Dynamic 如何从动态下拉菜单选择传递id？

您可以使用javascript来实现这一点

JAVASCRIPT

<?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>

HTML

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

值1
价值2
价值3
价值4

例如：

<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>