如何调用 Ajax 请求以在页面加载时使用 servlet 从数据库中获取记录并显示在 jsp 上

Question

我在数据库中有不同的字段，我需要显示 jsp 中的所有记录，但是当我向 servlet 发出 ajax 请求时，它会将所有结果绑定到所有字段。我希望名字应该与名字绑定，姓氏应该与姓氏绑定。目前它与 frstname 和 firstanamelastname 绑定。

我已尽力解决我的问题，但我认为，问题出在我提出的 ajax 请求上。

   <html>
   <head></head>
   <body>  
    <div class="form-row">
   <div class="col-md-9">
   <div class="form-row pad-left">
   <div class="col-md-6 mb-1">
   <label for="validationCustomUsername"><b>Birth Name:</b> 
   <span id='birthName'></span>                                                     
   </div>
   <div class="col-md-6 mb-3">
  <label for="validationCustomUsername"><b>Initiated Name:</b> 
  <span id='initiatedName'></span> 
  </div>
  </div>

<!-- SECOND ROW STARTS HERE -->
    <div class="form-row pad-left">
      <div class="col-md-6 mb-1">
    <label for="validationCustomUsername"><b>Place Of Birth: </b> 
   <span id='placeOfBirth'></span> 
   </div>
   </div>
    <div class="form-row pad-left">
     <div class="col-md-6 mb-1">
   <label for="validationCustomUsername"><b>Caste:</b>  
  <span id='caste'></span>
    </div>

     </body>

     </html>

          Servlet Code

       protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    response.setContentType("application/json");

    int userID = UserDetails.getInstance().getLastRegisteredID();
    Connection con = DBConnection.connectDB();
    String query = "Select * from PERSONS inner join 
          PersonsDetails on persons.PersonID=PersonsDetails.PersonId "
            + "where PERSONS.PersonID="+userID;
    try {
        ResultSet rs = DBConnection.getDBResultSet(con, query);
        UserDetails user = new UserDetails();
        while(rs.next()) {
            String birthName =rs.getString("BirthName");
            String initiatedName =rs.getString("InitiatedName");
            String placeOfBirth =rs.getString("PlaceOfBirth");
            String caste =rs.getString("Caste");



            response.getWriter().write(birthName);
            response.getWriter().write(initiatedName);
            response.getWriter().write(placeOfBirth);
            response.getWriter().write(caste);

        }
    } catch (SQLException e) {
        e.printStackTrace();
    }finally {
        DBConnection.closeDBConnection(con);
    }


}

         Ajax Call
     function userHomeDetails(){
var username = $('#username');
var url = "http://localhost:8080/IskconDevotteeMarriage/page/UserHome"
    $(document).ready(function(){
    var url=url
        $.post("../UserHomeController", function(responseText) {
            /*document.getElementById('birthName').innerHTML ="birthName"*/
                $('#birthName').html(responseText);
                $('#initiatedName').html(responseText);
                $('#placeOfBirth').html(responseText);
                $('#caste').html(responseText);
                alert(responseText);
        });
        });

}

Answer 1

您可以使用 JSONObject，首先添加 json jar 文件，然后在您的 servlet class 中创建 JSONObject 的对象，如下所示：

JSONObject ob= new JSONObject();

然后把你的parameter像下面这样:

try {
        ob.put("birthName",birthName);
        ob.put("initiatedName",initiatedName);
       ob.put("placeOfBirth",placeOfBirth);
         ob.put("caste",caste);
    } catch (JSONException e) {
        e.printStackTrace();
    }

现在，将上述参数传递给您的 ajax 调用，如下所示：

 response.getWriter().write(obj);

在您的 ajax 调用集 dataType: "json" 和您的 function(responseText) 中，您可以获得如下参数：

document.getElementById('birthName').value = responseText.birthName;//setting values to span with id birthName
document.getElementById('initiatedName').value = responseText.initiatedName;
document.getElementById('placeOfBirth').value = responseText.placeOfBirth;
document.getElementById('caste').value = responseText.caste;

希望对您有所帮助！

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