PHP的下拉菜單

Question

我使用下面的代碼將數據插入到mysql數據庫

<?php 
$refselect = $_POST['refselect']; 
$refname = $_POST['refname']; 
$refemail = $_POST['refemail']; 
$refcnt  = $_POST['refcnt']; 
$refdes  = $_POST['refdes']; 
$referror = $cberror = "<h1>Data not Added</h1><br/><br/><h3>Please Follow The Instructions</h3>"; 
$urefdb  = "INSERT INTO refdb(reftype,refname,refemail,refcnt,refdes) VALUES ('$refselect','$refname','$refemail','$refcnt','$refdes');"; 

include_once("db.php"); 

if ($refselect == "Select Type") die ("$referror"); 

if (empty ($refname)) die ("$referror"); 

if (mysql_query("$urefdb")) 

{ 
echo "<h3>One Record Updated Successfully with the following Details </h3><br/>"; 
echo "Reference Type =$refselect <br/><br/>"; 
echo "Reference Name = $refname <br/><br/>"; 
echo "Reference E-Mail = $refemail <br/><br/>"; 
echo "Reference Description = $refdes <br/><br/>"; 

} 
else 
{ 
echo mysql_error(); 
} 

?> 

「refselect」數據是從下拉頁面菜單來了，我想，當我通過這種形式添加數據另一個表格位於另一個頁面圖片「refname」從數據庫中，當我更新「refdb」時，下拉菜單相應地選取數據

現在該做什麼？

Answer 1

你必須填寫的下拉菜單中，使用MySQL的選擇像其他頁面：

$request = mysql_query("SELECT refselect FROM refdb");

然後通過所有的結果重複填寫您的組合框：

 
echo "<SELECT>"; 
while ($drow = mysql_fetch_assoc($request)) 
{ 
    echo "<OPTION>" . $drow['refselect'] . "</OPTION>"; 
} 
echo "</SELECT>"; 

或者這樣

東西