jianbo.hc 发表于 2013-1-29 13:36:02

Jquery异步请求

<script type="text/javascript">

jQuery(document).ready(function(){
jQuery("#button_submit").click(function(){

var app=jQuery("#app").val();
var host=jQuery("#hostname").val();
var paths=jQuery("#path").val();
   
var uri4="logQueryAjax.htm?app="+app+"&host="+host+"&path="+paths;
       jQuery.getJSON(uri4,function(data){
                jQuery.each(data, function(i, field){
    var oDiv = document.getElementById("processor");
                  oDiv.innerHTML = "processor is "+field;
                        
});
            });

   
   });
});
</script>


@RequestMapping("logQueryAjax.htm")
public String ajaxRespone(final HttpServletRequest request,
final HttpServletResponse response, ModelMap result) {
   String selectedApp = request.getParameter("app");
   String json = "";
/****/
   json = JsonUtils.jsonEncode(logPathes);将对象转为json串
   response.setContentType("application/json");
   PrintWriter out = null;
   try {
out = response.getWriter();
} catch (IOException e) {
    logger.info("io exception ", e);
}
   out.print(json);
   out.flush();
   out.close();
   return null;
}
注释:点击按钮,异步请求action,将结果使用json回传显现在页面上。
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