【发布时间】:2018-02-18 21:24:23
【问题描述】:
这是我的代码(可怕的):
<?php
include 'connect/con.php';
$result = mysqli_query($con,"SELECT id, vidTitle FROM newsvid");
$result1 = mysqli_query($con,"SELECT imgCover, vidSD FROM newsvid");
$result2 = mysqli_query($con,"SELECT published FROM newsvid");
echo "<table width=\"600\" border=\"1\"><tbody>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo '<td width=\"10%\"><a href="details.php?id='.$row['id'].'">'.$row['id'].'</a></td>';
echo "<td width=\"90%\">" . $row['vidTitle'] . "</td>";
echo "</tr>";
}
echo "</tbody></table>";
echo "<table width=\"600\" border=\"1\"><tbody>";
while($row = mysqli_fetch_array($result1)) {
echo "<tr>";
echo "<td width=\"40%\">" . $row['imgCover'] . "</td>";
echo "<td width=\"60%\">" . $row['vidSD'] . "</td>";
echo "</tr>";
}
echo "</tbody></table>";
echo "<table width=\"600\" border=\"1\"><tbody>";
while($row = mysqli_fetch_array($result2)) {
echo "<tr>";
echo "<td >" . $row['published'] . "</td>";
echo "</tr>";
}
echo "</tbody></table>";
mysqli_close($con);
?>
</body>
</html>
问题是如何在这个布局中显示来自数据库的数据:
--------------------------------
-id----------vidTitle-----------
--------------------------------
-imgCover------vidSD------------
--------------------------------
----------published-------------
所以每次我添加更多数据时,我之前展示的另一个块将在现有块下添加。 ..................................................... ......................................
【问题讨论】:
标签: php html mysqli html-table