問題描述

我正在嘗試制作一個連接到我的數據庫的表，以便我可以使用該數據庫中的信息.

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

我認為我的方法是正確的，但目前表格只顯示第一行.任何人對發生的事情有任何想法嗎?

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>

推薦答案

這可能是因為您沒有正確打開 HTML 表格標簽，這很容易被忽略.還在白色語句中循環結果集:

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

這篇關于使用 MySQLi 和 PHP 的 HTML 表的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！