使用Linq按属性对对象进行分组列表？

Question

我有一个对象:

public class SiteInfo
        {
            public string Title { get; set; }
            public string URL { get; set; }
            public string Type { get; set; }     
        }

我正在使用以下代码创建一个列表： var sites = new List();

        foreach (SPWeb site in web.GetSubwebsForCurrentUser())
        {
            string sitetype = getConfigurationKey(site, "siteType");
            //If sites have a site type then add to list
            if (sitetype != "*ERROR*" && sitetype != "*KEYNOTFOUND*")
            {
                SiteInfo s = new SiteInfo();
                s.Title = site.Title;
                s.URL = site.Url;
                s.Type = sitetype;

                sites.Add(s);
            }
        }
        //sort list by type
        sites.Sort((x, y) => string.Compare(x.Type, y.Type));

        // serialize and send..    
        JavaScriptSerializer serializer = new JavaScriptSerializer();
        StringBuilder sbJsonResults = new StringBuilder();
        serializer.Serialize(sites, sbJsonResults);
etc.....

不过我想做的是在将它们序列化之前，按类型对站点进行分组。是否可以使用LINQ或其他方法实现这一点。

- BlueBird

---

我会先查看所有LINQ方法的列表，看看它们是否与分组功能有关。如果这样做，GroupBy应该会引起你的注意。-1分，因为明显没有付出研究的努力。 - Servy

Answer 1

听起来你想要类似于这样的东西：

// No need to sort sites first
var grouped = sites.OrderBy(x => x.Type)
                   .GroupBy(x => x.Type);

那么只需对 grouped 进行序列化。但是，我不知道在 JSON 中 IGrouping 会长成什么样子......而且每种情况下都会出现类型。你可能想要这样做：

var grouped = sites.OrderBy(x => x.Type)
                   .GroupBy(x => x.Type)
                   .Select(g => new { Type = g.Key,
                                      Sites = g.Select(site => new {
                                                           site.Title,
                                                           site.URL
                                                       } });

我认为那会给你一个更好的JSON结构。

- Jon Skeet

Answer 2

这个

标签

var sites = new List<SiteInfo>()
{
    new SiteInfo(){Title="1",Type="a",URL="http://aaaa"},
    new SiteInfo(){Title="2",Type="b",URL="http://bbbb"},
    new SiteInfo(){Title="3",Type="a",URL="http://aaaa"},
    new SiteInfo(){Title="4",Type="b",URL="http://bbb"},
};

var json = new JavaScriptSerializer().Serialize(sites.GroupBy(s => s.Type));

将产生

[
    [{"Title":"1","URL":"http://aaaa","Type":"a"},{"Title":"3","URL":"http://aaaa","Type":"a"}],
    [{"Title":"2","URL":"http://bbbb","Type":"b"},{"Title":"4","URL":"http://bbb","Type":"b"}]
]

或者

var json = new JavaScriptSerializer().Serialize(sites.GroupBy(s => s.Type)
                                                      .ToDictionary(x=>x.Key,x=>x));

会产生

{
    "a":[{"Title":"1","URL":"http://aaaa","Type":"a"},{"Title":"3","URL":"http://aaaa","Type":"a"}],
    "b":[{"Title":"2","URL":"http://bbbb","Type":"b"},{"Title":"4","URL":"http://bbb","Type":"b"}]
}

- I4V

Answer 3

这是一个简单的控制台应用程序，可以实现你想要的功能：

static void Main(string[] args)
{
    List<SiteInfo> sites = new List<SiteInfo>()
    {
        new SiteInfo() { Title = "Site A", Type = "Whatever 2" },
        new SiteInfo() { Title = "Site B", Type = "Whatever 1" },
        new SiteInfo() { Title = "Site C", Type = "Whatever 1" },
        new SiteInfo() { Title = "Site D", Type = "Whatever 3" },
        new SiteInfo() { Title = "Site E", Type = "Whatever 3" }
    };

    var sitesGroupedByType =
        sites.GroupBy(s => s.Type).Select(g => new { Type = g.Key,
                                    Sites = g.Select(site => new
                                    {
                                          site.Title,
                                          site.URL
                                    })});

    foreach (var siteTypeGroup in sitesGroupedByType.OrderBy(g => g.Type))
    {
        foreach(var site in siteTypeGroup.Sites)
        {
            Console.WriteLine(string.Format("Type => {0}, Title => {1}",
                              siteTypeGroup.Type, site.Title));
        }
    }

    Console.ReadKey();
}

输出：

Type => Whatever 1, Title => Site B
Type => Whatever 1, Title => Site C
Type => Whatever 2, Title => Site A
Type => Whatever 3, Title => Site D
Type => Whatever 3, Title => Site E

- Leniel Maccaferri