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，但是当我单击按钮时它不起作用。 我尝试用 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。有什么帮助吗？

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：

                    <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新的连接链接，而不是数据库名。

new_link

如果使用相同的参数对mysql_connect() 进行第二次调用，则不会建立新链接，而是返回已打开链接的链接标识符。 >new_link 参数修改此行为并使mysql_connect() 始终打开一个新链接，即使之前使用相同参数调用了mysql_connect()。在 SQL 安全模式下，该参数被忽略。

我建议改用mysqli_*，因为不推荐使用mysql。

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/>';
    }
}

Answer 2

试试这个.. 工作 100% =)

<?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($_GET['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");

    if (mysql_num_rows($result) > 0) {
        // output data of each row
        while($row = mysql_fetch_assoc($result)) {
            echo "name: " . $row["name"]. "<br>";
        }
    } else {
        echo "0 results";
    }

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

?>

<form method="get" action="<?php echo $_SERVER['PHP_SELF']; ?>">
    <button type="submit" id="asc" name="asc" value="asc">ASC</button>
    <button type="submit" id="asc" name="desc" value="desc">DESC</button>
</form>

最好使用 PDO 就是这样..

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


    try
    {
        $conn = new PDO("mysql:host=".$servername.";dbname=".$dbname, $username, $password);
        $conn->setAttribute(PDO::ATTR_DEFAULT_FETCH_MODE, PDO::FETCH_ASSOC);
        $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
    }
    catch (PDOException $e)
    {
        die($e->getMessage());
    }


    try
    {
        if (isset($_GET['asc']))
            $result = $conn->prepare("SELECT * FROM websites ORDER BY name ASC");
        else
            $result = $conn->prepare("SELECT * FROM websites ORDER BY name DESC");

        $result->execute();

        if($result->rowCount())
        {
            while($r = $result->fetch(PDO::FETCH_OBJ))
            {
                echo 'Name:' . $r->name . '<br/>';
            }
        }
        else echo 'no record found!';

    }
    catch (PDOException $e)
    {
        die($e->getMessage());
    }

?>

使用mysql查看数据库中的数据

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

    <?php

    while ($row = mysql_fetch_assoc($result)) {
        ?>
        <tr>
            <td><?php echo $row["id"]; ?></td>
            <td><?php echo $row["name"]; ?></td>
            <td><?php echo $row["url"]; ?></td>
            <td><?php echo $row["description"]; ?></td>
            <td><?php echo "<img src='".$row["image"]."'>"; ?></td>
        </tr>
        <?php

    }
    ?>
</table>