这是我正在处理的代码演示，我在 MySQL 表中有链接，并希望在网站表上显示它们，但我不想显示链接，而是显示与 Excel 中的超链接相同的名称。示例：如果我只使用 href=www.google.com链接我的代码：<!DOCTYPE HTML><html><head><title>Pagina teste</title><div align="center"><font size="+3">Lista de Websites</font></div></head><body><style> table, td, th {border: 1px solid black; border-collapse: collapse; } tr:nth-child(even) {background-color: #dcdddd;}</style><table><tr><th><font size="+2">Website Name</font></th><th><font size="+2">Link</font></th></tr><?php$conn = mysqli_connect("localhost", "root", "", "test");// Check connectionif ($conn->connect_error) {die("Connection failed: " . $conn->connect_error);}if(isset($_GET['order'])){ $order = $_GET['order'];}else{ $order = 'websitename';}if(isset($_GET['sort'])){ $sort = $_GET['sort'];}else{ $sort = 'ASC';}$sql = "SELECT websitename, link FROM websites ORDER BY $order $sort";$result = $conn->query($sql);if ($result->num_rows > 0) {// output data of each rowwhile($row = $result->fetch_assoc()) {echo "<tr><td>" . $row["websitename"]. "</td><td>" . $row["link"]. "</td></tr>";}echo "</table>";} else { echo "0 results"; }$conn->close();?></table></body></html>
3 回答
湖上湖
TA贡献2003条经验 获得超2个赞
中国商店方法中的公牛，用标签标记包围输出：
echo "<tr>
<td>" . $row["websitename"]. "</td>
<td><a href=\"" . $row["link"]. "\">" . $row['websitename'] . "</a></td>
</tr>";
慕哥9229398
TA贡献1877条经验 获得超6个赞
尝试这个
<!DOCTYPE HTML>
<html>
<head>
<title>Pagina teste</title>
<div align="center">
<font size="+3">Lista de Websites</font>
</div>
</head>
<body>
<style>
table, td, th {border: 1px solid black; border-collapse: collapse; }
tr:nth-child(even) {background-color: #dcdddd;}
</style>
<table>
<tr>
<th><font size="+2">Website Name</font></th>
<th><font size="+2">Link</font></th>
</tr>
<?php
$conn = mysqli_connect("localhost", "root", "", "test");
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if(isset($_GET['order'])){
$order = $_GET['order'];
}else{
$order = 'websitename';
}
if(isset($_GET['sort'])){
$sort = $_GET['sort'];
}else{
$sort = 'ASC';
}
$sql = "SELECT websitename, link FROM websites ORDER BY $order $sort";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<tr>
<td>".$row["websitename"]."</td>
<td><a href=".$row["link"]." target='_blank'>" . $row["link"]. "</a></td>
</tr>";
}
echo "</table>";
} else { echo "0 results"; }
$conn->close();
?>
</table>
</body>
</html>
蝴蝶不菲
TA贡献1810条经验 获得超4个赞
您可以在表格内使用锚链接。
<?php
echo "<tr>
<td>" . $row["websitename"]. "</td>
<td><a href=".$row["link"]." target='_blank'>" . $row["link"]. "</a></td>
</tr>";
?>
- 3 回答
- 0 关注
- 120 浏览
添加回答
举报
0/150
提交
取消