Question

我有一个PHP脚本，可以动态显示我的数据库中的“课程”选项

<?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>';
    }


    ?>

我的HTML（scorecard.php）页面中有一个下拉列表。

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

我想知道是否有人知道在我的下拉列表中显示此数据的脚本或方法。

感谢您的帮助

Answer 1

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

将其更改为

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

以及你希望它显示在哪里

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

Answer 2

您必须在while代码

中运行select循环

这是更新后的代码

<?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>

如何加载php文件以动态加载下拉列表

2 个答案: