在Rally WebServices API中,如果我想跨过一个故事层次, 有必要对父故事进行查询, 然后从返回的故事中抓取儿童收藏, 然后对每个孩子进行循环查询, 直到进程达到叶节结果 。
问题 - 是否有一种手法可以不用迭接, 使用回溯 API 中的单一查询来做到这一点?
在Rally WebServices API中,如果我想跨过一个故事层次, 有必要对父故事进行查询, 然后从返回的故事中抓取儿童收藏, 然后对每个孩子进行循环查询, 直到进程达到叶节结果 。
问题 - 是否有一种手法可以不用迭接, 使用回溯 API 中的单一查询来做到这一点?
这是回溯性API的最好特点之一。
假设你们有这种等级制度:
任务12的文件是这样的:
{
ObjectID: 12,
_Type: "Task",
WorkProduct: 777,
_ItemHierarchy: [444, 555, 666, 777, 12],
...
}
因此,当您对一个有数组值的字段提交查询时(如 _Itrole Shirarchy),该查询将匹配数组中的任何成员。
要从 444 中获得所有从您找到的条款, 将包含 < code\ templech Shirarchy: 444 < /code> 。 请看这如何匹配任务 12 的 < code\ templech Shirchy code > 值?
为了从333得到所有从你发现的条款中衍生出来的东西, 将会包括
仅仅为了得到444(所有故事)中的故事,你发现的条款将包括:
_ItemHierarchy: 444,
_Type: "HierarchicalRequirement"
仅为获取 < 坚固> 叶子 < /坚固 > 故事,只需添加以下条款 < code> children: null 。
_ 集成项目结构也一直延伸到集成项目。
已被索引化,因此这些查询应非常高效。
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