java.lang.IllegalArgumentException: contains a path separator

Question

I have a filename in my code as :

String NAME_OF_FILE="//sdcard//imageq.png";
FileInputStream fis =this.openFileInput(NAME_OF_FILE); // 2nd line

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";

score 98 · Answer 1 · answered Jun 24 '13 at 20:34

98

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.

answered Jun 24 '13 at 20:34

runix

989
6
2

1

Given that one can use `FileInputStream` like this, why choose to use `openFileInput`? – Al Lelopath Dec 19 '13 at 22:32
1

This should be the correctly marked answer because it provides a clear example. – marienke Jul 21 '17 at 12:15
Simple solutions are the best one. Thanks. – Danger Mar 07 '18 at 06:30

Stephan · Accepted Answer · 2017-01-30T16:19:25.267

66

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.

edited Jan 30 '17 at 16:19

answered May 11 '11 at 11:36

Stephan

7,360
37
46

59

Please provide some example for this – Anand Savjani Apr 25 '16 at 10:09
14

Your answer is confusing; OP is using a FileInputStream – rootmeanclaire Jan 23 '17 at 01:10
8

provide example – Lutaaya Huzaifah Idris May 03 '17 at 12:14
2

Please expound on the answer. – IgorGanapolsky Nov 03 '17 at 13:58
1

this is the wrong answer - please see the answer below – bharal May 30 '18 at 00:39

score 33 · Answer 3 · edited Nov 07 '17 at 11:12

33

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

edited Nov 07 '17 at 11:12

Milad Faridnia

9,113
13
65
78

answered May 11 '11 at 11:34

reflog

7,587
1
42
47

score 5 · Answer 4 · edited Jan 03 '18 at 12:28

5

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"));

edited Jan 03 '18 at 12:28

jwpfox

5,124
11
45
42

answered Jan 03 '18 at 12:06

Swaran Singh

127
2
3

score 2 · Answer 5 · answered May 11 '12 at 14:53

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);

What ? `File f = new File(fileDirPath); f.mkdirs();` Please edit — Mr_and_Mrs_D, May 05 '13 at 00:19

score 1 · Answer 6 · answered Feb 19 '19 at 13:25

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());
        }

score 0 · Answer 7 · edited Mar 06 '16 at 07:11

0

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);

edited Mar 06 '16 at 07:11

Prune

76,765
14
60
81

answered Mar 06 '16 at 05:50

Muhammad syukur

11
1

Same answer as the previous one. – Anis LOUNIS aka AnixPasBesoin Jan 14 '17 at 17:09

Wraithious · Answer 8 · 2017-07-15T01:49:01.753

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
 }
}

score 0 · Answer 9 · answered Oct 05 '22 at 01:58

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())
}

java.lang.IllegalArgumentException: contains a path separator

9 Answers9

Linked

Related