首先我们来构造几个map集合。
假设如下代码 都是在ssh配置环境下搭建好，(至少struts2开发环境搭建好)
(1).java 代码
下面的student对象包含的字段为
private Long id;
private String num;
private String name;
private String sex;
private Integer age;
Action中的代码
private Map<String,String> map;
private Map<String,Student> studentMap;
private Map<String,String[]> arrayMap;
private Map<String,List<Student>> listMap; // 实现 四个map对象的get 和set方法。
map=new HashMap<String,String>();
map.put("1", "one");
map.put("2", "two");
studentMap=new HashMap<String,Student>();
studentMap.put("student1",new Student(new Long(1),"20034140201","张三1","男",25));
studentMap.put("student2",new Student(new Long(2),"20034140202","张三2","女",26));
studentMap.put("student3",new Student(new Long(3),"20034140202","张三3","男",27));
arrayMap=new HashMap<String,String[]>();
arrayMap.put("arr1", new String[]{"1","2003401","leejie","male","20"});
arrayMap.put("arr2", new String[]{"2","2003402","huanglie","male","25"});
arrayMap.put("arr3", new String[]{"3","2003403","lixiaoning","male","21"});
listMap=new HashMap<String,List<Student>>();
List<Student> list1=new ArrayList<Student>();
list1.add(new Student(new Long(1),"20034140201","张三1","男",25));
list1.add(new Student(new Long(2),"20034140202","张三2","男",25));
list1.add(new Student(new Long(3),"20034140203","张三3","男",25));
listMap.put("class1", list1);
List<Student> list2=new ArrayList<Student>();
list2.add(new Student(new Long(1),"20034140301","李四1","男",20));
list2.add(new Student(new Long(2),"20034140302","李四2","男",21));
list2.add(new Student(new Long(3),"20034140303","李四3","男",22));
list2.add(new Student(new Long(4),"20034140304","李四4","男",23));
listMap.put("class2", list2);
(2).通过上述java代码我们已经构建好了4个map集合。 接下来的重头戏就是如何通过strut2的标签来获取map集合中的值。
<b>1.map中的value为String字符串</b><br>
<s:iterator value="map" id="column">
<s:property value="#column"/><br> //这里获取到的值为key=value 即：键值对
key: <s:property value="key"/><br> //这里的key为内置的，我们只要在value中写上key 即会有值
value:<s:property value="value"/><br> //同样这里的value也为内置的
</s:iterator>
<b>2.map中的value为Student对象</b>
<s:iterator value="studentMap" id="column">
<tr>
<td><s:property value="#column"/></td>
<td><s:property value="key"/></td>
<td><s:property value="value"/></td> //这里的value返回的是一个student对象
<td><s:property value="value.id"/></td> //这里获取student对象中的属性值
<td><s:property value="value.num"/></td>
<td><s:property value="value.name"/></td>
<td><s:property value="value.sex"/></td>
<td><s:property value="value.age"/></td>
</tr>
</s:iterator>
遍历studentMap 还可以用下面方式，跟上面方式效果是一样的
<b>2.map中的value为Student对象</b>
<s:iterator value="studentMap" id="column">
<tr>
<td><s:property value="#column"/></td>
<td><s:property value="key"/></td>
<s:iterator value="value">
<td><s:property value="id"/></td>
<td><s:property value="num"/></td>
<td><s:property value="name"/></td>
<td><s:property value="sex"/></td>
<td><s:property value="age"/></td>
</s:iterator>
</tr>
</s:iterator>
<b>3.map中的value为String数组</b>
<s:iterator value="arrayMap" id="column">
<tr>
<td><s:property value="#column"/></td> <!--同时取出键和值-->
<td><s:property value="value[0]"/></td>
<td><s:property value="value[1]"/></td>
<td><s:property value="value[2]"/></td>
<td><s:property value="value[3]"/></td>
<td><s:property value="value[4]"/></td>
</tr>
</s:iterator>
<b>4.map中的value为list集合</b>
<s:iterator value="listMap" id="column">
<s:set name="total" value="#column.value.size"/> //注意<s:set 标签的用法
<s:iterator value="#column.value" status="s"> //这里#column.value 还是一个student的list集合，因而需要再次迭代一次
<tr>
<s:property value="#s.first"/> //判断是不是集合中的第一个对象
<s:if test="#s.first"><td rowspan="${total}"><s:property value="#column.key"/></td></s:if>
<td><s:property value="id"/></td>
<td><s:property value="num"/></td>
<td><s:property value="name"/></td>
<td><s:property value="sex"/></td>
<td><s:property value="age"/></td>
</tr>
</s:iterator>
</s:iterator>
转自：http://www.2cto.com/kf/201310/252442.html
