-
C#教程之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:
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; });
}
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:
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
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;