Getting an error when trying to open a FileInputStream to load Map from file with .ser extension.
Constructor where I create new File and invoke method that loads map from file:
protected DriveatorImpl() {
accounts = new ConcurrentHashMap<String, Client>();
db = new File("database.ser"); // oddly this does not create a file if one does not exist
loadDB();
}
#SuppressWarnings("unchecked")
private void loadDB() {
try {
fileIn = new FileInputStream(db);
in = new ObjectInputStream(fileIn);
accounts = (Map<String, Client>) in.readObject();
in.close();
fileIn.close();
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} catch (ClassNotFoundException e) {
e.printStackTrace();
}
}
I've tried to create file manually and put it in same package with class, but it does not help. What's going on?!
Thank You!
You provide a relative path for the file. That means program will look for the file relative to the working directory.
Depending on how you run the program it will be the directory you run it from (if run from Shell/Cmd) or whatever is configured in the project settings (if run from the IDE). For the latter, it depends on the IDE but usually it's the project root directory.
More info on working directory: https://en.wikipedia.org/wiki/Working_directory
More info on relative path: https://en.wikipedia.org/wiki/Path_(computing)#Absolute_and_relative_paths
Regarding creation of the file, it would create non-existing file if you were to write to it. When you read it, it expects it to exist. That means you have to create empty file (if one does not exist) before reading or simply treat exception as empty content.
The path to the file you have given might be wrong for IDE it can take relative path but from the command line, it will take the absolute path.
Related
This piece of code throws a FileNotFoundException, i'm sure the file exists in my working directory, am i doing something wrong?
private void generateInvoiceNumber(){ //uses reads previous invoice number and increments it.
try {
File invoiceFile = new File("./Invoices/invoiceFile.txt");
FileWriter writer = new FileWriter(invoiceFile,false);
Scanner getter = new Scanner(invoiceFile);
this.invoiceNumber = getter.nextInt();
writer.write(++invoiceNumber);
writer.close();
getter.nextInt();
getter.close();
}
catch (IOException e) {
System.out.println("An error occurred.");
e.printStackTrace();
}
}
My tip:
Print (in your code) the current path location.
Then you can use this path in order to find the exact path you should use in order to access your file.
Maybe you should put more concrete absolute path:
File invoiceFile = Paths.get ("C:","Invoices", "invoiceFile.txt").toFile();
or if you trying to get from current path:
File invoiceFile = Paths.get (".","Invoices", "invoiceFile.txt").toFile();
And you can check your . path:
System.out.println(new File(".").getCanonicalPath());
Which operating system you are using?
It’s better to use paths when you are constructing a path to your file like
File file = Paths.get (".","Invoices", "invoice.txt").toFile();
corrected " symbols and default root "." which is your folder where app started.
I followed Where to put own properties file in an android project created with Android Studio? and I got an InputStream which reads from my .properties file successfully. However, I can't write to that .properties file, as there is no similar method to getBaseContext().getAssets().open ("app.properties") which returns an OutputStream. I have also read Java Properties File appending new values but this didn't seem to help me, my guess is my file name for the file writer is wrong but I also tried "assets\userInfo.properties" which also doesn't work.
My .properties file is in src\main\assets\userInfo.properties
Properties props = new Properties();
InputStream inputStream = null;
try{
inputStream = getBaseContext().getAssets().open("userInfo.properties");
props.load(inputStream);
props.put("name", "smith");
FileOutputStream output = new FileOutputStream("userInfo.properties"); //this line throws error
props.store(output, "This is overwrite file");
String name = props.getProperty("name");
Log.d(TAG, "onCreate: PROPERTIES TEST NAME CHANGE: " + name);
} catch (IOException e) {
e.printStackTrace();
} finally {
try {
inputStream.close();
} catch (IOException e) {
e.printStackTrace();
}
}
Current code throws this error:
java.io.FileNotFoundException: userInfo.properties (Read-only file system)
You can't write to the assets folder, as it is inside the APK which is read-only.
Use internal or external storage instead
You can't write to the assets folder. If you want to update your properties file, you'll have to put them some place else. If you want the initial version in the assets or raw folder, just copy it to the default files dir when the app is first used, then read from/write to it there.
