Question

我正在从网址检索XML Feed然后解析它。我需要做的是将内部存储到手机中，这样当没有互联网连接时，它可以解析保存的选项，而不是现场选项。

我面临的问题是我可以创建url对象，使用getInputStream来获取内容，但它不会让我保存它。

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();

我可以如何保存输入流，以便稍后加载。

干杯

Answer 1

在这里，输入是你的inputStreamReader。然后使用相同的文件（名称）和FileInputStream来读取将来的数据。

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();
}

Answer 2

简单功能

尝试使用这个简单的功能整齐地将其包裹起来：

// 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();
        }
    }
}

感谢Jordan LaPrise和他的answer。

Answer 3

较短的版本：

OutputStream out = new FileOutputStream(file);
fos.write(IOUtils.read(in));
out.close();
in.close();

Answer 4

这是一个处理所有异常的解决方案，它基于以前的答案：

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();
        }
    }
}

Answer 5

Kotlin 版本（经过测试，不需要库）：

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

我们使用use函数将在最后自动关闭两个流。

Answer 6

在应用程序的 build.gradle 文件中，添加到 dependencies 下：

    implementation 'commons-io:commons-io:2.5'

在您的代码中：

import org.apache.commons.io.FileUtils;

// given you have a stream, e.g.
InputStream inputStream = getContext().getContentResolver().openInputStream(uri);

// you can now write it to a file with
FileUtils.copyToFile(inputStream, new File("myfile.txt"));

Answer 7

有IOUtils的方式：

copy(InputStream input, OutputStream output)

它的代码类似于：

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;
}

Android - 将输入流存储在文件中

7 个答案:

简单功能