如何加载php文件动态加载下拉

Question

我有一个PHP脚本，动态显示从我的数据库

<?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）页面下拉的“课程”选项。

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

我想知道有没有人知道脚本或获取这些数据显示在我的下拉列表中的方式。

感谢所有帮助

Answer 1

你必须运行在select标签

这里内环路更新代码

<?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>';

改变它

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

和其中u希望它显示

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