File source=new File(fname1);
System.out.println("souce name "+fname1);
File dest = new File("F:\\BackupFiles",source.getName());
try
{
FileUtils.moveFile(source, dest);
source.delete();
}
catch (IOException ex)
{
Logger.getLogger(FileCompare.class.getName()).log(Level.SEVERE, null, ex);
}
System.out.println("file moved successfully...");
the above code throws exception
"java.io.IOException: Failed to delete original file 'C:\xampp\htdocs\eyeOS\eyeos\users\ajkani\files\html.txt' after copy to 'F:\BackupFiles\html.txt' "
and i tried to delete the file after copied it to the destination but unable to delete.
i tried deleteOnExit() method instead of delete() but nothing works.
i have used md5 algorithm to check the similarity of two files.
if the files are not same.i want to move the files to destination directory.
From above code, it seems that you want to move one file from one directory to another.
As per this assumption, you can use below code.
String sourcePath = "D:\\other\\new.xls";
File source = new File(sourcePath);
System.out.println("souce name " + sourcePath);
File destDirPath = new File("D:\\");
try {
FileUtils.moveFileToDirectory(source, destDirPath, false);
} catch (IOException e) {
e.printStackTrace();
}
System.out.println("file moved successfully...");
This will definitely help you.
I am trying to createNewFile() in java.I have written down the following example.I have compiled it but am getting a run time error.
import java.io.File;
import java.io.IOException;
public class CreateFileExample
{
public static void main(String [] args)
{
try
{
File file = new File("home/karthik/newfile.txt");
if(file.createNewFile())
{
System.out.println("created new fle");
}else
{
System.out.println("could not create a new file");
}
}catch(IOException e )
{
e.printStackTrace();
}
}
}
It is compiling OK.The run time error that I am getting is
java.io.IOException: No such file or directory
at java.io.UnixFileSystem.createFileExclusively(Native Method)
at java.io.File.createNewFile(File.java:947)
at CreateFileExample.main(CreateFileExample.java:16)
some points here
1- as Victor said you are missing the leading slash
2- if your file is created, then every time you invoke this method "File.createNewFile()" will return false
3- your class is very platform dependent (one of the main reasons why Java is powerful programming language is that it is a NON-PLATFORM dependent), instead you can detect a relative location throw using the System.getProperties() :
// get System properties :
java.util.Properties properties = System.getProperties();
// to print all the keys in the properties map <for testing>
properties.list(System.out);
// get Operating System home directory
String home = properties.get("user.home").toString();
// get Operating System separator
String separator = properties.get("file.separator").toString();
// your directory name
String directoryName = "karthik";
// your file name
String fileName = "newfile.txt";
// create your directory Object (wont harm if it is already there ...
// just an additional object on the heap that will cost you some bytes
File dir = new File(home+separator+directoryName);
// create a new directory, will do nothing if directory exists
dir.mkdir();
// create your file Object
File file = new File(dir,fileName);
// the rest of your code
try {
if (file.createNewFile()) {
System.out.println("created new fle");
} else {
System.out.println("could not create a new file");
}
} catch (IOException e) {
e.printStackTrace();
}
this way you will create your file in any home directory on any platform, this worked for my windows operating system, and is expected to work for your Linux or Ubuntu as well
You're missing the leading slash in the file path.
Try this:
File file = new File("/home/karthik/newfile.txt");
That should work!
Actually this error comes when there is no directory "karthik" as in above example and createNewFile() is only to create file not for directory use mkdir() for directory and then createNewFile() for file.
I've got an XML file that is parsed and written in my application. From within my IDE (Eclipse) I simply address it like this:
Reading:
private String xmlFile = "file.xml";
and then I build the document:
doc = sax.build(xmlFile);
Writing is done like this:
writer = new FileWriter("file.xml");
Runs great so far, but after bundling my application, the file is no longer accessible.
What exactly do I have to do to make it accessible from within an application bundle?
I'm absolutely not familiar with the classpath or whatever I need for that, so please go easy on me!
Thank you very much!
To read the file you can create a BufferedReader as follows:
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(
getClass().getResourceAsStream("file.xml")));
You can write the file as before, but keep track of where it's written to - then you can re-read the file from there if necessary.
