MySQL使用PHP將一個表連接到另一個表

Question

我需要一點幫助，因為我不知道如何將一個表連接到另一個表。 我想在機器表中從客戶表中輸出所有者的姓名，而不是輸出客戶 ID。 如果是 full_name，則在 clients 表所有者的名稱中。

<table class="tbl-full" id="clients">
        <tr>
            <th>ID</th>
            <th>VIN</th>
            <th>Model</th>
            <th>Client ID</th>
            <th>Desc</th>
            <th>Make</th>
            <th>Year</th>
            <th>Actions</th>
        </tr>
        <?php
        $sql = "SELECT * FROM machines ORDER BY year DESC";
        $res = mysqli_query($conn, $sql);

        if ($res == TRUE) {
            $count = mysqli_num_rows($res);
            $sn = 1;

            if ($count > 0) {
                while ($rows = mysqli_fetch_assoc($res)) {
                    $id = $rows['id'];
                    $vin = $rows['vin'];
                    $client_id = $rows['client_id'];
                    $model = $rows['model'];
                    $description = $rows['description'];
                    $make = $rows['make'];
                    $year = $rows['year'];
        ?>
                    <tr>
                        <td><?php echo $sn++; ?></td>
                        <td><?php echo $vin; ?></td>
                        <td><?php echo $model; ?></td>
                        <td><?php echo $client_id; ?></td>
                        <td><?php echo $description; ?></td>
                        <td><?php echo $make; ?></td>
                        <td><?php echo $year; ?></td>
                        <td>
                            <a href="<?php echo SITEURL; ?>/update-machine.php?id=<?php echo $id; ?>" class="main-btn">Update Machine</a>
                            <a href="<?php echo SITEURL; ?>/delete-machine.php?id=<?php echo $id; ?>" class="danger-btn">Delete Machine</a>
                        </td>
                    </tr>

Answer 1

$sql = "SELECT m.*, c.full_name AS 'client_full_name'
FROM machines m
LEFT JOIN clients c ON c.id = m.client_id
ORDER BY m.year DESC";

然后

$client_full_name = $rows['client_full_name'];

最后

<td><?php echo $client_full_name; ?></td>

Answer 2

您的查詢應該是這樣的

SELECT M.*,C.full_name 
FROM machines as m 
INNER JOIN clients as c ON m.id = c.client_id 
ORDER By m.year DESC

然后使用

$client_name = $row['client_name']

然后回顯客戶端名稱

<td><?php echo $client_name; ?></td>