php - 使用 PHP 对 MYSQL 表进行排序

Question

我有一个按 ASC 顺序对名称进行排序的表格，但是当我单击该按钮时它不起作用。 我试着用 2 个按钮做同样的事情，并检查了一些可用的代码，但它根本不起作用。有帮助吗？

PHP 代码:

<?php
    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "myfeeds";

    $conn = mysql_connect($servername, $username, $password, $dbname);
    if (!$conn) {
        die("Connection failed");
    }

    $db = mysql_select_db("myfeeds", $conn);
    if (!$db) {
        die("Can't select database");
    }

    if (isset($_POST['asc'])) {
        $result = mysql_query("SELECT * FROM websites ORDER BY name ASC");
    } else {
        $result = mysql_query("SELECT * FROM websites ORDER BY name DESC");

    }

    if (!$result) {
        die("Failed to show queries from table");
    }


    $num = mysql_numrows($result);
    mysql_close();
    ?>

这是按钮:

SORT BY:  
            <form method="POST" action="<?php echo $_SERVER['PHP_SELF']; ?>">
                <button type="submit" id="asc" name="asc">ASC</button>
            </form>

表格:

                    <table cellpadding="0">
                    <tr>
                        <th>ID</th>
                        <th>Name</th>
                        <th>URL</th>
                        <th>Description</th>
                        <th>Logo</th>
                    </tr>

                    <?php
                    $i = 0;

                    while ($i < $num) {
                        $f5 = mysql_result($result, $i, "id");
                        $f1 = mysql_result($result, $i, "name");
                        $f2 = mysql_result($result, $i, "url");
                        $f3 = mysql_result($result, $i, "description");
                        $f4 = mysql_result($result, $i, "image");
                        ?>
                        <tr>
                            <td><?php echo $f5; ?></td>
                            <td><?php echo $f1; ?></td>
                            <td><?php echo $f2; ?></td>
                            <td><?php echo $f3; ?></td>
                            <td><?php echo "<img src='$f4'>"; ?></td>
                        </tr>
                        <?php
                        $i++;
                    }
                    ?>
                </table>

Answer 1

根据manual ，mysql_connect的第四个参数应该是一个新的连接链接，不是数据库名。

new_link

If a second call is made to mysql_connect() with the same arguments, no new link will be established, but instead, the link identifier of the already opened link will be returned. >The new_link parameter modifies this behavior and makes mysql_connect() always open a new link, even if mysql_connect() was called before with the same parameters. In SQL safe mode, this parameter is ignored.

我建议改用 mysqli_*，因为 mysql 已被弃用。

当然，不要忘记在查询后获取行。

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "myfeeds";

$conn = mysqli_connect($servername, $username, $password, $dbname);

$order = isset($_POST['asc']) ? 'ASC' : 'DESC';
$sql = "SELECT * FROM websites ORDER BY name $order";
$query = mysqli_query($conn, $sql);

$num = $query->num_rows;
if($num > 0) {
    while($row = mysqli_fetch_assoc($query)) {
        echo $row['name'] . '<br/>';
    }
}