如何將下拉列表中的選定值插入另一個 MySQL 表

Question

我正在嘗試制作一個下拉列表，允許用戶 select 一個他們需要的部件，所以在選擇了他們需要的所有部件並將其提交到 go 到 MySQL 數據庫之后。 但是在選擇提交后沒有任何東西插入到我的數據庫中。

---

我的代碼和連接：

    <?php
    $servername = "localhost";
    $username = "root";
    $password = "";
    $db = "userregistration";

    // Create connection
    $conn = mysqli_connect($servername, $username, $password, $db);

    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    }
    echo 'good connection';
    ?>
<form action="trying.php" method=" POST">
  <table border=" 1">

        <thead>
            <tr>
                <th>Component</th>
                <th>Item Name</th>
                <th>Price </th>
            </tr>
        </thead>

                    <tbody>
                <tr>
                    <td>CPU</td>
                    <td>
                        <?php
                        //Retrieving CPU table
                        $query = $conn->query("SELECT * FROM cpu");
                        echo '<select name="cpu" class="cpu"  onChange = $("#cpuprice").val($(this).find("option:selected").attr("cpuprice"))>';                    
                        while ($obj = mysqli_fetch_assoc($query)) {
                            echo '<option cpuprice = ' . $obj['price'] . ' cpuname=' . $obj['cpuname'] . ' >' . $obj['cpuname'] . '</option> /n';
                        }
                        echo '</select>';
                        ?>
                    </td>
                    <td>
                        <output id="cpuprice" disabled value="">
                    </td>
                </tr>
            </tbody>

            <tbody>
                <tr>
                    <td>GPU</td>
                    <td>
                        <?php
                        //Retrieving GPU table
                        $query = $conn->query("SELECT * FROM gpu");
                        echo '<select name="gpu" class ="gpu"  onChange = $("#gpuprice").val($(this).find("option:selected").attr("gpuprice"))>';
                        while ($obj = mysqli_fetch_assoc($query)) {
                            echo '<option  gpuprice = "' . $obj['price'] . '" gpuname = "' . $obj['gpuname'] . '">' . $obj['gpuname'] . '</option>';
                        }
                        echo '</select>';
                        ?>
                    </td>
                    <td>
                        <output class="form-control prc" id="gpuprice" disabled value="">
                    </td>
                </tr>
            </tbody>

    </table>

    <input class="submit" type="submit" />
</form>

---

我試過了，但它不起作用，單擊提交后回顯添加錯誤並且沒有任何內容插入我的數據庫

<?php

if (!empty($_POST["cpu"]) && !empty($_POST["gpu"])) {
    $cpu = isset($_POST["cpu"]);
    $gpu = isset($_POST["gpu"]);

    $qstr = "INSERT INTO trycombuild(cpuname, gpuname) VALUES ('$cpu' , '$gpu')";
    $query = mysqli_query($conn, $qstr);
} else
    echo 'adding error';


?>

---

我試圖回顯$cpu和$gpu ，它說未定義的變量

---

我也試過這個：

if (!empty($_POST['cpu']) && !empty($_POST['gpu'])) {
    $cpu = $_POST['cpu'];
    $gpu = $_POST['gpu'];

    $qstr = $conn->prepare("INSERT INTO trycombuild (cpuname, gpuname) VALUES (?, ?)");
    $qstr->bind_param("ss", $cpu, $gpu);
    $qstr->execute();
    $sqtr->close();
}

Answer 1

您的 html 不正確。 刪除不必要的空間並嘗試

<tbody>
        <tr>
            <td>CPU</td>
            <td>
                <?php
                //Retrieving CPU table
                $query = $conn->query("SELECT * FROM cpu");
                echo '<select name="cpu" class="cpu"  onChange = $("#cpuprice").val($(this).find("option:selected").attr("cpuprice"))>';
                echo "<option>---select your CPU---</option>/n";
                while ($obj = mysqli_fetch_assoc($query)) {
                    echo '<option cpuprice = ' . $obj['price'] . ' cpuname=' . $obj['cpuname'] . ' >' . $obj['cpuname'] . '</option> /n';
                }
                echo '</select>';
                ?>
            </td>
            <td>
                <output id="cpuprice" disabled value="">
            </td>
        </tr>
    </tbody>

    <tbody>
        <tr>
            <td>GPU</td>
            <td>
                <?php
                //Retrieving GPU table
                $query = $conn->query("SELECT * FROM gpu");
                echo '<select name="gpu" class ="gpu"  onChange = $("#tpuprice").val($(this).find("option:selected").attr("gpuprice"))>';
                echo "<option>---select your GPU---</option>";
                while ($obj = mysqli_fetch_assoc($query)) {
                    echo '<option  gpuprice = "' . $obj['price'] . '" gpuname = "' . $obj['gpuname'] . '">' . $obj['gpuname'] . '</option>';
                }
                echo '</select>';
                ?>
            </td>
            <td>
                <output class="form-control prc" id="tpuprice" disabled value="">
            </td>
        </tr>
    </tbody>