php - 正确的连接语法

Question

合并这些语句的正确语法是什么？我不确定要使用哪个连接函数

<?php
  $tag_shows_result = mysql_query("SELECT * 
                                     FROM tags 
                                    WHERE tagname = '$n' 
                                      AND `show` > 0");
  while ($row = mysql_fetch_array($tag_shows_result)) {
    $shows_to_tag_result = mysql_query("SELECT * 
                                          FROM shows 
                                         WHERE id = ".$row['show']." 
                                      ORDER BY name ASC");
    while ($row = mysql_fetch_array($shows_to_tag_result)) {
?>
&nbsp;<a href="./show.php?id=<?php echo $row['id']; ?>" title="<?php echo $row['name']; ?>"><img src="./images/shows/<?php echo $row['id']; ?>.jpg" width="150" height="150" border="0" alt="<?php echo $row['name']; ?>" /></a>   
<?php } } ?>

得到它在这里工作是正确的格式

<?php
$tag_shows_result2 = mysql_query("SELECT * FROM tags JOIN shows ON tags.show = shows.id WHERE tagname='$n' AND `show` > 0 ORDER BY shows.name ASC");
while ($row = mysql_fetch_array($tag_shows_result2))
{
?>
&nbsp;<a href="./show.php?id=<?php echo $row['id']; ?>" title="<?php echo $row['name']; ?>"><img src="./images/shows/<?php echo $row['id']; ?>.jpg" width="150" height="150" border="0" alt="<?php echo $row['name']; ?>" /></a>   
<?php } ?>

Answer 1

无需花哨:

SELECT show.* FROM tags
JOIN shows ON (tags.show = show.id)
WHERE tags.tagname = ?
ORDER BY show.name ASC

或者，更简单:

SELECT * FROM shows WHERE id IN (
     SELECT show FROM tags WHERE tagname = ?
) ORDER BY name ASC