I'm writing a fairly simple Java program, and some options I require from the user give them the ability to read from a list of text for data comparisons.
What I would like to do is to give them the ability to simply define the file if it is in the current working path, and assume so unless they provide a separate path to their file. I'd like to use the same approach to generate a output file wherever they define.
So I start by checking if the file even exists (which does not work if you do not define a full path):
File f1 = new File(inputFile);
if(f1.exists())
System.out.println("Exists");
else if(!f1.exists())
{
System.out.println("Cannot find file " + inputFile);
System.exit(1);
}
So I could create the file with new File(path/inputFile), but I can break that fairly easily.
I want them to be able to do any of the following:
program.exe -m -1 inputFile.txt outputFile.txt
program.exe -m -1 C:\Users\tmp\Desktop\inputFile.txt outputFile.txt
program.exe -m -1 inputFile.txt C:\outputFile.txt
Any suggestions on where I might make my next step?
you can try something like
String currentUserPath = System.getProperty("user.dir")
to get the current path from where the application is being ran. Then you can check if the user provided on args[0] a full path, something like:
String inputPath = args[0];
inputPath = inputPath.trim().toLowerCase(); // as you are using windows case doesn't matter
if (!inputPath.startsWith("c:"))
inputPath = currentUserPath + inputPath;
and you could do something similar for the outputFile
Related
tl;dr I'm more used to writing command-line scripts that can just output based on the current working directory, so I'm unsure what directory to use for output files in a program that will be launched from a JAR.
Program Description:
My program builds an HTML file from data given to it from the rest of the program, and then is supposed to write it to a file that we'll call "Output.html" for simplicity.
Relevant Code:
public void outputHTML()
{
String output = buildHTML();
// Expanded to explain my confusion better
String fileDirectory = ""; // ???
String fileName = "Output.html";
String fullPath = fileDirectory + "\\" + fileName;
try (BufferedWriter writer = new BufferedWriter(new FileWriter(fullPath)))
{
writer.write(output);
writer.close();
} catch (IOException e)
{
System.out.println("File not found.");
e.printStackTrace();
}
}
Problem
I don't know what to put the file directory as. Usually I run my programs from the command line and use ".\\Output.txt" as my output path, but I don't know where to put it if it's being run from a JAR.
The desired file structure is as follows:
Encompassing Folder
Program.jar
output
Output.html
Or alternatively (not sure if this makes it easier to understand or harder):
main\
main\Program.jar
main\output\
main\output\Output.html
Everything I can find on SE only relates to reading files that are both immutable and internal, but I'm trying to output a non-static file to a location outside of my jar.
Can anyone help with this? Thanks!
Misc Details
I'm using Eclipse without Gradle currently, because I don't know what Gradle is and new things are scary. If this particular problem would be easier to solve with Gradle, let me know and I'll look up more about it.
EDIT:
Added syntax highlighting to code block.
Formatted everything a bit better
Changed title to be more descriptive
You can use an absolute path: e.g. fileDirectory = "\\project\\test\\main\\output";
using normal slash should also work even on Windows ("/project/test/main/output")
Or use a relative path - this will start from the current working directory (user directory), the one where the JVM was started in - e.g. fileDirectory = "main\\output";
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();
I am working in Netbeans IDE.
What I want to do is:
Get The directory of the Current Java Application (Ex: "F:\PadhooWorld")
Join a file name to it. (Ex: "\Somestuff.txt")
Check if that File exists (Ex: "F:\PadhooWorld\Somestuff.txt")
Do a if.. else activity
When I tam trying to Join Directory + Filename, it is throwing lots of error messages like Path cannot be converted to string etc . Searching the net the whole day, doesn't yield any simple usable solution
Please specify a very simple solution.
EDIT
I have only 2 lines of code as yet
String AppPath = System.getProperty("user.dir");
String fullPath = AppPath + "\Surabhi.txt";
The First Line resolves alright
The Second line (I tried different variations) No Luck. It is underlined in red. Error hints say stuffs like 'Path cannot be converted to string'..
