Develop Reference

How to create a file in java without random numbers getting added to the filename and delete after use?

I have to create a temp file in the /tmp directory the code that I am using is adding random numbers to the filename. I have to use the name of the file in order to do something. With the random number, I am not able to use that file. The code I wrote is :
File dir = new File("/tmp");
String prefix = "temp";
String suffix = ".txt";
File tempFile = File.createTempFile(prefix, suffix, dir);
After using the file with the correct file name I also have to delete it how can I do that?

If you need to access the File again, you can store the path of the file to be accessed at another time.
You can get the absolute path file using:
tempFile.getAbsolutePath();
To answer your question about deleting the file after you are finished using it, you can either use the detete() or deleteOnExit() methods of File.

If your code needs a predictable filename, and you want that file to be cleaned up automatically when the program ends, don’t use a temp file (they have a random name - it’s just how they work) but rather just use deleteOnExit() with a regular file:
File file = new File("some/filename.ext");
file.deleteOnExit();

How to create executable file in Java using Netbeans

I made a cache simulator program for a homework, I decided to use java. I want to create an executable jar file that will work on any system, but the problem is that my program gathers data from an external text file. How can I include that text file inside the jar so that there won't be any problem when executing file? By the way, I am using NetBeans IDE.

If you don't need to write to the file, copy into the src directory. You will no longer be able to access like a File, but instead will need to use Class#getResource, passing it the path from the top of the source tree to where the file is stored.
For example, if you put it in src/data, then you'd need to use getClass().getResource("/data/..."), passing it what ever name the file is...
Clean and build...

Yes and I said Yes. To really make your jarfiles along with the text files. Please ensure that links to the folder on which the text files is where properly coded and well linked.
The three Examplary Method below should get you working irrespective of any IDEs. Please rate this and give me a shout if you still need further help.......Sectona
Method 1
Step 1:- Locate your folder that contain your java file by using cd command.
Step 2:- Once your enter your folder location then view your java file by dir
command.
Step 3:- Compile your java file using
javac file.java
Step 4:- view class file by type dir command.
Step 5:- Now you want to create a manifest file.
I)Go to folder<br>
II)Right-click->New->Text Document
III)open text document. Type main class name ,
Main-Class: main-class-name
IV)Save this file your wish like MyManifest.txt
Step 6:- To create executable jar file type
jar cfm JarFileName.jar MyManifest.txt JavaFike1.class JavaFile2.class
Step 7:- Now you see the Executable jar file on your folder. Click the file to
Run.
Step 8:- To run this file via command prompt then type
java -jar JarFileName.jar
Step 9:- You done this..........Sectona
Method 2
The basic format of the command for creating a JAR file is:
jar cf jar-file input-file(s)
The options and arguments used in this command are:
The c option indicates that you want to create a JAR file.
The f option indicates that you want the output to go to a file rather than to stdout.
jar-file is the name that you want the resulting JAR file to have. You can use any filename for a JAR file. By convention, JAR filenames are given a .jar extension, though this is not required.
The input-file(s) argument is a space-separated list of one or more files that you want to include in your JAR file. The input-file(s) argument can contain the wildcard * symbol. If any of the "input-files" are directories, the contents of those directories are added to the JAR archive recursively.
Method 3
import java.io.*;
import java.util.jar.*;
public class CreateJar {
public static int buffer = 10240;
protected void createJarArchive(File jarFile, File[] listFiles) {
try {
byte b[] = new byte[buffer];
FileOutputStream fout = new FileOutputStream(jarFile);
JarOutputStream out = new JarOutputStream(fout, new Manifest());
for (int i = 0; i < listFiles.length; i++) {
if (listFiles[i] == null || !listFiles[i].exists()|| listFiles[i].isDirectory())
System.out.println();
JarEntry addFiles = new JarEntry(listFiles[i].getName());
addFiles.setTime(listFiles[i].lastModified());
out.putNextEntry(addFiles);
FileInputStream fin = new FileInputStream(listFiles[i]);
while (true) {
int len = fin.read(b, 0, b.length);
if (len <= 0)
break;
out.write(b, 0, len);
}
fin.close();
}
out.close();
fout.close();
System.out.println("Jar File is created successfully.");
} catch (Exception ex) {}
}
public static void main(String[]args){
CreateJar jar=new CreateJar();
File folder = new File("C://Answers//Examples.txt");
File[] files = folder.listFiles();
File file=new File("C://Answers//Examples//Examples.jar");
jar.createJarArchive(file, files);
}
}

You can keep any file in classpath and read as class path resource. Sample code is given below.
InputStream in = this.getClass().getClassLoader().getResourceAsStream("yourinputFile.txt");
Your jar will be class path, that means you can keep your file in root folder of java source which will get added to jar file while building it.

Write to a .txt file in a package

I would like to write to a .txt file that is inside a package. I can get it to read from the exact location the .txt file is stored but not from inside the package. I'm assuming it is using class loaders but I cannot seem to get it to work.
Here is what I have so far.
public void writeFile(String fileLocation) {
Writer output = null;
File file = new File(fileLocation);
try {
output = new BufferedWriter(new FileWriter(file));
output.append("WRITING TEST");
output.close();
} catch (IOException ex) {
System.out.println("Couldn't write to file.");
}
}
Then I use this in another class to write.
WriteFile writeFile = new WriteFile();
writeFile.writeFile("src/com/game/scores.txt");
I understand that if using class loaders you remove "src/" because that will no longer exist when the program is compiled in a .jar.

