如何在异步httpclient java 11中将多个异步get请求的响应写入单个文件？

Question

我可以像这样在 java 11 中使用 httpclient 下载单个媒体文件

public class Httptest {
    
    private static HttpClient client = HttpClient.newBuilder().build();
            
    public static void main(String[] args) throws Exception {
        File fts = new File("P:/sample.ts");  //Destination of downloaded file
        fts.createNewFile();
        URI url = new URI("File url here"); //File Url
        
        HttpRequest request = HttpRequest.newBuilder()   //Creating HttpRequest using Builder class
                .GET()
                .uri(url)
                .build();
        Path file = Path.of("P:/samp.ts");
        //BodyHandlers class has methods to handle the response body
        // In this case, save it as a file (BodyHandlers.ofFile())
        HttpResponse<Path> response = client.send(request,BodyHandlers.ofFile(file)); 
    }
}

以上代码片段从 url 下载 .ts 文件。并且下载正常

现在，我有 url 的列表，List<URI> urls。我对 url 的列表进行了异步调用，并且还通过添加 Executor service 来确保并发调用。我卡住的地方是，如何将响应列表写入单个文件。

到目前为止我写的代码：

public class httptest{
    
   // Concurrent requests are made in 4 threads
   private static ExecutorService executorService = Executors.newFixedThreadPool(4); 

   //HttpClient built along with executorservice
   private static HttpClient client = HttpClient.newBuilder() 
            .executor(executorService)
            .build();
    
   public static void main(String[] args) throws Exception{
        File fts = new File("P:/Spyder_directory/sample.ts");
        fts.createNewFile();
        List<URI> urls = Arrays.asList(
                         new URI("Url of file 1"),
                         new URI("Url of file 2"),
                         new URI("Url of file 3"),
                         new URI("Url of file 4"),
                         new URI("Url of file 5"));
        
        
        List<HttpRequest> requests = urls.stream()
                .map(HttpRequest::newBuilder)
                .map(requestBuilder -> requestBuilder.build())
                .collect(toList());
        Path file = Path.of("P:/Spyder_directory/sample.ts");
        List<CompletableFuture<HttpResponse<Path>>> results = requests.stream()
                .map(individual_req -> client.sendAsync(individual_req,BodyHandlers.ofFile(file)))
                .collect(Collectors.toList());
   }
}

执行结束时创建的文件sample.ts没有请求的响应。如果你明白我的问题的要点，谁能为这个问题提出替代解决方案。

Answer 1

一种可能性是使用 HttpResponse.BodyHandlers.ofByteArrayConsumer 和 Consumer<Optional<byte[]>> 将字节写入文件。这将使您可以控制文件的打开方式，允许您附加到现有文件而不是每次都创建一个新文件。

请注意，如果这样做，则不应使用 sendAsync，因为请求将同时发送，因此也会同时接收响应。如果您仍想并发发送请求，则需要缓冲响应并在将它们写入文件时强加一些同步。

如何在异步httpclient java 11中将多个异步get请求的响应写入单个文件？

How to write responses of multiple asynchronous get requests to a single file in asynchronous httpclient java 11?

java

httpclient

java-11

asynchttpclient

sendasync