Question

我试图根据从数据库中获取的数组，将带有数据库变量的li标签动态地加载到页面中。 li标签的数量将是可变的。我正在努力实现这一目标。请您提供任何帮助

$packages_menu = array();
    $sql = ("SELECT provider, technology_name FROM technology ORDER BY technology_id ASC");
    $result = mysqli_query($conn, $sql);
    while($row = mysqli_fetch_assoc($result)) {

        $packages_menu[] = $row;

        }

我有什么？

<li><?php echo $packages_menu; ?></li>

这是数组

Array
(
    [0] => Array
        (
            [provider] => openserve
            [technology_name] => fibre
        )

    [1] => Array
        (
            [provider] => openserve
            [technology_name] => adsl
        )

    [2] => Array
        (
            [provider] => trusc
            [technology_name] => fibre
        )

    [3] => Array
        (
            [provider] => trusc
            [technology_name] => wireless
        )

    [4] => Array
        (
            [provider] => sonic
            [technology_name] => wireless
        )

)

Answer 1

您有两种解决方案：收集完所有行之后，可以使用另一个循环输出数据。看起来可能是这样的：

foreach($packages_menu as $row) {
   echo '<li>' . $row['provider'] . '</li>';
}

如果收集所有数据并输出不是直接接连发生，这可能是一个很好的解决方案。

如果两者之间没有代码，则无需使用中间的行数组。因此，即使这不是一个很好的解决方案（如第一个解决方案那样，它应保持PHP逻辑并输出更分离的效果），也可以这样做：

$sql = ("SELECT provider, technology_name FROM technology ORDER BY technology_id ASC");
$result = mysqli_query($conn, $sql);
while($row = mysqli_fetch_assoc($result)) {
    echo '<li>' . $row['provider'] . '</li>';
}

Answer 2

Now, I'm not sure exactly how you're trying to display it, but you can see from the code below how to put your data in side of a list of tags. Then just return that list in the middle of your opening and closing `<ul>` tags.

$list_items = ""; //declare outside of the loop    
    while($row = mysqli_fetch_assoc($result)) {
          $list_items += "<li>" . $row['provider'] . "</li>";
          $list_items += "<li>" . $row['technology_name'] . "</li>";                

                }

如何根据数组将带有数据库变量的<li>标签动态加载到我的页面中？

2 个答案: