将变量id链接到其配置文件页面

Question

我有一个页面显示了数据库中的一些信息。我想在每一行添加一个链接(比如把名字作为链接(，我可以点击它，它会把我带到一个显示该行其余信息的页面(比如个人资料页面(。我正在考虑制作一个链接，将id传递到配置文件页面，以便配置文件页面可以收集信息。

我相信这很简单，但我就是不明白。我如何使链接显示在只发送该行id号的每一行中？因为我宁愿不必进入每一行并制作一个特殊的链接。

这是我的代码：

<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT id, FirstName, LastName, Phone, Email FROM Contacts";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["id"]. "
- Name: " . $row["FirstName"]. " 
" . $row["LastName"]. " 
" . $row["Phone"]. " 
" . $row["Email"]. " <br>";
}
} else {
echo "0 results";
}
$conn->close();
?>```

Answer 1

根据您的示例，您可以这样做：

<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT id, FirstName, LastName, Phone, Email FROM Contacts";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while ($row = $result->fetch_assoc()) {
echo "id: " . $row["id"] . " - Name: <a href='profile.php/?id=" . $row["id"] . "> " . $row["FirstName"] . "</a> " . $row["LastName"] . " " . $row["Phone"] . " " . $row["Email"] . " <br/>";
}
} else {
echo "0 results";
}
$conn->close();
?>

该profile.php页面将检查id是否设置为isset，并基于id模拟器将数据查询为：php&MYSQL：从id=$id 中选择

未经过测试，请确保清除任何用户生成的变量。