如何在RxJava中传输多个参数

Question

我使用RxJava和Okhttp下载文件。例如，我需要用于下载文件的URL和用于创建名称的名称。但我一次只能得到一个参数。

有没有办法同时获取网址和名称？

api.getRatedWallpapers()
            .flatMap(Observable::from)
            .map(wallpaper -> wallpaper.detail)
            .filter(details -> details != null && details.size() != 0)
            .flatMap(Observable::from)
            .filter(detail1 -> detail1.width > 1920)
            .map(detail -> detail.il_file)
            .map(s -> new Request.Builder().url(s).build())
            .map(client::newCall)
            .map(call -> {
                try {
                    return call.execute();
                } catch (IOException e) {
                    e.printStackTrace();
                }
                return null;
            })
            .map(response -> {
                try {
                    return response.body().byteStream();
                } catch (IOException e) {
                    e.printStackTrace();
                }
                return null;
            })
            .map(BufferedInputStream::new)
            .subscribe(bufferedInputStream -> {

                        //Here I should use the name in the Detail instance.
                        File file = new File(random.nextInt(10000) + ".jpg");
                        try {
                            FileOutputStream fos = new FileOutputStream(file);
                            byte[] buffer = new byte[512];
                            int hasRead;
                            while ((hasRead = bufferedInputStream.read(buffer)) != -1) {
                                fos.write(buffer, 0, hasRead);
                            }
                            fos.close();
                            bufferedInputStream.close();
                        } catch (IOException e) {
                            e.printStackTrace();
                        }
                    },
                    Throwable::printStackTrace,
                    () -> System.out.println("Completed"));

Answer 1

使用专用的包装类组合两者，或使用Zip运算符。