PHP中基于dropdownlist选择的显示表

选择任何项目名称时不显示我的表。我猜onchange函数工作不正常，但我无法找出问题所在。 代码如下：

   <div class="span6">
                                                <?php $sql = "SELECT * from login_projects WHERE login_id='".$record['login_id']."'";
                                                 $res_sql = mysql_query($sql); ?>
                                                <label>Project Name <span class="f_req">*</span></label>
                                                <!--<input type="text" name="city" class="span8" />-->
                                                <select name="project_name" onchange="get_list_onnet()" id="project_name" class="span8">
                                                <option value="">--</option>
                                                <?php while($rec_sql = mysql_fetch_array($res_sql)){ ?>
                                                <option value="<?php echo $rec_sql['project_id']; ?>">
                                                <?php echo $rec_sql['project_name']; ?></option>
                                                <?php } ?>

                                                </select>
                                            </div>

功能：

<script>
                    function get_list_onnet(){
                    var project_name=$("#project_name").val();
                    $.ajax
                    ({
                    type: "POST",
                    url: "ajax.php",
                    data: {action: 'get_list_onnet',list_onnet:project_name},
                    success: function()
                    {
                    document.getElementById("dt_a").style="block";  
                    $("#dt_a").html(html);
                    } 
                    });
                    };

            </script>
<script>
                $(document).ready(function() {
                    //* show all elements & remove preloader
                    setTimeout('$("html").removeClass("js")',1000);
                });
            </script>

Php页面：

function get_list_onnet(){  
        $list_onnet=$_POST['list_onnet'];

        $sql_list_onnet=mysql_query("SELECT * from projects,project_wise_on_net_codes 
                                where projects.project_id = project_wise_on_net_codes.project_id AND
                                 project_wise_on_net_codes.project_id='$list_onnet'");
        $row1 = mysql_num_rows($sql_list_onnet);                         
        if($row1>0)
        {                        
        echo "<tr><th>id</th><th>Project Name</th><th>Country Code</th><th>On-net prefix</th>
        <th>Action</th></tr>";
        $k = 1; while($row_list_onnet=mysql_fetch_array($sql_list_onnet))
        {
            $project3 = $row_list_onnet['project_name'];
            $countrycode1 = $row_list_onnet['country_code'];
            $prefix1 = $row_list_onnet['on_net_prefix'];
            $id_proj = $row_list_onnet['project_id'];
            $on_prefix = $row_list_onnet['on_net_prefix'];

        echo "<tr><td>".$k."</td><td>".$project3."</td><td>".$countrycode1."</td>
        <td>".$prefix1."</td><td><a href='process/update_process_onnet.php?ID=".$id_proj."&Onnet=".$on_prefix."'>Delete</a></td>
        </tr>";
        $k++;

        }
        }

        else
        {   

            echo "<script>alert('No Record Found')</script>";
        }


        }

---

问题是它总是在else条件下运行，并且表中没有显示任何内容。

停止使用不推荐的mysql_*函数；改用PDO/MySQLi。不要将JavaScript DOM操作与jQuery混合使用；只使用其中一个。问题可能显示在浏览器的开发人员控制台中。无法理解该怎么办。。