将每个mySQL结果显示为另一个页面php的链接

Question

<?php
$servername = "localhost";
$username = "root";
$password = "admin";
$dbname = "sbsuite";


$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT * from sys_frm WHERE ParamType='Title';";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    while($row = $result->fetch_assoc()) {
        echo $row["Data"] . "<br>";
    }
} else {
    echo "0 results";
}
$conn->close();
?

这是给我两个结果的代码..

1. 客户端
2. 服务器

我想将它们作为链接插入另一个带有图像的php页面，取决于我们点击它去另一个php页面...

感谢您的帮助！

Answer 1

这是一个简单的HTML。将client/server php和jpg替换为您想要的内容。

<a href="client.php"><img src="client.jpg" alt="Client" /></a><br />
<a href="server.php"><img src="server.jpg" alt="Server" /></a>

使用PHP

<?php
while ($row = $result->fetch_assoc()) {
    ?>
<a href="<?php echo $row["Data"]; ?>.php"><img src="<?php echo $row["Data"]; ?>.jpg" alt="<?php echo $row["Data"]; ?>" /></a><br />
<?php
}