index.jsp:
<html>
<head>
<title></title>
<script src="js/jquery.js"></script>
<script type="text/javascript">
function aja(){
$.ajax({
url:"/ajax/aja.do",
type:"get",
dataType:"json",
success:
function(msg){
alert(msg);
}
error:
function(xhr,msg){
alert(msg);
}
})
}
</script>
</head>
<body>
<input type = "button" id="button" value="查看全部信息" onclick="aja()"/>
</body>
</html>ajax.jsp:
<%
List list = new ArrayList();
list = (List)request.getAttribute("li");
%>
action文件:
package com.ajax.action;
import java.util.*;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.apache.struts.action.Action;
import org.apache.struts.action.ActionForm;
import org.apache.struts.action.ActionForward;
import org.apache.struts.action.ActionMapping;
import com.ajax.dao.UserinfoDAO;
public class ajaaction extends Action {
private UserinfoDAO uf;
public void setUf(UserinfoDAO uf) {
this.uf = uf;
}
@Override
public ActionForward execute(ActionMapping mapping, ActionForm form,
HttpServletRequest request, HttpServletResponse response)
throws Exception {
List li = new ArrayList();
li = uf.findAll();
request.setAttribute("li", li);
return mapping.findForward("success");
}
}
为什么一直提示我网页有错误??
大家帮忙看下,
坐等
<html>
<head>
<title></title>
<script src="js/jquery.js"></script>
<script type="text/javascript">
function aja(){
$.ajax({
url:"/ajax/aja.do",
type:"get",
dataType:"json",
success:
function(msg){
alert(msg);
}
error:
function(xhr,msg){
alert(msg);
}
})
}
</script>
</head>
<body>
<input type = "button" id="button" value="查看全部信息" onclick="aja()"/>
</body>
</html>ajax.jsp:
<%
List list = new ArrayList();
list = (List)request.getAttribute("li");
%>
action文件:
package com.ajax.action;
import java.util.*;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.apache.struts.action.Action;
import org.apache.struts.action.ActionForm;
import org.apache.struts.action.ActionForward;
import org.apache.struts.action.ActionMapping;
import com.ajax.dao.UserinfoDAO;
public class ajaaction extends Action {
private UserinfoDAO uf;
public void setUf(UserinfoDAO uf) {
this.uf = uf;
}
@Override
public ActionForward execute(ActionMapping mapping, ActionForm form,
HttpServletRequest request, HttpServletResponse response)
throws Exception {
List li = new ArrayList();
li = uf.findAll();
request.setAttribute("li", li);
return mapping.findForward("success");
}
}
为什么一直提示我网页有错误??
大家帮忙看下,
坐等
但是你dataType:"json",期待返回的类型是json类型的啊,而你action中根本就没返回数据
所以index.jsp会提示错误吧
建议楼主先做一个简单的,一个jsp和action的处理,记得action里要out.write(json);
这样前台才能接收到json