Question

我编辑我的帖子，因为假设3方法从数据库表中返回行表是：docTags，标签和docs docTags是中间表一个文档（文档）可以有许多标记，一个标记可以属于许多文档我怎么得到：TagKey =“123”其中docTags.Add（新配置（1,1））; 课程 {

    static void Main(string[] args)
    {

    }
}

public class Configuration
{
    public  int  DocID { get; set; }
    public int TagID { get; set; }
    public string Name { get; set; }
    public string DocKey { get; set; }
    public string TagKey { get; set; }

    public Configuration(int _docId,string _name, string _docKey)
    {
        DocID = _docId;
        Name = _name;
        DocKey = _docKey;
    }
    public Configuration(int _tagId,string name, string _dockey,string _tagKey)
    {
        TagID = _tagId;
        Name = name;
        TagKey = _tagKey;
    }
    public Configuration(int _tagId, int _docId)
    {
        TagID = _tagId;
        DocID = _docId;
    }
    public static List<Configuration> getDocType()
    {
        List<Configuration> docs = new List<Configuration>();
        docs.Add(new Configuration(1,"contract", "xxx"));
        docs.Add(new Configuration(2,"Action", "yyy"));
        return docs;
    }

    public static List<Configuration> getTagName()
    {
        List<Configuration> tags = new List<Configuration>();
        tags.Add(new Configuration( 1,"contractid", "123"));
        tags.Add(new Configuration(2,"SuperDuper", "332123"));
        tags.Add(new Configuration(22, "rama", "yyy"));
        tags.Add(new Configuration(32, "aktiv",  "123456"));
        tags.Add(new Configuration(42, "data ","xx764fhx"));
        return tags;

    }
    public static List<Configuration> getDocTags()
    {
        List<Configuration> docTags = new List<Configuration>();
        docTags.Add(new Configuration(1, 2));
        docTags.Add(new Configuration(1, 1));
        docTags.Add(new Configuration(1, 22));
        docTags.Add(new Configuration(2, 2));
        docTags.Add(new Configuration(2, 32));
        return docTags;
    }        
}

}

Answer 1

您需要Join：

var commonUsers = (from t in tags
                  join d in docs on t.DocKey equals d.DocKey
                  select t)
                  .Distinct();  // to remove duplicates

或另一种方式

var commonUsers = tags.Where(t => docs.Any(d => d.DocKey == t.DocKey));

第二种方法较短，但第一种方法可能表现更好。

Answer 2

var newList = docs.Intersect(tags);

您可能需要覆盖Equals运算符才能生效。

Answer 3

以下是使用Intersect执行此操作而不重载Equals方法的方法。您需要创建一个新类作为IEqualityComparer

传入

class ConfigurationComparer : IEqualityComparer<Configuration>
{
    //You can change which string comparer fits your needs best.
    private readonly StringComparer comparer = StringComparer.CurrentCulture;

    public bool Equals(Configuration x, Configuration y)
    {
        return comparer.Equals(x.DocKey,y.DocKey);
    }

    public int GetHashCode(Configuration obj)
    {
         return comparer.GetHashCode(obj.DocKey);
    }

}

然后你只需做

var newList = tags.Intersect(docs, new ConfigurationComparer());

Answer 4

您可以使用Contains方法并将其应用于具有lambda表达式的过滤器，以使用一个或多个条件从另一个列表中提取项目，以获取样本：

var result = needThis.Where(x => needThis2.Contains(x)).ToList();

或者使用扩展方法Intersect作为示例：

var result = needThis.Intersect(needThis2);

相交两个列表＆lt;＆gt;

4 个答案: