将带有sql数据的php变量传递给带有链接的新php文件

我有一个博客设置，其中journal.php显示了数据库中博客的片段。它创建一个链接，以便在单击时应打开博客.php并显示完整的博客和评论部分。我不知道如何在日记中获取链接.php打开所选博客的 id，然后将其传递给博客.php并显示该博客。

期刊.php

<?php   // retreive post
     include('php/config.php');
    include ('php/function.php');
    dbConnect();
    $blog_query = mysql_query(
    'SELECT * 
    FROM Blog
    ORDER BY DATE DESC');

    function getIntro($str, $count = 200, $id=1) { 
        return preg_replace('/'s+?('S+)?$/', '', substr(nl2br($str), 0, 
$count)) . '<a href="blog.php?page=' . $ID . '">Read more...</a>';
        }
    while($row = mysql_fetch_array($blog_query)):
    $date = date_create($row['DATE']);
    $str = $row['CONTENT'];
    $ID = $row['ID'];

        $new_string = getIntro($str);
    ?>
    <div class="post">
        <h2><?php echo $row['TITLE']?></h2>
        <h3><?php echo date_format($date, 'l, F j, Y')?></h3>
        <p><?php echo $new_string?></p>
    </div>
</div>      
<?php endwhile ?>   

博客.php

<?php
    include('php/config.php');
    include ('php/function.php');               
    $_GET[$ID]; //Get ID from link in journal.php
    dbConnect();
    $data = mysql_query("SELECT * FROM Blog WHERE ID = '$ID'") or die(mysql_error()); 
    $info = mysql_fetch_array( $data ); 
    $BID = $info['ID'];
    $date = date_create($info['DATE']);
     ?>
    <div class="post">
        <h2><?php echo $info['TITLE']?></h2>
        <h3><?php echo date_format($date, 'l, F j, Y')?></h3>
        <p><?php echo $info['CONTENT']?></p>
    <h2>Comments.....</h2>
    <div class="comment-block">