This question already has answers here:
Test if file exists
(7 answers)
Closed 9 years ago.
I'm tried to implement following if condition to skip the downloading process if the file found. But the if condition doesn't seem to be working and without if condition this works fine. can someone please help to solve this. I think, I used the if condition improperly.
File file = new File( Environment.getExternalStorageDirectory().getAbsolutePath() + "/Android/data/com.sample.project/files/images/");
if(file.getPath().equals(false)){
String url = "http://www.sample.com/example.zip";
new DownloadTask().execute( url );
}
#Julien is right. Actually you may try using exists() function of File class.
if (file.exists()){
//true
}else{
//false
}
There are many similar questions which may help:
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Hope it helps.
Related
This question already has answers here:
java.io.FileNotFoundException: the system cannot find the file specified
(8 answers)
Closed 4 years ago.
My program get a filename in parameter but if i want to use it i need the full directory path or i get filenotfoundexception.
For example:
My program got sample.txt in parameter from C:\Users\me\documents.
File file = new File(args[0]);
FileReader fr = new FileReader(file);
That throw filenotfoundexception.
So what should i use to locate the file?
I saw so many similar question but i didnt find solution :(
I tried to use getResources and getPath but nothing.
You can open a file with just the filename if that file exists in the same directory as of your source code. If the file is located at any random location then you need to give the complete path of the file along with its name.
Eg: c:\documents\sample.txt
Or another thing that you can try is recursively going through all folders present in your file system and locate the file. However, this is will be a very horrible solution.
File fileName = new File("myfile.txt");
if(!fileName.exists()) {
fileName.createNewFile();
}
FileOutputStream oFile = new FileOutputStream(fileName, false);
add this code if your file is not on the location this will create a one for you then you wont be getting filenotfound exception at the latter part
This question already has answers here:
How do I check if a file exists in Java?
(19 answers)
Closed 7 years ago.
I would like to know how to open a file in java for writing, but I only want to open it if it exists already. Is that possible? I've searched around and found two suggestions so far which both don't meet my needs. One was to open a file for appending. The problem there is that if the file doesn't exist it will create it. The second was to use file.exists() first. That of course is not a valid solution as the file may exist when I call file.exists() but then not exist when I go to open the file. I want something similar to the Windows API OpenFile() in which I can pass OPEN_EXISTING flag such that the call will fail is the file doesn't exists. Does anything like that exist in java?
Only editing question because it was marked duplicate. Not sure why. I thought I was pretty specific about the other answers and why they were not sufficient.
So I'll restate, I want to open a file for writing. I want the open to fail if the file doesn't already exist.
exists() returns true if the file path is a valid directory even if the file isn't there. You can get around this by using:
File f = new File(filePathString);
if(f.exists() && !f.isDirectory()) {/*Do whatever*/}
or use:
File f = new File(filePathString);
if f.isFile() {/*Do whatever*/}
Just catch the FileNotFoundException that is thrown:
try (FileInputStream fis = new FileInputStream(file))
{
// ...
}
catch (FileNotFoundException exc)
{
// it didn't exist.
}
Solutions involving File.exists() and friends are vulnerable to timing-window problems and therefore cannot be recommended. They also merely duplicate what the OS already has to do, but at the wrong time, so they are a waste of time and space. You have to deal with IOExceptions anyway, so do that.
No there is nothing like that In Java.... Core library
You should be able to wrap your logic in a if statement that names use of the file.exists() method. If you do the check just before opening the file the you would be extremely unlucky if someone has deleted the file in-between. The method that checks if the file exists and the code to open the file and lock it should run in milliseconds..
Eg
If (file.exists() {
//Your Code goes here..
} else {
System.out.Print("missing file");
}
This question already has answers here:
Get path of Android resource
(3 answers)
How can I write a Drawable resource to a File?
(2 answers)
Closed 8 years ago.
I am writing an Android app and I need to do something I would consider incredibly simple, yet am having the hardest time figuring out.
How can I obtain the path to a JPG file so that I can upload it to a server?
The file is snoopy.jpg and is in the res/drawable folder
I've tried:
File sourceFile = new File("android.resource://com.appname.something/drawable/snoopy");
But, that's not a file
I've tried:
File sourceFile = new File("drawable/snoopy");
and that doesn't work.
I tried putting snoopy.jpg in the root of the app directory and attempting:
File sourceFile = new File("/snoopy.jpg");
And that still didn't work.
Any help would be greatly appreciated!
Try this :
InputStream is = getResources().openRawResource(R.drawable.snoopy);
You can open an InputStream from your drawable resource using the above code. Also, before doing any file operation, you need to check if file.exist() and if it returns false then you need to create a file via f.createNewFile();
This question already has answers here:
Java FileNotFoundException even though file exists
(2 answers)
Closed 9 years ago.
I do this and write some content to this file.
fileSymbol = new File("D:\\TempFiles\\SymbolFile.xml");
fileSymbol.createNewFile();
prSymbol = new PrintWriter(fileSymbol);
while(//condition goes here){
prSymbol.write(text);
}
Then I try to read from this file as,
FileReader fr = new FileReader(fileSymbol);
BufferedReader br = new BufferedReader(fr);
String line = br.readLine();
When I try to run, it gives the following exception
java.io.FileNotFoundException: fileSymbol (The system cannot find the file specified)
When I try
fileSymbol.canRead();
fileSymbol.exists();
returns true for both.
I do not understand why I get this error though I have already created the file. When I check the location manually, I am able to see my created file.
PS: I have just provided the code snippet of my code.
Add the following before you read the file
prSymbol.close();
It is not required to call flush() before close .Reference JavaDoc - close()
java.io.FileNotFoundException: fileSymbol (The system cannot find the file specified)
That message, if you have transcribed it accurately, can only mean that you are using "fileSymbol" as the file name, where previously you were using "D:\\TempFiles\\SymbolFile.xml".
NB calling createNewFile() the line before you create an output stream or writer to the same file is a complete waste of time.
This question already has answers here:
How to create a folder in Java?
(8 answers)
Closed 3 years ago.
I tried to use the File class to create an empty file in a directory like "C:/Temp/Emptyfile".
However, when I do that, it shows me an error : "already made folder Temp". Otherwise, it won't create one for me.
So, how do I literally create folders with java API?
Looks file you use the .mkdirs() method on a File object: http://www.roseindia.net/java/beginners/java-create-directory.shtml
// Create a directory; all non-existent ancestor directories are
// automatically created
success = (new File("../potentially/long/pathname/without/all/dirs")).mkdirs();
if (!success) {
// Directory creation failed
}
You can create folder using the following Java code:
File dir = new File("nameoffolder");
dir.mkdir();
By executing above you will have folder 'nameoffolder' in current folder.