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List<>中Find的用法小结

  • 时间:2021-03-20 03:12 编辑: 来源: 阅读:
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摘要:List<>中Find的用法小结
I've been looking for help on how to find objects in Generics with List.Find() method .... and ... take a look what I have found. In the follow example, I created a simple class:
[u]复制代码[/u] 代码如下:
public class Person {        private int _id;        private string _name;        public int ID {  get{ return _id;} set{ _id = value;}}        public int Name {  get{ return _name;} set{ _name= value;}}        public Person(int id, string name)        {              _id = id;              _name = name;        } }
In the example, there's a simple class with two private attributes. Now we're going to create a typed List of this object and take advantage of the Find() method
[u]复制代码[/u] 代码如下:
public void CreateAndSearchList() {       //create and fill the collection       List<Person> myList = new List<Person>();       myList.Add(new Person(1, "AndreySanches"));       myList.Add(new Person(2, "AlexandreTarifa"));       myList.Add(new Person(3, "EmersonFacunte"));      //find a specific object      Person myLocatedObject = myList.Find(delegate(Person p) {return p.ID == 1; }); }
备注:在list和array集合中搜索元素经常使用该方法,主要技术是泛型委托 list用法注意:如果增加一个对象,必须重新new一个对象,看下面的例子:
[u]复制代码[/u] 代码如下:
person p=new pewson(); for(int i=0;i<5;i++) { p.ID=i; p.Name="xxxx"; list.add(p); } for(int i=0;i<5;i++) { person p=new person(); p.ID=i; p.Name="xxxx"; list.add(p); }
上面有区别吗?在输出list的值是有区别了,第一个list里面存放的都是一样的,结果和最后一个一样,都是同一个人对象,第二个list达到预期的效果,存储不同的人对象。这是为什么?原因很简单,一个list对象中存储的都是对一个共有的对象进行引用,所以以最后改变的值为准。 对象的排序:本文主要阐述对存储datatable的list进行按TableName排序 1、新建一个sort类
[u]复制代码[/u] 代码如下:
public class SceneSort:IComparer<DataTable>     {        public int Compare(DataTable obj1, DataTable obj2)        {            int tableNameLength1;            int tableNameLength2;            if (obj1 == null)            {                if (obj2 == null)                    return 0;                else                    return -1;            }            else            {                if (obj2 == null)                    return 1;                else                {                    tableNameLength1=obj1.TableName.Length;                    tableNameLength2=obj2.TableName.Length;                    if (Convert.ToInt32(obj1.TableName.Substring(2,tableNameLength1-2))>Convert.ToInt32(obj2.TableName.Substring(2,tableNameLength2-2)))                        return 1;                            //大于返回1                    else if (Convert.ToInt32(obj1.TableName.Substring(2,tableNameLength1-2))<Convert.ToInt32(obj2.TableName.Substring(2,tableNameLength2-2)))                        return -1;                           //小于返回-1                    else                        return 0;                            //相等返回0                }            }        }
2 排序: List<DataTable> ldt = new List<DataTable>(); SceneSort ss=new SceneSort (); tablelist.sort(ss);即可对list进行排序;
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