如何从动态下拉菜单选择中传递 id？

【问题标题】：How to pass id from a dynamic dropdown menu selection?如何从动态下拉菜单选择中传递 id？
【发布时间】：2011-12-03 13:22:32
【问题描述】：

<?php 

  include ("Includes/DB_config.php");

    //Define dropdown menu
      echo "<select name=\'ProjectName\'>\n";
      echo "<option value=\"NULL\">Select your project</option>\n";

      //Select data from the database
        $result = mysql_query("SELECT ProjectID,ProjectName FROM tb_project");

      //Fetch data & Populate dropdown menu


        while($row=mysql_fetch_array($result))   
          { 
            $pjt_id = $row['ProjectID'];
            $pjt_name = $row['ProjectName'];
            echo "<option value=\"$pjt_id\">$pjt_name</option>\n";    
          }

      echo "</select>";


?>

<a href="/www/gdis_sys_test/Project_Setup_Display.php?id=<?php echo $pjt_id ?>">Edit</a>

【问题讨论】：

标签： dynamic drop-down-menu selection

【解决方案1】：

您可以使用 javascript 来实现这一点。

JAVASCRIPT

function upperCase(x)
{
    var e = document.getElementById("editButton");
    e.href = "/www/gdis_sys_test/Project_Setup_Display.php?id=" + x;
}

HTML

<select onchange="upperCase(this.value)">
    <option value="1">Value 1</option>
    <option value="2">Value 2</option>
    <option value="3">Value 3</option>
    <option value="4">Value 4</option>
</select>
<a id="editButton" href="/www/gdis_sys_test/Project_Setup_Display.php?id=1">Edit</a>

示例：http://jsfiddle.net/brenjt/WaA92/

【讨论】：