jsp Ajax获取工作日志不工作_Ajax_Jsp_Tomcat6

jsp Ajax获取工作日志不工作

ajax jsp

jsp Ajax获取工作日志不工作,ajax,jsp,tomcat6,Ajax,Jsp,Tomcat6,我需要刷新浏览器的某些部分。因此，我使用Ajax。当我使用GET方法时，我可以在服务器上看到数据，但当我使用POST时，我无法看到数据。有什么建议吗 -----------searchPan.jsp的代码------------------------- <html> <% try { System.out.println("url is " + request.getRequestURI()); System.out.pr

我需要刷新浏览器的某些部分。因此，我使用Ajax。当我使用GET方法时，我可以在服务器上看到数据，但当我使用POST时，我无法看到数据。有什么建议吗

-----------searchPan.jsp的代码-------------------------

<html>
  <%
  try
  {
          System.out.println("url is " + request.getRequestURI());
          System.out.println("pan is " + request.getQueryString());
          System.out.println("pan is " + request.getParameter("pan"));
  }
  catch(Exception E)
  {
          System.out.println("Error happened " + E);
  }
          //promo_card cardobj = new promo_card();
          //out.println(cardobj.getData());
  %>
</html>

xmlhttp=new XMLHttpRequest();
xmlhttp.onreadystatechange=function()
{
 if(xmlhttp.readyState==0)
         document.getElementById('status').innerHTML='Uninitialized';
 else if(xmlhttp.readyState==1)
         document.getElementById('status').innerHTML='Loading...';
 else if(xmlhttp.readyState==2)
         document.getElementById('status').innerHTML='Loaded';
 else if(xmlhttp.readyState==3)
         document.getElementById('status').innerHTML='Interactive';
 else if(xmlhttp.readyState == 4 && xmlhttp.readyState == 200)
         document.getElementById("div1").innerHTML = xmlhttp.responseText;
}
if(xmlhttp)
{
  var queryString = 'searchPan.jsp?pan=' + document.getElementById('pan_number').value;
  xmlhttp.open("GET", queryString, true);
  xmlhttp.send();
}

xmlhttp.open("POST", "searchPan.jsp", true);
  var queryString = 'searchPan.jsp';
  queryString = 'pan=' + document.getElementById('pan_number').value;
  alert(queryString);
  xmlhttp.send(queryString);

当我使用GET方法时，我得到以下输出

----------------------邮政编码----------------------------

<html>
  <%
  try
  {
          System.out.println("url is " + request.getRequestURI());
          System.out.println("pan is " + request.getQueryString());
          System.out.println("pan is " + request.getParameter("pan"));
  }
  catch(Exception E)
  {
          System.out.println("Error happened " + E);
  }
          //promo_card cardobj = new promo_card();
          //out.println(cardobj.getData());
  %>
</html>

xmlhttp=new XMLHttpRequest();
xmlhttp.onreadystatechange=function()
{
 if(xmlhttp.readyState==0)
         document.getElementById('status').innerHTML='Uninitialized';
 else if(xmlhttp.readyState==1)
         document.getElementById('status').innerHTML='Loading...';
 else if(xmlhttp.readyState==2)
         document.getElementById('status').innerHTML='Loaded';
 else if(xmlhttp.readyState==3)
         document.getElementById('status').innerHTML='Interactive';
 else if(xmlhttp.readyState == 4 && xmlhttp.readyState == 200)
         document.getElementById("div1").innerHTML = xmlhttp.responseText;
}
if(xmlhttp)
{
  var queryString = 'searchPan.jsp?pan=' + document.getElementById('pan_number').value;
  xmlhttp.open("GET", queryString, true);
  xmlhttp.send();
}

xmlhttp.open("POST", "searchPan.jsp", true);
  var queryString = 'searchPan.jsp';
  queryString = 'pan=' + document.getElementById('pan_number').value;
  alert(queryString);
  xmlhttp.send(queryString);

添加这一行，上次它帮助了我

xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");

请尝试以下代码

xmlhttp.open("POST", "searchPan.jsp", true);
xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
var queryString = 'pan=' + document.getElementById('pan_number').value;
xmlhttp.send(queryString);

哇！成功了。url为/cardsit/promo/card/searchPan.jsp pan为空pan为123456789