如何使鏈接顯示數據庫中的特定數據？

Question

我正在為我的最終項目建立一個產品網站，並且想要建立一個鏈接，例如（如果單擊網頁上的處理器，它將僅在另一頁上顯示數據庫中的處理器列表。同樣應該發生當我單擊其他類別（如主板，圖形卡等）時。 例如，我用方法發布做了一個按鈕，然后在另一頁上我寫了以下代碼：

parts.php

    <form action="display.php" method="post">


        <div class="partbox">
            <a href="#">
            <img src="./img/processor.jpg" alt="Go to Processor">
            </a>
            <br>
            <input type="submit" class="btnlink" value="Processor" name="processor"</input>
        </div>

        <div class="partbox">
            <a href="#">
            <img src="./img/motherboard.jpg" alt="Go to Motherboard">
            </a>
            <br>
            <input type="submit" class="btnlink" value="Motherboard" name="motherboard"</input>
        </div>

等等...

display.php

//When Clicking on Processor

if(isset($_POST['processor'])){

$sql = "SELECT * FROM products WHERE category = 'Processor' ";

$result = mysqli_query($conn, $sql);
$row = mysqli_fetch_assoc($result);

echo "<h1>" . $row["category"] . "</h1>";

while($row = mysqli_fetch_assoc($result)) {
    include('info.php');
    echo $info;
}
}

//When clicking on Motherboard



if(isset($_POST['motherboard'])){

$sql = "SELECT * FROM products WHERE category = 'Motherboard ' ";
$result = mysqli_query($conn, $sql);
$row = mysqli_fetch_assoc($result);

echo "<h1>" . $row["category"] . "</h1>";

while($row = mysqli_fetch_assoc($result)) {
    include('info.php');
    echo $info;
}
}

等等...

好吧，我不知道通過鏈接是否可行。 還是可以這樣保留它？ 恐怕我花了很長時間才想到可能是最簡單的方法。

Answer 1

parts.php

    <div class="partbox">
        <a href="display.php?category_name=Processor">
        <img src="./img/processor.jpg" alt="Go to Processor">
        </a>
        <br>
    </div>

    <div class="partbox">
        <a href="display.php?category_name=Motherboard">
        <img src="./img/motherboard.jpg" alt="Go to Motherboard">
        </a>
        <br>
    </div>

display.php

// based on the link clicked, $_GET['category_name'] will have the 
//values Processor or Motherboard
if(isset($_GET['category_name'])){

    $sql = "SELECT * FROM products WHERE category = $_GET['category_name'] ";

    $result = mysqli_query($conn, $sql);
    $row = mysqli_fetch_assoc($result);

    echo "<h1>" . $row["category"] . "</h1>";

    while($row = mysqli_fetch_assoc($result)) {
        include('info.php');
        echo $info;
    }
}