如何在网站上显示存储在 MySQL 数据库表中的网站链接，但显示为要点击的超链接？
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
$conn = mysqli_connect("localhost", "root", "", "test");
// Check connection
if ($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>" . $row["link"]. "</td>
</tr>";
}
echo "</table>";
} else { echo "0 results"; }
$conn->close();
?>
</table>
</body>
</html>
千万里不及你3回答
-
湖上湖
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
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
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>"; ?>
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