendant to the working directory:
program/files/example.file
c)A root folder:
home/mydocuments/files/example.file (Linux; in Windows replace home/ with C:)
1) Get the right path:
a)String path = "res/files/example.file";
b)String path = "../projects/files/example.file"
b2)String path = "../../program/files/example.file"
c)String path = "/home/mydocuments/files/example.file"
Basically, if it is a root folder, start the path name with a leading slash.
If it is a sub folder, no slash must be before the path name. If the sub folder is not descendant to the working directory you have to cd to it using "../". This tells the system to go up one folder.
2) Create a File object by passing the right path:
File file = new File(path);
3) You are now good to go:
BufferedReader br = new BufferedReader(new FileReader(file));

Writing multiple lines to a file

I am trying to write timestamps to a file when clicking on a JButton. Once .close() is invoked the data is written to the file and any other writes throw an error. How can I write the data without have to create a new FileWriter and overwriting the previous line?

Instead of closing, which does this implictly, you call flush() on the FileWriter object. That keeps it open, but forces the data which has been buffered to be written to the file.
Don't forget to close() when you are done writing though.

You can either keep the writer open between clicks and close it at some other time (perhaps on form exit), or you can create a new FileWriter for each click and have it append to contents already in the file.
FileWriter writer = new FileWriter("output.txt", true); //true here indicates append to file contents
If you choose to keep the writer open between clicks, then you might want to call .flush() on each button press to make sure the file is up to date.

Try this,
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileWriter;
import java.io.IOException;
public class Wr {
public static void main(String[] args) throws IOException {
File f = new File("viv.txt");
FileWriter fw = new FileWriter(f, true);
BufferedWriter bw = new BufferedWriter(fw);
bw.write("Helloooooooooo");
bw.close();
}
}

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