我是AJAX / Spring MVC的新手。我有一个显示团队名称的索引页面。在选择一个团队时,我想在同一页面中动态显示玩家名称i:e index.jsp 。这是我的代码。
<script type="text/javascript">
$('#teamName').on('change', function(){
$('#playerNames').show();
function getPlayersList(){
$.ajax({
type : "GET",
url: "Player/getPlayersList",
contentType : "application/json",
dataType : 'json',
contentType: 'application/x-www-form-urlencoded',
success : function(json) {
$.each(json, function(key, value) {
if (value != null && key == 'playersList') {
var players = String(value);
var array = players.split(",");
for (var i = 0; i < array.length; i++) {
if(array[i] != ''){
players.append($('<option/>').val(array[i]).text(array[i]));
}
}
}
});
}
});
}
});
</script>
控制器类
@Controller
public class Player {
MappingJackson2JsonView view;
@ResourceMapping("/")
public View getPlayersList(){
view = new MappingJackson2JsonView();
List<String> playerList = new ArrayList<>();
playerList.add("Shehwag");
playerList.add("Sachin");
playerList.add("Dhoni");
playerList.add("Virat");
view.addStaticAttribute("playersList", playerList);
return view;
}
}
的web.xml
<servlet>
<servlet-name>player</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>player</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
玩家servlet.xml中
<context:component-scan base-package="controller" />
<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix" value="/WEB-INF/" />
<property name="suffix" value=".jsp" />
</bean>
我在哪里做错了?请帮忙。