我有一個HTML表在PHP鏈接從SQL變量
<?php
$result = mysqli_query($con,"SELECT * FROM recepten ORDER BY datum DESC");
echo "<table border='1' class='ms-list8-main'>
<tr>
<th class='ms-list8-top'>Link</th>
<th class='ms-list8-top'>Naam</th>
<th class='ms-list8-top'>Beschrijving</th>
<th class='ms-list8-top'>Datum</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td><a href='$url'>link</td>";
echo "<td class='ms-list8-even'>" . $row['naam'] . "</td>";
echo "<td class='ms-list8-even'>" . $row['beschrijving'] . "</td>";
echo "<td class='ms-list8-even'>" . $row['datum'] . "</td>";
echo "</tr>";
}
echo "</table>";
$url = $row['url'];
?>
鏈接echo "<td><a href='$url'>link</td>";
不起作用
鏈接的名字是叫「網址」 SQL表
鏈接在每一行不同
我怎樣才能得到這個工作?
不起作用?你遇到了什麼錯誤? – Pascamel
我收到了一個空白頁 –
我先回答了這個問題...如何投票? – Gavin