選擇下拉菜單後，如何填充數據庫中的文本字段

Question

這是我的代碼，用於打印擁有ID_no的人的姓氏，該人員位於下拉選項中。

<?php 
$servername = "localhost"; 
$username = "root"; 
$password = ""; 
$dbname = "auth"; 

// Create connection 
$conn = new mysqli("localhost", "root", "", "auth"); 
// Check connection 
if ($conn->connect_error) { 
die("Connection failed: " . $conn->connect_error); 
} 

$sql = "SELECT id_no,fn,sn,mn from employee ORDER BY id_no asc"; 
$result = $conn->query($sql); 
?> 

<select name='id_no' onchange="changeSelect(this.value)"> 
    <?php while ($row = $result->fetch_array()) { ?> 

<option value="<?php echo($row['sn']); ?>" > 

     <?php echo($row['id_no']); ?> 

    </option> 
    <?php } ?> 

</select> 

<div id="demo"></div> 
<script> 
function changeSelect(value) 
{ 
    document.getElementById("demo").innerHTML = value 
} 
</script> 

---

• 我的問題是，我這裏將把<input type="text">啄：所以一旦id_no上被選中它會自動填充該字段我想是在文本框中它的輸出？

Answer 1

<?php echo"<input type=text name=fullname size=50 value=$row[sn],$row[fn],$row[mn] required>"; 

是的，我做到了:)