ajax处理返回json值 ajax后台处理返回json值示例代码-

2013-10-2 23:33 发布者: Timiky 阅读:524


很多的新手朋友们不知道ajax如何处理返回json值,在本文将为大家详细介绍下
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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;
}
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