ajax后台处理返回json值示例代码
编程学习 2021-07-05 10:37www.dzhlxh.cn编程入门
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
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;
}