android - Android - 将输入流存储在文件中

Question

我正在从 url 检索 XML 提要，然后对其进行解析。我需要做的是将其存储在手机内部，这样当没有互联网连接时，它可以解析保存的选项而不是实时选项。

我面临的问题是我可以创建 url 对象，使用 getInputStream 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();

有什么想法可以保存输入流，以便以后加载。

干杯

score 117 · Accepted Answer

在这里，输入是你的inputStream. 然后使用相同的文件（名称）并FileInputStream在将来读取数据。

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();
}

score 35 · Accepted Answer

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.

score 24 · Accepted Answer

Kotlin版本（经过测试，无需库）：

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()
        }
    }
}

我们利用了use最后会自动关闭两个流的功能。

即使发生异常，流也会正确关闭。

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

score 7 · Accepted Answer

一个较短的版本：

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

score 7 · Accepted Answer

这是一个处理所有异常并基于先前答案的解决方案：

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();
        }
    }
}

score 3 · Accepted Answer

在您的应用程序build.gradle文件中添加dependencies：

    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"));

score 1 · Accepted Answer

有IOUtils的方式：

copy(InputStream input, OutputStream output)

它的代码与此类似：

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;
}

score 0 · Accepted Answer

你可以使用谷歌番石榴

 import com.google.common.io.ByteStreams;

代码：

 try (FileOutputStream fos = new FileOutputStream(new File("C:\\example.txt"))){
      ByteStreams.copy(inputStream, fos)
 }

android - Android - 将输入流存储在文件中

8 回答 8

Simple Function

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