开发者

java.lang.IllegalArgumentException: contains a path separator

I have a filename in my code as :

String NAME_OF_FILE="//sdcard//imageq.png";
FileInputStream fis =this.openFileInput(NAME_OF_FILE); // 2nd li开发者_开发百科ne

I get an error on 2nd line :

05-11 16:49:06.355: ERROR/AndroidRuntime(4570): Caused by: java.lang.IllegalArgumentException: File //sdcard//imageq.png contains a path separator

I tried this format also:

String NAME_OF_FILE="/sdcard/imageq.png";


The solution is:

FileInputStream fis = new FileInputStream (new File(NAME_OF_FILE));  // 2nd line

The openFileInput method doesn't accept path separators.

Don't forget to

fis.close();

at the end.


This method opens a file in the private data area of the application. You cannot open any files in subdirectories in this area or from entirely other areas using this method. So use the constructor of the FileInputStream directly to pass the path with a directory in it.


openFileInput() doesn't accept paths, only a file name if you want to access a path, use File file = new File(path) and corresponding FileInputStream


I got the above error message while trying to access a file from Internal Storage using openFileInput("/Dir/data.txt") method with subdirectory Dir.

You cannot access sub-directories using the above method.

Try something like:

FileInputStream fIS = new FileInputStream (new File("/Dir/data.txt"));


You cannot use path with directory separators directly, but you will have to make a file object for every directory.

NOTE: This code makes directories, yours may not need that...

File file= context.getFilesDir();
file.mkdir();

String[] array=filePath.split("/");
for(int t=0; t< array.length -1 ;t++)
{
    file=new File(file,array[t]);
    file.mkdir();
}

File f=new File(file,array[array.length-1]);

RandomAccessFileOutputStream rvalue = new RandomAccessFileOutputStream(f,append);


String all = "";
        try {
            BufferedReader br = new BufferedReader(new FileReader(filePath));
            String strLine;
            while ((strLine = br.readLine()) != null){
                all = all + strLine;
            }
        } catch (IOException e) {
            Log.e("notes_err", e.getLocalizedMessage());
        }


File file = context.getFilesDir(); 
file.mkdir();
String[] array = filePath.split("/"); 
for(int t = 0; t < array.length - 1; t++) {
    file = new File(file, array[t]); 
    file.mkdir();
}
File f = new File(file,array[array.length- 1]); 
RandomAccessFileOutputStream rvalue = 
    new RandomAccessFileOutputStream(f, append);


I solved this type of error by making a directory in the onCreate event, then accessing the directory by creating a new file object in a method that needs to do something such as save or retrieve a file in that directory, hope this helps!

 public class MyClass {    

 private String state;
 public File myFilename;

 @Override
 protected void onCreate(Bundle savedInstanceState) {//create your directory the user will be able to find
    super.onCreate(savedInstanceState);
    if (Environment.MEDIA_MOUNTED.equals(state)) {
        myFilename = new File(Environment.getExternalStorageDirectory().toString() + "/My Directory");
        if (!myFilename.exists()) {
            myFilename.mkdirs();
        }
    }
 }

 public void myMethod {

 File fileTo = new File(myFilename.toString() + "/myPic.png");  
 // use fileTo object to save your file in your new directory that was created in the onCreate method
 }
}


I did like this

var dir = File(app.filesDir, directoryName)
if(!dir.exists()){
    currentCompanyFolder.mkdir()
}
var directory = app.getDir(directoryName, Context.MODE_PRIVATE)
val file = File(directory, fileName)
file.outputStream().use {
    it.write(body.bytes())
}
0

上一篇:

下一篇:

精彩评论

暂无评论...
验证码 换一张
取 消

最新问答

问答排行榜