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Android - Store inputstream in file

I am retrieveing an XML feed from a url and then parsing it. What I need to do is also store that internally to the phone so that when there is no internet connection it can parse the saved option rather than the live one.

The problem I am facing is that I can create the url object, use getInputStream to get the contents, but it will not let me save it.

URL url = null;
InputStream inputStreamReader = null;
XmlPullParser xpp = null;

url = new URL("http://*********");
inputStreamReader = getInputStream(url);

ObjectOutput out = new ObjectOutputStream(new FileOutputStream(new File(getCacheDir(),"")+"cacheFileAppeal.srl"));

//--------------------------------------------------------
//This line is where it is erroring.
//--------------------------------------------------------
out.writeObject( inputStreamReader );
//--------------------------------------------------------
out.close();

Any ideas how I can go about saving the input stream so I can load it later.

Cheers

like image 373
Dobes Avatar asked Jun 01 '12 16:06

Dobes


4 Answers

Simple Function

Try this simple function to neatly wrap it up in:

// Copy an InputStream to a File.
//
private void copyInputStreamToFile(InputStream in, File file) {
    OutputStream out = null;

    try {
        out = new FileOutputStream(file);
        byte[] buf = new byte[1024];
        int len;
        while((len=in.read(buf))>0){
            out.write(buf,0,len);
        }
    } 
    catch (Exception e) {
        e.printStackTrace();
    } 
    finally {
        // Ensure that the InputStreams are closed even if there's an exception.
        try {
            if ( out != null ) {
                out.close();
            }

            // If you want to close the "in" InputStream yourself then remove this
            // from here but ensure that you close it yourself eventually.
            in.close();  
        }
        catch ( IOException e ) {
            e.printStackTrace();
        }
    }
}

Thanks to Jordan LaPrise and his answer.

like image 45
Joshua Pinter Avatar answered Nov 19 '22 22:11

Joshua Pinter


Here it is, input is your inputStream. Then use same File (name) and FileInputStream to read the data in future.

try {
    File file = new File(getCacheDir(), "cacheFileAppeal.srl");
    try (OutputStream output = new FileOutputStream(file)) {
        byte[] buffer = new byte[4 * 1024]; // or other buffer size
        int read;

        while ((read = input.read(buffer)) != -1) {
            output.write(buffer, 0, read);
        }

        output.flush();
    }
} finally {
    input.close();
}
like image 105
Volodymyr Lykhonis Avatar answered Nov 19 '22 22:11

Volodymyr Lykhonis


Kotlin version (tested and no library needed):

fun copyStreamToFile(inputStream: InputStream, outputFile: File) {
    inputStream.use { input ->
        val outputStream = FileOutputStream(outputFile)
        outputStream.use { output ->
            val buffer = ByteArray(4 * 1024) // buffer size
            while (true) {
                val byteCount = input.read(buffer)
                if (byteCount < 0) break
                output.write(buffer, 0, byteCount)
            }
            output.flush()
        }
    }
}

We take advantage of use function which will automatically close both streams at the end.

The streams are closed down correctly even in case an exception occurs.

https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.io/use.html
https://kotlinlang.org/docs/tutorials/kotlin-for-py/scoped-resource-usage.html

like image 26
vovahost Avatar answered Nov 19 '22 22:11

vovahost


A shorter version:

OutputStream out = new FileOutputStream(file);
fos.write(IOUtils.read(in));
out.close();
in.close();
like image 7
Tuan Chau Avatar answered Nov 19 '22 21:11

Tuan Chau