Question

我正在使用引导程序4和mysql，并试图将包含多个记录的表显示到列表组中，这样我应该能够单击每个记录并导航到另一页。我希望您能提供有关代码的帮助。提前Tnx。

Answer 1

尝试以下代码：

        <!-- Brand and toggle get grouped for better mobile display -->
        <div class="navbar-header">
            <button type="button" class="navbar-toggle" data-toggle="collapse" data-target="#bs-example-navbar-collapse-1">
                <span class="sr-only">Toggle navigation</span>
                <span class="icon-bar"></span>
                <span class="icon-bar"></span>
                <span class="icon-bar"></span>
            </button>
            <a class="navbar-brand" href="index.php">Home</a>
        </div>


        <!-- Collect the nav links, forms, and other content for toggling -->
        <div class="collapse navbar-collapse" id="bs-example-navbar-collapse-1">
            <ul class="nav navbar-nav">

              <?php 

$query = "SELECT * FROM categories LIMIT 3";
$result = mysqli_query($connection,$query);

while($row = mysqli_fetch_assoc($result)) {
   $title = $row['title'];
   $id = $row['id'];

    echo "<li><a href='category.php?category={$id}'>{$title}</a></li>";
}

?>



            </ul>
        </div>
        <!-- /.navbar-collapse -->
    </div>
    <!-- /.container -->
</nav>

Answer 2

放置此数据或通过使其成为<table><tr>..</tr></table>

中的函数或类来调用它

<?php 

    $query = "SELECT * FROM data";
    $result = mysqli_query($con, $query);

    while($row = mysqli_fetch_assoc($result)) {
       $title = $row['title'];
       $link = $row['link'];

        echo "<td><a href='category.php?data={$link}'>{$title}</a></td>";
    }

?>

如何在引导程序中将mysql表记录显示为导航

2 个答案: