问题描述

我正在尝试制作一个连接到我的数据库的表，以便我可以使用该数据库中的信息.

I'm trying to make a table connected to my database so I can use the information in that database.

我认为我的方法是正确的，但目前表格只显示第一行.任何人对发生的事情有任何想法吗?

I think I'm on the right way, but currently the table only displays the first row. Anyone got any ideas on what's going on?

         <table id="fairtable">
            <tr>
                <td>Fair Name</td>
                <td>Date</td>
                <td>Are we there?</td>
                <td>Website</td>
            </tr>
            <?php 
            $sql = "SELECT `name`,`date`,`present`,`website` FROM `dates`";
            $results = mysqli_query($conn,$sql);

            while($rowitem = mysqli_fetch_array($results)) {
            echo "<tr>";
                echo "<td>" . $rowitem['name'] . "</td>";
                echo "<td>" . $rowitem['date'] . "</td>";
                echo "<td>" . $rowitem['present'] . "</td>";
                echo "<td>" . $rowitem['website'] . "</td>";
            echo "</tr>";
            }
            ?>
        </table>

推荐答案

这可能是因为您没有正确打开 HTML 表格标签，这很容易被忽略.还在白色语句中循环结果集:

Perhaps this is because you have not opened your HTML table tags correctly, its an easy thing to overlook. Also loop the result set inside a white statement:

$sql = "SELECT `name`,`date`,`present`,`website` FROM `dates`";
$results = mysqli_query($conn,$sql);
echo "<table>"; //begin table tag...
//you can add thead tag here if you want your table to have column headers
 while($rowitem = mysqli_fetch_array($results)) {
    echo "<tr>";
    echo "<td>" . $rowitem['name'] . "</td>";
    echo "<td>" . $rowitem['date'] . "</td>";
    echo "<td>" . $rowitem['present'] . "</td>";
    echo "<td>" . $rowitem['website'] . "</td>";*/
    echo "</tr>";
}
echo "</table>"; //end table tag

这篇关于使用 MySQLi 和 PHP 的 HTML 表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！