MySQL PHP CMS创建URL(需要编辑)

时间:2011-10-20 17:16:09

标签: php mysql url content-management-system

我为这个问题的长度道歉,并且想提前感谢。

以下是相关代码片段。我有一个MySQL表,它存储来自基于本教程的自定义PHP MySQL CMS程序的信息:cms tutorial。正如您在列表页面和显示管理功能中所看到的,URL是使用基本URL和ID以及标题构建的。 ID是MySQL表中的主键,用于提取要在页面上显示的所有必要信息。我如何重写下面的代码,以便它只在URL而不是ID中显示标题,但仍然将ID作为在页面中显示信息的唯一标识符?

来自functions.php:

'// Display center bottom column
function centerbottomcolumn() {
if ($_GET['id']) {
    $pageID = (int) $_GET['id'];

    $result = mysql_query("SELECT centerbottomcolumn FROM pages WHERE id='$pageID'");
    $row = mysql_fetch_array($result);

    echo $row['centerbottomcolumn'];
} else {
    $result = mysql_query("SELECT value FROM settings WHERE name='homePage'");
    $row = mysql_fetch_array($result);

    $pageID = $row['value'];

    $result = mysql_query("SELECT centerbottomcolumn FROM pages WHERE title='$pageID'");
    $row = mysql_fetch_array($result);

    echo $row['centerbottomcolumn'];
}

}

// List the pages
function listPages() {
// List the home page first
$result = mysql_query("SELECT value FROM settings WHERE name='homePage'");
$row = mysql_fetch_array($result);

$homeID = $row['value'];

$result = mysql_query("SELECT title FROM pages WHERE id='$homeID'");
$row = mysql_fetch_array($result);

$homeTitle = $row['title'];

echo "<li><a href='" . BASE_URL . "/index.php'>$homeTitle</a></li>";

// List the rest of the pages
$result = mysql_query("SELECT id, title FROM pages");

while ($row = mysql_fetch_array($result)) {
    // Do not list the home page twice
    if ($row['title'] != $homeID) {
        $pageID = $row['title'];
        $pageTitle = $row['title'];

        echo "<li><a href='" . BASE_URL . "/?id=$id&title=$pageTitle'>$pageTitle</a></li>";
    }
}

}

// Display admin table
function displayAdmin() {
// Find the home page ID
$result = mysql_query("SELECT value FROM settings WHERE name='homePage'");
$row = mysql_fetch_array($result);

$homeID = $row['value'];

// Display a table
$result = mysql_query("SELECT id, title, date FROM pages");

echo '<table width="961">';
echo '<tr height="50">
    <th align="left">ID</th>
    <th align="left">Title of the Page</th>
    <th align="left">Date Created</th>
    <th align="left">Actions</th>
    </tr>';

while ($row = mysql_fetch_array($result)) {
    $id = $row['id'];
    $title = $row['title'];
    $date = date('M d, Y', $row['date']);

    echo "<tr>
        <td>$id</td>
        <td><a href='". BASE_URL . "/id=$id&title=$title'>$title</a>";
    if ($id == $homeID) {
        echo ' <strong>(Home Page)</strong>';
    }
    echo "</td>
        <td>$date</td>
        <td><a href='edit.php?id=$id'>Edit</a></td><td>
        <a href='confirm.php?id=$id'>Delete</a></td><td>
        <a href='sethome.php?id=$id'>Set as Home</a>";
}
echo '</table>';

}

// Get array with page IDs
function getArray() {
$result = mysql_query("SELECT id FROM pages");

$IDs = array();
$i = 0;
while ($row = mysql_fetch_array($result)) {
    $IDs[$i] = $row['id'];
    $i++;
}
return $IDs;

}

From index.php:

(above head tag)
require_once 'functions.php';

connect();

$IDs = getArray();

(from body)
<?php if (in_array($_GET['id'], $IDs) || !$_GET): ?>
<?php centerbottomcolumn(); ?>
<?php else: ?>

        <!-- Show a not found error -->
        <p>Not found</p>

    <?php endif; ?>'

1 个答案:

答案 0 :(得分:1)

您希望页面根据给定的ID显示信息,但您不希望将ID传递给URL查询字符串中的页面?您可以将数据发布到目标页面而不是查询字符串。您可以创建一个新ID,它是真实ID的代理。但一般情况下,如果没有输入,您不能让页面根据用户输入(在这种情况下单击带有ID的链接)执行某些操作。如果你想隐藏ID,因为它应该是秘密的,或者你不希望人们猜测它们,那么你可以用某种随机guid替换你的ID,这些guid映射到数据库中的真实(秘密)ID