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使用mysql-php和ajax的动态导航_Php_Jquery_Mysql_Ajax - Fatal编程技术网

使用mysql-php和ajax的动态导航

使用mysql-php和ajax的动态导航,php,jquery,mysql,ajax,Php,Jquery,Mysql,Ajax,我已经使用mysql php创建了导航。现在我正在尝试使用ajax将值传递到另一个页面。请帮助我 <ul id="all_navs"> <?php $sql = mysql_query("select * from sub_category_master where main_category =1"); while($row = mysql_fetch_array($sql)) {?> <li id="s1"> <

我已经使用mysql php创建了导航。现在我正在尝试使用ajax将值传递到另一个页面。请帮助我

   <ul id="all_navs">

  <?php $sql = mysql_query("select * from sub_category_master where main_category =1");
   while($row = mysql_fetch_array($sql))
   {?>

   <li id="s1">

   <a href="#tab_deals" onclick="showUsers('<?php echo $row['sub_category_id']; ?>')"  id="insert">
   <?php echo $row['sub_category_name']; ?></a>




   </li>

   <?php }?> 


   </ul>

如果我理解你的问题,为什么不试试这样的方法:

      <ul id="all_navs">
      <?php $sql = mysql_query("select * from sub_category_master where main_category =1");
      while($row = mysql_fetch_array($sql)){
      echo '<li id="s1">';
      echo '<a href="#tab_deals" onclick="showUsers('.$row['sub_category_id'].')"  id="insert">';
      echo $row['sub_category_name'].'</a>';
      echo '</li>';
      }
      ?> 
      </ul>

如果您告诉我们问题出在哪里,会有更大帮助。

是我还是您没有超链接?我看到的只是同一页上一个ID的链接,告诉它调用showUsersvalue。