php - 有没有更有效的方法来编写这 4 个 MYSQL 查询以使用更少的查询？

Question

我当前的查询。我可以将它们组合得更少以获得相同的结果吗？我正在尝试将有关路径的特定信息放入名为 $trail 的数组中。

$result = mysql_query("SELECT meta_value FROM wp_postmeta WHERE `post_id` = $id AND `meta_key` = 'location_city' ");
$result = mysql_fetch_array($result);
$trail[city] = $result[0];

$result = mysql_query("SELECT meta_value FROM wp_postmeta WHERE `post_id` = $id AND `meta_key` = 'length' ");
$result = mysql_fetch_array($result);       
$trail[length] = $result[0];

$result = mysql_query("SELECT meta_value FROM wp_postmeta WHERE `post_id` = $id AND `meta_key` = 'location_state' ");
$result = mysql_fetch_array($result);
$trail[state] = $result[0];

$result = mysql_query("SELECT meta_value FROM wp_postmeta WHERE `post_id` = $id AND `meta_key` = 'difficulty' ");
$result = mysql_fetch_array($result);
$trail[difficulty] = $result[0];

$result = mysql_query("SELECT guid FROM wp_posts WHERE `post_parent` = $id AND `post_type` = 'attachment' ");
$result = mysql_fetch_array($result);
$trail[image] = $result[0];

Answer 1

这将适用于元值:

SELECT meta_value, meta_key
...
WHERE (post_id = $id) AND (meta_key IN ('location_city', 'length', ...)

guid 查询最好单独运行。如果您坚持减少到单个查询调用，那么联合就可以了

SELECT 'meta' AS source, meta_value, meta_key
WHERE ...

UNION ALL

SELECT 'guid' AS source, guid, null
FROM ...
WHERE post_parent = $id and post_type = 'attachment'

然后使用这个派生的“源”字段来确定哪些值属于哪里。