PHP/SQL顯示鏈接結果

Question

我創建了一個鏈接

echo '<a href="description.php?id='.$dash.'" >'."<strong>$name</strong>".'</a>'; 

其作品，但帶我到一個空白頁（description.php）

這是我試過的代碼，不會顯示任何信息。

if (isset($_GET['$name'])) { 
$result = $_GET['$name']; 

$sql = "SELECT PedalID, Manufacturer, ProductName FROM PedalDirectory"; 
$result = $conn->query($sql); 

if ($result->num_rows > 0) { 
    // output data of each row 
    while($row = $result->fetch_assoc()) { 
     echo "id: " . $row["PedalID"]. " - Name: " . $row["ProductName"]. " " . $row["Manufacturer"]. "<br>"; 
    } 
} else { 
    echo "0 results"; 
} 
$conn->close(); 

} 

任何人都可以提供一些見解，爲什麼此代碼不顯示結果？

Answer 1

看起來像你的鏈接發送$_GET['id']但你引用$_GET['$name']

你的代碼應該是

<?php 
if (isset($_GET['id'])) { 
    $result = $_GET['id']; 

    $sql = "SELECT PedalID, Manufacturer, ProductName FROM PedalDirectory"; 
    $result = $conn->query($sql); 

    if ($result->num_rows > 0) { 
     // output data of each row 
     while($row = $result->fetch_assoc()) { 
      echo "id: " . $row["PedalID"]. " - Name: " . $row["ProductName"]. " " . $row["Manufacturer"]. "<br>"; 
     } 
    } else { 
     echo "0 results"; 
    } 
    $conn->close(); 
} 
?>