php - 通过 php 和 jquery 从 mysql 数据库获取结果时出现问题

Question

我尝试实现以下几点:

我通过 php 生成一个包含两个字段(关于培训类(class))的表格 - 类(class)名称和位置。 该表有几行，我实现了这样的:

        <table border='0' id='testing'>
            <thead>
                <tr class='small_title_bold'>
                    <th>Formation</th><th align='center'>Niveau</th><th align='center'>Ville</th>
                </tr>
            </thead>
            <tbody>        
              <?php
              db_connect();

              if (isset($etablissement)){
                $sql_find_id_etablissement = "select id_etablissement, subdesignation, address1, address2, zip, city, town, department, phone, fax, www from etablissements where designation = '$etablissement' and id_region='$id_region'";
                $result_find_id_etablissement = mysql_query($sql_find_id_etablissement, $connection) or die('error');
                while ($row_find_id_etablissement = mysql_fetch_row($result_find_id_etablissement)){
                $id_etablissement  = $row_find_id_etablissement[0];
                $subdesignation = $row_find_id_etablissement[1];            
                $address1 = $row_find_id_etablissement[2];
                $address2 = $row_find_id_etablissement[3];          
                $zip = $row_find_id_etablissement[4];
                $city = $row_find_id_etablissement[5];
                $town = $row_find_id_etablissement[6];          
                $department = $row_find_id_etablissement[7];            
                $phone = $row_find_id_etablissement[8];
                $fax = $row_find_id_etablissement[9];
                $www = $row_find_id_etablissement[10];

                print "<tr class='small_subtitle_bold'><td colspan='3' align='center'>".$id_etablissement."-". $subdesignation."</td></tr>";        
                $sql_find_master = "select id_master, designation, master_level from masters where id_etablissement = '$id_etablissement' order by designation, master_level";
                $result_find_master = mysql_query($sql_find_master, $connection) or die('error');

                $count=mysql_num_rows($result_find_master);

                for($i = 0; $i < $count; $i++) {                
                    $row_find_master = mysql_fetch_row($result_find_master);
                    $id_master = $row_find_master[0];   
                    $designation = $row_find_master[1]; 
                    $level = $row_find_master[2];
                    if($i % 2) {
                        print "<tr class='univ' bgcolor='#A4D2FD'>";
                        print "<td><a class='univ' href='master-details.php?master-id=".$id_master."&etablissement-id=".$id_etablissement."' onclick='popup('master-details.php?master=".$designation."')>".$designation."</a></td><td class='small_text' align='center'>".$level."</td><td class='small_text' align='center'>".$town."</td>";
                        print "</tr>";
                    }
                    else {
                        print "<tr class='univ' bgcolor='#FFFFFF'>";
                        print "<td><a class='univ' href='#' onClick='javascript:swapContent(".$id_master.",".$id_etablissement.")'>".$designation."</a></td><td class='small_text' align='center'>".$level."</td><td class='small_text' align='center'>".$town."</td>";
                        print "</tr>";          
                    }
                }
            }
        }
        ?>
        </tbody>
    </table>

当用户单击一行时，我想运行一个 mysql 查询并在同一页面的单独位置显示结果(类(class)的详细信息)，而无需重新加载页面。

我在 head 部分放置了以下脚本:

<script language="JavaScript" type="text/javascript">
function swapContent(id_master, id_etablissement) {
var url = "master-details.php";
$.post(url, {id_master: id_master, id_etablissement: id_etablissement} ,function(data) {
          var content = $( data ).find( '#content', function () {alert('Query was performed.');} );
      $( "#queryResults" ).empty().append( content );
});
}
</script>

我希望我的 jquery 脚本使用参数 id_master 和 id_etablissement 运行名为 master-details 的 php 文件，并在名为 queryResults 的 DIV 部分中返回结果。

但是，当我点击表格中的一行时，没有任何反应。

如有任何帮助，我们将不胜感激。

Answer 1

像这样:

jQuery("tr").live("click", function(){
   $.post///what you already have.
});

你只需要一种方法来告诉它在你点击时做某事。