I cannot RUN the code.
It sounds like you're overthinking it. You can just create a File object with the file name you want (the path to the current directory will be used by default) and then call exists() on it:
File f = new File("filename.txt");
System.out.println(f.getAbsolutePath()); //Just for debug if you want to check the path
if(f.exists()) {
//Whatever
}
Alternatively, if you want to specify the path as well as the file name:
String AppPath = System.getProperty("user.dir");
String fileName = "Surabhi.txt";
File f = new File(AppPath, fileName); //f.getAbsolutePath() will give the concatenated name
if(f.exists()) {
//Whatever
}
I have a programming mini competition tomorrow and we will be required to create our program on a flash drive given. The judges won't edit our code so it runs and I am worried that the flash drive letter will change and then my program won't be able to locate the text file it needs to read in.
I have always used paths for my flash drive like this:
FileReader file = new FileReader("E:/BPA/Crypto/input.txt");
Is there a way for me to guarantee my program will be able to read in the text file despite if the letter name for my flash drive isn't the same on the judges computer as it was on mine? Thanks!
You may
Put the file inside your sources
Use Class.getResourceAsStream(String name) to get InputStream of the file
For example, if you have class x.y.z.A
Copy input.txt to src folder into x/y/z package
Get corresponding InputStreamReader as InputStreamReader fileStream = new InputStreamReader(A.class.getResourceAsStream("input.txt"));
If you aren't sure what drive the file will be you could do something like this
char drive = 'A';
String filePath = ":/BPA/Crypto/input.txt";
while(drive != 'Z')
{
try{
Scanner readFromFile = new Scanner(new File(drive + filePath));
readFromFile.close(); //add this if you simply want the path or drvie letter
break;
}catch(FileNotFoundException error)
{
System.out.println("Drive: " + drive + " did not contained file in " + drive + filePath);
}
drive += 1;
}
Basically the idea is to attempt to open the file for reading from different drives starting at A up until Y. Obviously you can go further but I am going to assume that drives A-Y would safely exhaust all the possible drives on where ever you are running your software.
By the time you get our of the While loop the variable "drive" will contain the correct letter of the drive you want. You can modify it to be a function that returns the letter, or perhaps the file path, or simply use it once whenever you try to read from the text file. Up to you.
I need to execute a command from a program. The command line is ok, I tried it in the terminal, but it doesn't work in the program.
I add a copy from my code:
File dir = new File("videos");
String[] children = dir.list();
if (children == null) {
// Either dir does not exist or is not a directory
System.out.print("No existe el directorio\n");
} else {
for (int i=0; i<children.length; i++) {
// Get filename of file or directory
String filename = children[i];
//Recojo el momento exacto
System.out.print("\n" +filename);
Process p = Runtime.getRuntime().exec("exiftool -a -u -g1 -j videos/"+filename+">metadata/"+filename+".json");
}
The program must get the name of all of the files in a folder (filename) and extract the metadata of theese videos, writting them on a .json files in the folder 'metadata'.
Where is the problem?
The problem is, the redirection character (>) is a shell-based construct, not an executable. So unless you're running this command through something like bash (which you're not), it's going to be interpreted as a literal character argument to your exiftool invocation.
If you want to get this to work, you have two options:
Get bash to do it - pass the whole command line as an argument to bash -c. This might need some heroic escaping, although in your case it looks OK.
Do the redirection yourself within Java. Invoke the command without the redirected output (i.e. everything up to the > sign), then read from the process' outputstream and write all the contents to the appropriate file.
The latter approach sounds like more work initially, but when you consider that you need to always read a Process' output anyway (see the javadocs, second paragraph), it's actually very little extra on top of that. You're simply sending this output to a file instead of throwing it away.
If you have Java 7, it's easier:
Process p = new ProcessBuilder()
.command("exiftool", "-a", "-u", "-g1", "-j",
new File("videos", filename).toString())
.redirectOutput(new File("metadata", filename + ".json"))
.start();
This falls under "solution 2", but the runtime library takes care of the boilerplate.