It is not possible to write or update a file inside jar. Since jar itself is a file.
Please refer this link.
Write To File Method In JAR

You could use a class in that package to give you the location of the folder.
Try something like
public URL getPackageLocation() {
return getClass().getResource(".");
}
This should give you the location of the folder from which this method is being called from.

From the comments you already know that you cann't write to a file, which resides in a JAR file. At best what you can do, is creating your file, relative to the path where the JAR is located like bellow:
mylocation
|-- my-jar.jar
|-- com
|--game
|--myfile.txt

I would like to write to it to update the scores in my game as the
user goes through the levels.
While it might be possible to write to the JAR file, I don't recommend it for this use case. Just write it somewhere at:
Path userdir = Paths.get(System.getProperty("user.home"), ".myApp", "<my app version>");

You can't write into Jar file. Writing into Jar file is not recommendable. You can write outside the jar file.
Please refer this and this stack overflow question for more details.

java.io.FileNotFoundException (File not found) using Scanner. What's wrong in my code?

I've a .txt file ("file.txt") in my netbeans "/build/classes" directory.
In the same directory there is the .class file compiled for the following code:
try {
File f = new File("file.txt");
Scanner sc = new Scanner(f);
}
catch (IOException e) {
System.out.println(e);
}
Debugging the code (breakpoint in "Scanner sc ..") an exception is launched and the following is printed:
java.io.FileNotFoundException: file.txt (the system can't find the
specified file)
I also tried using "/file.txt" and "//file.txt" but same result.
Thank you in advance for any hint

If you just use new File("pathtofile") that path is relative to your current working directory, which is not at all necessarily where your class files are.
If you are sure that the file is somewhere on your classpath, you could use the following pattern instead:
URL path = ClassLoader.getSystemResource("file.txt");
if(path==null) {
//The file was not found, insert error handling here
}
File f = new File(path.toURI());

The JVM will look for the file in the current working directory.
Where this is depends on your IDE settings (how your program is executed).
To figure out where it expects file.txt to be located, you could do
System.out.println(new File("."));
If it for instance outputs
/some/path/project/build
you should place file.txt in the build directory (or specify the proper path relative to the build directory).

Try:
File f = new File("./build/classes/file.txt");

Use "." to denote the current directory
String path = "./build/classes/file.txt";
File f = new File(path);

File Object loads, looking for match in its current directory.... which is Directly in Your project folder where your class files are loaded not in your source ..... put the file directly in the project folder

How to pass a text file as a argument?

Im trying to write a program to read a text file through args but when i run it, it always says the file can't be found even though i placed it inside the same folder as the main.java that im running.
Does anyone know the solution to my problem or a better way of reading a text file?

Do not use relative paths in java.io.File.
It will become relative to the current working directory which is dependent on the way how you run the application which in turn is not controllable from inside your application. It will only lead to portability trouble. If you run it from inside Eclipse, the path will be relative to /path/to/eclipse/workspace/projectname. If you run it from inside command console, it will be relative to currently opened folder (even though when you run the code by absolute path!). If you run it by doubleclicking the JAR, it will be relative to the root folder of the JAR. If you run it in a webserver, it will be relative to the /path/to/webserver/binaries. Etcetera.
Always use absolute paths in java.io.File, no excuses.
For best portability and less headache with absolute paths, just place the file in a path covered by the runtime classpath (or add its path to the runtime classpath). This way you can get the file by Class#getResource() or its content by Class#getResourceAsStream(). If it's in the same folder (package) as your current class, then it's already in the classpath. To access it, just do:
public MyClass() {
URL url = getClass().getResource("filename.txt");
File file = new File(url.getPath());
InputStream input = new FileInputStream(file);
// ...
}
or
public MyClass() {
InputStream input = getClass().getResourceAsStream("filename.txt");
// ...
}

Try giving an absolute path to the filename.
Also, post the code so that we can see what exactly you're trying.

When you are opening a file with a relative file name in Java (and in general) it opens it relative to the working directory.
you can find the current working directory of your process using
String workindDir = new File(".").getAbsoultePath()
Make sure you are running your program from the correct directory (or change the file name so that it will be relative to where you are running it from).

If you're using Eclipse (or a similar IDE), the problem arises from the fact that your program is run from a few directories above where the actual source is located. Try moving your file up a level or two in the project tree.
Check out this question for more detail.

The simplest solution is to create a new file, then see where the output file is. That is the correct place to put your input file into.

If you put the file and the class working with it under same package can you use this:
Class A {
void readFile (String fileName) {
Url tmp = A.class.getResource (fileName);
// Or Url tmp = this.getClass().getResource (fileName);
File tmpFile = File (tmp);
if (tmpFile.exists())
System.out.print("I found the file.")
}
}
It will help if you read about classloaders.

say I have a text file input.txt which is located on the desktop
and input.txt has the following content
i came
i saw
i left
and below is the java code for reading that text file
public class ReadInputFromTextFile {
public static void main(String[] args) throws Exception
{
File file = new File(
"/Users/viveksingh/desktop/input.txt");
BufferedReader br
= new BufferedReader(new FileReader(file));
String st;
while ((st = br.readLine()) != null)
System.out.println(st);
}
}
output on the console:
i came
i saw
i left