如何在反应通量流中重新分配变量？

Question

我正在向网络服务发送请求，将结果转换为大的 csv 并将 csv 行保存到数据库中。

由于请求长时间运行（10-20 秒），我想并行化请求。 我将所有数据收集在一个包含转换后的 csv 行的 StringBuilder 中。

问题：如果在 csv 中达到我的 1000 行块，我如何将数据取出以进行持久化，而任何其他并发响应将写入新的 StringBuilder？

因为，流的最终变量无法重新初始化。

final StringBuilder sb = new StringBuilder();
AtomicInteger count = new AtomicInteger();

Flux.fromIterable(requests)
    .flatMap(req -> {
        return webClientService.send(req); //assume long running response
    }, 8) //send 8 requests in parallel, as response takes up to 10s
    .map(rsp -> {
        //convert response to csv values and add to StringBuilder
        int c = addCsv(sb, rsp);
        if (count.addAndGet(c) > 1000) {
            //TODO how can I assign a new StringBuilder,
            //so that all further finished responses will append the csv to the new builder?
            //same problem with the counter.
            
            databaseWriter.write(sb.build()); //writes the content so far to db, but not threadsafe so far
        }
        return c;
    })
    .blockLast();
    

Answer 1

也许您可以尝试完全避免副作用，例如类似于：

.map(x -> toCsv(x))
.reduce((a, b) -> {
    if (length(a) < 1000) {
        return concat(a, b);
    }
    databaseWriter.write(a);
    return b;
})
.doOnNext(x -> databaseWriter.write(x))

Answer 2

在我看来，您可以使用内置运算符来实现相同的结果：

      Flux.fromIterable(requests)
            .flatMap(req -> webClientService
                    .send(req)
                    .subscribeOn(Schedulers.boundedElastic()), 8)// subscribeOn to subscribe from different threads
            .map(resp -> converToCsvLine(resp)) //make some transformations on the respnse
            .window(1000) //split incoming data into 1000 lines
            .flatMap(stringFlux -> stringFlux.collect(Collectors.joining("\n")))// collect last 1000
            .flatMap(s -> Mono.fromRunnable(() -> writeToDb(s))) //do some logic on the collected 1000 lines
            .blockLast();