如何加載php文件動態加載下拉

Question

我有一個PHP腳本，動態顯示從我的數據庫

<?php 


    $db_host = 'localhost'; 
    $db_user = 'root'; 
    $db_pass = ''; 
    $db_name = ''; 

    $con = mysqli_connect($db_host,$db_user,$db_pass, $db_name); 
    if (!$con) { 
    die('Could not connect: ' . mysqli_error($con)); 
    } 


    $sql = "SELECT courseID, name FROM courses"; 

    $result = mysqli_query($con, $sql) or die("Error: ".mysqli_error($con)); 

    while ($row = mysqli_fetch_array($result)) 
    { 
    echo'<option value="'.$row['courseID'].'">'.$row['name'].'</option>'; 
    } 


    ?> 

我在我的HTML（scorecard.php）頁面下拉的「課程」選項。

<form> <select id="selectCourse" > <option value = "">Select Course</option></select></form> 

我想知道有沒有人知道腳本或獲取這些數據顯示在我的下拉列表中的方式。

感謝所有幫助

Answer 1

你必須運行在select標籤

這裏內環路更新代碼

<?php 

$db_host = 'localhost'; 
$db_user = 'root'; 
$db_pass = ''; 
$db_name = ''; 

$con = mysqli_connect($db_host,$db_user,$db_pass, $db_name); 
if (!$con) { 
die('Could not connect: ' . mysqli_error($con)); 
} 


$sql = "SELECT courseID, name FROM courses"; 

$result = mysqli_query($con, $sql) or die("Error: ".mysqli_error($con)); 

?> 

<form> 
    <select id="selectCourse" > 
    <option value = "">Select Course</option> 
    <?php while ($row = mysqli_fetch_array($result)) 
    { 
     echo'<option value="'.$row['courseID'].'">'.$row['name'].'</option>'; 
    } ?> 

</select> 

Answer 2

echo'<option value="'.$row['courseID'].'">'.$row['name'].'</option>';

改變它

$courses[] = '<option value="'.$row['courseID'].'">'.$row['name'].'</option>';

和其中u希望它顯示

foreach($courses as $c){ 
    echo $c; 
}