动态加载grid在ExtJs中如何实现，貌似有很多的朋友都不知道吧，下面有个不错的示例，希望对大家有所帮助
Ext3.3中文文档
数据表的结构是：数据表table > 记录record > 字段
store的结构是: Ext.data.Store > Ext.data.Record>Ext.dataDataField
store 首先驱动 DataProxy 加载数据 ，DataProxy加载完成会驱动 DataReader时行解析，最终获得Record对象。
1.bean :
复制代码 代码如下:
package com.leo.bean;
public class Person {
private String name;
private int age;
private String sex;
private String birthday;
public Person(String name, int age, String sex, String birthday) {
super();
this.name = name;
this.age = age;
this.sex = sex;
this.birthday = birthday;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public int getAge() {
return age;
}
public void setAge(int age) {
this.age = age;
}
public String getSex() {
return sex;
}
public void setSex(String sex) {
this.sex = sex;
}
public String getBirthday() {
return birthday;
}
public void setBirthday(String birthday) {
this.birthday = birthday;
}
}
2.action
复制代码 代码如下:
package com.leo.action;
import java.util.ArrayList;
import java.util.List;
import com.leo.bean.Person;
import com.opensymphony.xwork2.ActionSupport;
public class ExtjsAction extends ActionSupport {
private long results;
private List items;
public long getResults() {
return results;
}
public void setResults(long results) {
this.results = results;
}
public List getItems() {
return items;
}
public void setItems(List items) {
this.items = items;
}
public String execute() throws Exception {
this.results = 3;
Person p1 = new Person("张三", 29, "男", "1990-10-22");
Person p2 = new Person("李四", 28, "男", "1991-03-30");
Person p3 = new Person("王五", 27, "女", "1993-08-17");
this.items = new ArrayList
this.items.add(p1);
this.items.add(p2);
this.items.add(p3);
return SUCCESS;
}
}
3.struts-xml
复制代码 代码如下:
<?xml version="1.0" encoding="UTF-8"?>
"http://struts.apache.org/dtds/struts-2.1.dtd">
4.xml
复制代码 代码如下:
<?xml version="1.0" encoding="UTF-8"?>
5.jsp
复制代码 代码如下:
< page language="java" import="java.util.*" pageEncoding="UTF-8"
图示：
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