kit*_*its 13 javascript java ajax jquery json
我正在开发一个j2ee应用程序.在我的应用程序中,我有一个下拉列表(或选择元素).我想用JSON数据填充这个下拉列表作为Ajax响应.
以下是我的代码:
服务器端代码(json_source.java),它生成JSON响应.:
package demo.model;
import java.io.IOException;
import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.ResultSet;
import java.sql.Statement;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import com.google.gson.*;
/**
* Servlet implementation class json_source
*/
public class json_source extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
*/
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException
{
JsonArray data_json=new JsonArray();
Statement st_loginId=null;
ResultSet rs_loginId=null;
try
{
Connection con=null;
Class.forName("oracle.jdbc.OracleDriver");
/* Connection String for "OPERWH"(exadata) Database */
con=DriverManager.getConnection("jdbc:oracle:thin:*************","*****","*****");
con.setAutoCommit(true);
st_loginId=con.createStatement();
rs_loginId=st_loginId.executeQuery("select login_id \"LOGIN ID\" from user_access");
//System.out.println("entered in frame_login_code");
int login_val=0;
JsonObject json_response=new JsonObject();
while(rs_loginId.next())
{
login_val++;
JsonObject json=new JsonObject();
json.addProperty("value", "login"+login_val);
json.addProperty("text", rs_loginId.getString(1));
data_json.add(json);
}
System.out.println(data_json);
json_response.add("aaData", data_json);
response.setContentType("application/Json");
response.getWriter().write(json_response.toString());
System.out.println(json_response);
}
catch(Exception ex)
{
System.out.println("Exception occured during retrieval of Login_Id in ComboBox :"+ex);
ex.printStackTrace();
}
}
/**
* @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
*/
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
}
}
以及通过上述服务器端代码成功生成的JSON数据:
{
"aaData": [{
"value": "login1",
"text": "kapils"
}, {
"value": "login2",
"text": "davidn"
}, {
"value": "login3",
"text": "alanp"
}]
}
以下是生成ajax请求的客户端代码(source1.jsp):
(使用$ .ajax()):
<script type="text/javascript">
$(document).ready(function()
{
$('#id_trial').click(function() {
alert("entered in trial button code");
$.ajax({
type: "GET",
url:"/demo_trial_application/json_source",
dataType: "json",
success: function (data) {
$.each(data.aaData,function(i,data)
{
alert(data.value+":"+data.text);
var div_data="<option value="+data.value+">"+data.text+"</option>";
alert(div_data);
$(div_data).appendTo('#ch_user1');
});
}
});
});
});
</script>
<body>
<div id="div_source1">
<select id="ch_user1" >
<option value="select"></option>
</select>
</div>
<input type="button" id="id_trial" name="btn_trial" value="Trial Button..">
</body>
或使用($ .getJSON()):
$.getJSON("/demo_trial_application/json_source", function (data) {
$.each(data.aaData, function (i, data) {
var div_data = "<option value=" + data.value + ">" + data.text + "</option>";
alert(div_data);
$(div_data).appendTo('#ch_user1');
});
});
现在,当我单击按钮(#id_trial)时,服务器端代码成功执行,因此创建了JSON对象.但我没有在使用jQuery的ajax调用的Success参数的回调函数中获得"JSON响应".
除了jQuery的ajax调用之外,我还尝试使用$.getJSON函数来接收JSON响应.但是我没有得到JSON数据.
所以请告诉我我的代码是否有任何错误,以及如何使用上面的代码获取JSON数据并填充下拉列表.
我想使用ajax响应使用JSON数据填充我的下拉列表. 请帮我解决这个问题......我的申请非常紧急.
raj*_*wat 20
尝试更改jquery方法变量,它可能导致问题(即,您正在使用data来自ajax回调PLUS 的变量然后尝试将其分配给itemjquery方法中的对象 - 更改为obj):
$.ajax({
type: "GET",
url:"/demo_trial_application/json_source",
dataType: "json",
success: function (data) {
$.each(data.aaData,function(i,obj)
{
alert(obj.value+":"+obj.text);
var div_data="<option value="+obj.value+">"+obj.text+"</option>";
alert(div_data);
$(div_data).appendTo('#ch_user1');
});
}
});
});