RavenDB Map Reduce Transform折叠一个独特的字符串列表

时间:2013-01-07 13:27:56

标签: c# linq mapreduce ravendb

我的问题很短暂,我坚持下去。我有一系列文件,如:

{
...,
"Type": [
    "Type1",
    "Type2",
    "Type3",
    "Type4"
  ],
...
}

我认为我需要创建一个map / reduce索引来将这些值组合成一个字符串列表,例如,如果我有这两个文档:

{
...,
"Type": [
    "Type1",
    "Type3"
  ],
...
}

{
...
"Type": [
    "Type2",
    "Type3",
    "Type4"
  ],
...
}

我需要一个包含所有不同值的列表的结果,例如{“Type1”,“Type2”,“Type3”,“Type4”}

我不知道这样做,我尝试了几种方法但没有成功。

有人可以帮助我吗?

谢谢大家

我忘了说,我有大量的数据,比如现在超过1500个文件

public class ProjectTypeIndex : AbstractIndexCreationTask<Project, ProjectTypeIndex.ReduceResult>
{
    public class ReduceResult
    {
        public string ProjectType { get; set; }
    }

    public ProjectTypeIndex()
    {
        Map = projects => from project in projects
                                   select new
                                       {
                                           project.Type
                                       };
        Reduce = results => from result in results
                            group result by result.Type into g 
                            select new
                                {
                                    ProjectType = g.Key
                                };
    }
}

1 个答案:

答案 0 :(得分:2)

大家好我发现了一个解决方案,但我不知道是否是最好的方法,因为我仍然没有字符串列表只有结果,我解决了它通过foreach填充字符串列表,这里是索引创建。如果有人可以检查代码,我将非常感激。

public class ProjectTypeIndex : AbstractIndexCreationTask<Project, ProjectTypeIndex.ReduceResult>
{
    public class ReduceResult
    {
        public string ProjectType { get; set; }
    }

    public ProjectTypeIndex()
    {
        Map = projects => from project in projects
                                   from type in project.Type
                                   select new
                                       {
                                           ProjectType = type
                                       };

        Reduce = results => from result in results
                            group result by result.ProjectType into g
                            select new
                                {
                                    ProjectType = g.Key
                                };
    }
}