PHP/SQL 显示链接结果


PHP/SQL Display Link Results

>我创建了一个链接

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

这有效,但将我带到空白页(描述.php)

这是我尝试过的代码,没有显示任何内容。

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关于为什么此代码不显示结果的见解吗?

看起来您的链接正在发送$_GET['id']但您引用的是$_GET['$name']

你的代码应该是

<?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();
}
?>