当结果少于scrollSize设置时，Scroll SearchResponse不可迭代

我有一个这样的循环...

while (true) {
    scrollResp = client.prepareSearchScroll(scrollResp.getScrollId()).setScroll(new TimeValue(600000)).execute().actionGet();

    for (SearchHit hit : scrollResp.getHits()){
        // does this when totalHits > scrollSize, skips it otherwise
    }

    //Break condition: No hits are returned
    if (scrollResp.hits().hits().length == 0) {
        break;
    }
}

查询的scrollSize 设置为500。当scrollResp.getHits().totalHits()

一般来说，我期望每个查询有 > 500 个结果，但我需要它在任何一种情况下都能工作。不确定我是否可能错误地尝试迭代或什么。感谢您的反馈。

我相信你可以做这样的事情：

    // tested also with pageSize = 1
    final int pageSize = 100;
    SearchResponse firstScrollResponse = client
            .prepareSearch("db")
            .setTypes("collection")
            .setSearchType(SearchType.SCAN)
            .setScroll(TimeValue.timeValueMinutes(1))
            .setSize(pageSize)
            .execute()
            .actionGet();

    //get the first page of actual results
    firstScrollResponse = client.prepareSearchScroll(firstScrollResponse.getScrollId())
                                  .setScroll(TimeValue.timeValueMinutes(1))
                                  .execute()
                                  .actionGet();

    while (firstScrollResponse.getHits().hits().length > 0) {

        for (final SearchHit hit : firstScrollResponse.getHits().hits()) {
            // do smth with it
        }

        // update the scroll response to get more entries from the old index
        firstScrollResponse = client.prepareSearchScroll(firstScrollResponse.getScrollId())
                                      .setScroll(TimeValue.timeValueMinutes(1))
                                      .execute()
                                      .actionGet();
    }

技巧是首先从滚动中获取实际结果的第一页，因为第一个搜索响应不包含命中。