问题描述
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));
}
}
现在,我有网址列表List<URI> urls
。我对网址列表进行了异步调用,并通过添加Executor service
确保并发调用。
我受困的地方是如何将响应列表写入单个文件。
我到目前为止编写的代码:
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());
}
}
在执行结束时创建的文件sample.ts
没有所请求的响应。
如果您了解我的问题的要点,那么谁能为该问题提供替代解决方案。
解决方法
一种可能性是将HttpResponse.BodyHandlers.ofByteArrayConsumer与将字节写入文件的Consumer<Optional<byte[]>>
一起使用。这样一来,您可以控制文件的打开方式,从而可以将其追加到现有文件中,而不必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,因此也将同时接收到响应。如果您仍要同时发送请求,则需要缓冲响应并在将其记入文件时进行一些同步。