将变量 id 链接到其个人资料页面

Question

我有一个页面显示数据库中的一些信息。 我想在每一行中添加一个链接（例如将名字设为链接），我可以单击该链接将我带到显示该行信息的 rest 的页面（如个人资料页面）。 我正在考虑创建一个将 id 传递到个人资料页面的链接，以便个人资料页面可以收集信息。

我敢肯定这很简单，但我就是不明白。 如何使链接显示在仅发送该行 ID 号的每一行中？ 因为我宁愿不必将 go 放入每一行并制作特殊链接。

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();
?>

That profile.php page would check if the id is set with isset and query the data based on id simular to this: PHP & MYSQL: Select from where id=$id

未经测试并确保清理任何用户生成的变量。