Android - Store inputstream in file

2019-01-21 10:22发布

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

5条回答
地球回转人心会变
2楼-- · 2019-01-21 10:56

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.

查看更多
不美不萌又怎样
3楼-- · 2019-01-21 10:58

Here is a solution which handles all the Exceptions and is based on the previous answers:

void writeStreamToFile(InputStream input, File file) {
    try {
        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();
        }
    } catch (FileNotFoundException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    } finally {
        try {
            input.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}
查看更多
倾城 Initia
4楼-- · 2019-01-21 11:14

A shorter version:

OutputStream out = new FileOutputStream(file);
fos.write(IOUtils.read(in));
out.close();
in.close();
查看更多
Anthone
5楼-- · 2019-01-21 11:18

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

try {
    File file = new File(getCacheDir(), "cacheFileAppeal.srl");
    OutputStream output = new FileOutputStream(file);
    try {
        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 {
        output.close();
    }
} finally {
    input.close();
}
查看更多
祖国的老花朵
6楼-- · 2019-01-21 11:18

There's the way of IOUtils:

copy(InputStream input, OutputStream output)

The code of it is similar to this :

public static long copyStream(InputStream input, OutputStream output) throws IOException {
    long count = 0L;
    byte[] buffer = new byte[4096]; 
    for (int n; -1 != (n = input.read(buffer)); count += (long) n)
        output.write(buffer, 0, n);
    return count;
}
查看更多
登录 后发表回答