List<>中Find的用法小結。本站提示廣大學習愛好者：（List<>中Find的用法小結）文章只能為提供參考，不一定能成為您想要的結果。以下是List<>中Find的用法小結正文
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:
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
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; });
}
備注：在list和array聚集中搜刮元素常常應用該辦法，重要技巧是泛型拜托
list用法留意：假如增長一個對象，必需從新new一個對象，看上面的例子：
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);
}
下面有差別嗎？在輸入list的值是有差別了，第一個list外面寄存的都是一樣的，成果和最初一個一樣，都是統一小我對象，第二個list到達預期的後果，存儲分歧的人對象。這是為何？緣由很簡略，一個list對象中存儲的都是對一個共有的對象停止援用，所以以最初轉變的值為准。
對象的排序：本文重要論述對存儲datatable的list停止按TableName排序
1、新建一個sort類
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; //年夜於前往1
else if (Convert.ToInt32(obj1.TableName.Substring(2,tableNameLength1-2))<Convert.ToInt32(obj2.TableName.Substring(2,tableNameLength2-2)))
return -1; //小於前往-1
else
return 0; //相等前往0
}
}
}
2 排序：
List<DataTable> ldt = new List<DataTable>();
SceneSort ss=new SceneSort ();
tablelist.sort(ss);便可對list停止排序；
