How to make a Java HTTP async request?

Question

By executing the code below, I was expecting to have "passed" printed right away on console and then, 3 seconds later, postman's server delayed response ({"delay":"3"}), but instead I'm having to wait 3 seconds to see "passed" printed, after the server response. What am I doing wrong?

    HttpClient client = HttpClient.newHttpClient();
       HttpRequest request = HttpRequest.newBuilder()
             .uri(URI.create("https://postman-echo.com/delay/3"))
             .header("cache-control", "no-cache")
             .header("postman-token", "6bac0475-7cd1-f51a-945f-2eb772483c2c")
             .build();
       client.sendAsync(request, BodyHandlers.ofString())
             .thenApply(HttpResponse::body)
             .thenAccept(System.out::println)
             .join(); 

       System.out.println("passed");

If you want to print before the request completes, do the `println` before you call `join`. — Thilo, Feb 26 '20 at 11:33

score 2 · Answer 1 · answered Feb 26 '20 at 13:26

Thank to @Thilo's hints, I got the behavior I was looking for. What I had in mind before was something more like javascript, where you chain up your fluent calls and move ahead. In Java, you trigger the calls and later on, when you're ready to block, you call get() or join(). So, it became like that:

    HttpClient client = HttpClient.newHttpClient();
       HttpRequest request = HttpRequest.newBuilder()
             .uri(URI.create("https://postman-echo.com/delay/3"))
             .header("cache-control", "no-cache")
             .header("postman-token", "6bac0475-7cd1-f51a-945f-2eb772483c2c")
             .build();

       CompletableFuture cf = client.sendAsync(request, BodyHandlers.ofString())
             .thenApply(HttpResponse::body)
             .thenAccept(System.out::println); 

       System.out.println("passed");

       HttpResponse<String> result = (HttpResponse<String>)cf.join();

How to make a Java HTTP async request?

1 Answers1