ajax后台处理返回json值示例代码

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复制代码代码如下:

 
public ActionForward xsearch(ActionMapping mapping, ActionForm form, 
HttpServletRequest request, HttpServletResponse response) 
throws Exception { 
String parentId = request.getParameter("parentId"); 
String supplier = request.getParameter("supplier"); 
List itemList = new ArrayList(); 
if(parentId.equals("")){ 
parentId="0"; 
} 
Map map=new TawApTreeServlet().getTypeList(parentId, supplier); 

for (Iterator rowIt = map.keySet().iterator(); rowIt.hasNext();) { 
String id = (String) rowIt.next(); 
TawCommonsUIListItem uiitem = new TawCommonsUIListItem(); 
uiitem.setItemId(id); 
uiitem.setText((String)map.get(id)); 
uiitem.setValue(id); 
itemList.add(uiitem); 
} 

response.setContentType("text/xml;charset=UTF-8"); 

// 返回JSON对象 
response.getWriter().print(JSONUtil.list2JSON(itemList)); 
return null; 
} 