如何从动态下拉菜单选择中传递ID？

-1

<?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>

来源

2011-11-28 sid

您可以使用javascript来实现此目的。

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>

实施例：http://jsfiddle.net/brenjt/WaA92/

来源

2011-12-03 05:22:32 brenjt

如何从动态下拉菜单选择中传递ID？

回答

相关问题