kristenlee1218 2017-03-24 02:23 采纳率: 0%
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ajax如何获取servlet的数据

servlet已经取出了数据,显示在html上
servlet:
public void doGet(HttpServletRequest req, HttpServletResponse resp)
throws ServletException, IOException {
Connection conn = DBConnectUtils.getConnection();
PreparedStatement pstmt = null;
ResultSet rs = null;

    try {
        String sql = "select id,username from User";
        pstmt = conn.prepareStatement(sql);
        rs = pstmt.executeQuery();

        if (rs != null) {
            while (rs.next()) {
                 req.setAttribute("id", rs.getInt("id"));
                 req.setAttribute("username", rs.getString("username"));
                 req.getRequestDispatcher("showUser.html").forward(req,
                 resp);
            }
        }
    } catch (Exception e) {
        e.printStackTrace();
    } finally {
        DBCloseUtils.closeCSR(conn, pstmt, rs);
    }
}

ajax应该怎么写
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  • kun_hello 2017-03-24 04:02
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    这个你在循环里做的操作还是改改吧 ,另外 你要用ajax传值的话可以使用json字符串 吧要传的值转成json然后再传

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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