2

在这第二天挣扎。我有一个带有标签列表的选择选项表单。我想根据所选标签显示数据库中的文章。我的 index.jsp 中有一个 div“showArticles”,我想在其中显示文章。我为此目的使用 jquery 和 ajax。我编写了名为 test 的 Servlet,我只输出了一个简单的字符串,但我什至无法接收到文本,似乎我的 servlet 从未被调用过。servlet 在名为“servlets”的包中,我使用的是 NetBeans。

这是表格:

                            <select id="b_sub_tag" name="b_sub_tag">
                                <option value='${0}'>Subject</option>
                                <c:forEach var="item" items="${subtagList}">
                                    <option value='${item}'>${item}</option>
                                    <c:set var="i" value='${i+1}'> </c:set>
                                </c:forEach>
                            </select>

这是jQuery:

$(document).ready(function(){       
$("#b_sub_tag").change(function(){       
    var option_value = $(this).children('option:selected').val();

    $.ajax({
                   type: "POST",
                   url: "test",
                   data :"value="+option_value,
                   success: function(html) {
                               $("#showArticles").html(html);
                   }
           });
});

  });

这是 Servlet:

@WebServlet(name = "test", urlPatterns = {"/test"})
public class test extends HttpServlet {
protected void processRequest(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException {
    response.setContentType("text/html;charset=UTF-8");
    PrintWriter out = response.getWriter();
    try {
        /* TODO output your page here. You may use following sample code. */
        //response.getWriter().write("Omething");

    } finally {            
        out.close();
    }
}

@Override
protected void doGet(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException {
    processRequest(request, response);
}

@Override
protected void doPost(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException {
    processRequest(request, response);

    response.getWriter().write("Smething");

}

@Override
public String getServletInfo() {
    return "Short description";
}// </editor-fold>
}

我试过 PrintWriter out = response.getWriter(); out.println("某事");

同样,什么也没有。我在 doPost() 中写入输出,我尝试在 doGet 和 processRequest 中写入,但没有运气。

任何人都知道为什么这不起作用?


问题已解决。我已将代码更改如下:

 $.ajax({
        url : "test",
        type: 'GET',
        ***data: {value:option_value},***

        error : function(jqXHR, textStatus, errorThrown) {
            alert(textStatus);
        },
        success : function(html){
           $("#showArticles").html(html);
           }
        }
    });

小服务程序:

@Override
protected void doGet(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException {
    try {

        final InitialContext context = new InitialContext();
        statefulb = (StatefulBeanRemote)     context.lookup("java:comp/env/StatefulBeanRemote");


        response.setContentType("text/plain");
        response.setCharacterEncoding("UTF-8");
        response.getWriter().write("A text"); 

    } catch (NamingException ex) {
        Logger.getLogger(LoadArticlesByTag.class.getName()).log(Level.SEVERE, null, ex);
    }
}
4

2 回答 2

1

问题已解决。我已将代码更改如下:

$.ajax({ url : "test", type: 'GET', *data: {value:option_value},*

    error : function(jqXHR, textStatus, errorThrown) {
        alert(textStatus);
    },
    success : function(html){
       $("#showArticles").html(html);
       }
    }
});

小服务程序:

@Override protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { try {

    final InitialContext context = new InitialContext();
    statefulb = (StatefulBeanRemote)     context.lookup("java:comp/env/StatefulBeanRemote");


    response.setContentType("text/plain");
    response.setCharacterEncoding("UTF-8");
    response.getWriter().write("A text"); 

} catch (NamingException ex) {
    Logger.getLogger(LoadArticlesByTag.class.getName()).log(Level.SEVERE, null, ex);
}

}

于 2013-04-06T16:18:13.620 回答
0

尝试将您的网址更改为 /test:

$.ajax({
                   type: "POST",
                   url: "/test",
                   data :"value="+option_value,
                   datatype : "html",
                   success: function(html) {
                               $("#showArticles").html(html);
                   }
});
于 2013-04-05T15:08:56.837 回答