Peter,
I've run into a similar issue, and since you're new to this, I'll start with this: When you package a file in a jar, it is compressed, and so accessing it like you have in your original code will not work, as we cannot read the files compressed in the jar. Instead, you need to ask Java to pull out this resource as a stream (as many others have helpfully pointed out) which Java does know how to supply to you.
Now, actually outputting the file from there IS a pain. Here's some code I wrote a while back to do this, which is of course from some other source I found. Fill in the exceptions as needed! =)
InputStream in = null;
in = this.getClass().getResourceAsStream("/main.cc");
File outputFile = createMainOutputFile();
OutputStream out = null;
try {
out = new FileOutputStream(outputFile);
} catch (FileNotFoundException e) {
System.out.println(outputFile + " was not found!");
e.printStackTrace();
}
// Transfer bytes from in to out
byte[] buf = new byte[1024];
int len;
try {
while ((len = in.read(buf)) > 0) {
out.write(buf, 0, len);
}
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
try {
in.close();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
try {
out.close();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
We actually have to output the file piece by piece, sad I know, I really wish what you were hoping to do worked! It'd make life a lot easier. The only thing that may be confusing there is the createMainOutputFile() call, so here's the code for that too...
private File createMainOutputFile() {
File directoryPath = new File(mainOutputFolder);
directoryPath.mkdirs();
File newFile = new File (mainOutputFolder + "main.cc");
try {
newFile.createNewFile();
} catch (IOException e) {
System.out.println("failed To create new file.");
e.printStackTrace();
}
return newFile;
}
Best of Luck!
Edit: Now that I notice you're actually parsing it with an XML parser, I'll point out to you that if you're using a SAX parser (at least apache's) you'll find that it will actually accept an inputStream just like it would a file:
DocumentBuilderFactory docFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder docBuilder = null;
try {
docBuilder = docFactory.newDocumentBuilder();
} catch (ParserConfigurationException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
try {
//Here we supply the inputStream instead of the file...
doc = docBuilder.parse(inputStream);
} catch (SAXException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
Node xmlRoot = doc.getFirstChild();
Imagine that you have a package named myXmlDir put your file.xml file under this package (a directory actually)
Notice that this package is packaged with you class inside the jar and can be accessed by the classLoader.
See this code that return an input stream:
return SomeClass.class.getClassLoader().getResource("myXmlDir/file.xml").openStream();
you can also, instead of openStream(), activate the getFile() function
I can't comment, so my answer is referring to LES2's answer. The trick is that your file.xml must be on the classpath if you want to use a classloader.
Usually, it will be included in the classpath if it is in your JAR file. to check if it's there, open the jar with 7zip or any other zip program.
If the file is not inside the jar, you must ship it along with the jar and add it to the classpath manually when starting the app.
lets say file.xml and the jar are in the same directory. then
java -cp .:myjar.jar com.example.Main
should help.
Use the file as a property. You can than do with it whatever you want. To access it use
Class.getResourceAsStream()
or an equivalent method. Check out this link for a few hints: http://www.javaworld.com/javaworld/javaqa/2003-08/01-qa-0808-property.html?page=2
Edit:
I'm sorry to say that my suggestion isn't as good as I thought it was. I felt that I had done something as you want to once but I looked at the app and all I did was read the files.
I also did a bit of googleing and found this forum entry which is a bit old but considering the lacking progress on Sun's part I reckon it's essence is still valid: http://lists.apple.com/archives/Java-dev/2003/Feb/msg00876.html
Basically, if you want to write something, either require the user to extract the jar before launching or (my suggestion) create a directory in the same directory as the jar is in and write your file to that directory. If you really needed to, you could even build another jar with that file and all the files of the original jar, which would of course leave you with two archives.
Sorry for bringing your hopes up.
Best option to this common problem:
Absolute path for the location outside the jar/war.
The the app itself can be deployed anywhere, but the host should have a common file area such as "/common/files/someApplication"
Better yet is to have a configuration to specify (during build or at runtime) which path to use. E.g. dev build would use "/dev/myprojects/someApplication"
Not as good solution:
Deploy as exploded war file. Read and write is not a problem, but now you've cause a deployment dependency.
your input should be the absolute or relative path of the file....