简单易行的方法!

import android.os.FileUtils;

try (InputStream in = new FileInputStream(sourceFile); 
     OutputStream out = new FileOutputStream(destinationFile) ){
                
     FileUtils.copy(in, out); 

}catch(Exception e){
     Log.d("ReactNative","Error copying file: "+e.getMessage());
}

在kotlin中，只需:

val fileSrc : File = File("srcPath")
val fileDest : File = File("destPath")

fileSrc.copyTo(fileDest)

要复制文件并将其保存到目标路径，您可以使用下面的方法。

public static void copy(File src, File dst) throws IOException {
    InputStream in = new FileInputStream(src);
    try {
        OutputStream out = new FileOutputStream(dst);
        try {
            // Transfer bytes from in to out
            byte[] buf = new byte[1024];
            int len;
            while ((len = in.read(buf)) > 0) {
                out.write(buf, 0, len);
            }
        } finally {
            out.close();
        }
    } finally {
        in.close();
    }
}

在API 19+上，您可以使用Java自动资源管理:

public static void copy(File src, File dst) throws IOException {
    try (InputStream in = new FileInputStream(src)) {
        try (OutputStream out = new FileOutputStream(dst)) {
            // Transfer bytes from in to out
            byte[] buf = new byte[1024];
            int len;
            while ((len = in.read(buf)) > 0) {
                out.write(buf, 0, len);
            }
        }
    }
}

简单易行的方法!

import android.os.FileUtils;

try (InputStream in = new FileInputStream(sourceFile); 
     OutputStream out = new FileOutputStream(destinationFile) ){
                
     FileUtils.copy(in, out); 

}catch(Exception e){
     Log.d("ReactNative","Error copying file: "+e.getMessage());
}

这在Android O (API 26)上很简单，如你所见:

  @RequiresApi(api = Build.VERSION_CODES.O)
  public static void copy(File origin, File dest) throws IOException {
    Files.copy(origin.toPath(), dest.toPath());
  }

Kotlin扩展它

fun File.copyTo(file: File) {
    inputStream().use { input ->
        file.outputStream().use { output ->
            input.copyTo(output)
        }
    }
}