MySQL条件Where子句

基本上，当meta_key字段等于'location'时，我想添加一个额外的where子句，以基于该位置进行比较

我试过这两个Case和IF语句，但似乎都没用，有什么想法我哪里出了问题吗

$query = "
SELECT wp_posts.ID, wp_posts.post_title, wp_posts.post_type,
wp_postmeta.meta_key, wp_postmeta.meta_value, wp_postmeta.post_id,
wp_term_relationships.object_id, wp_term_relationships.term_taxonomy_id, 
wp_term_taxonomy.term_id,
wp_terms.term_id, wp_terms.name

FROM wp_posts

/* Join Post Meta Table */
JOIN wp_postmeta
ON wp_posts.ID = wp_postmeta.post_id

/* Join Term Relationships Table to get the taxonomy id */
JOIN wp_term_relationships
ON wp_posts.ID = wp_term_relationships.object_id

/* Join Term Taxonomy Table to get the Term ID */
JOIN wp_term_taxonomy
ON wp_term_relationships.term_taxonomy_id = wp_term_taxonomy.term_taxonomy_id

/* Finally Join to the Terms table to get the Category Name *sigh* */
JOIN wp_terms
ON wp_term_taxonomy.term_id = wp_terms.term_id

WHERE post_type = 'listing' AND
wp_postmeta.meta_key IN ('locations', 'email', 'phone-number')";

if(isset($location)) {

    $query .= "
    IF wp_postmeta.meta_key='location'
        THEN AND wp_postmeta.meta_value LIKE '".mysql_real_escape_string($location)."'
    END IF";

}

---

谢谢，如果我没有解释清楚，请告诉我

简单的解决方案是

WHERE 
    (meta_key = 'Location' AND Location = 'UK') OR meta_key <> 'Location'

在哪里
（meta_键='Location'和Location='UK'）或meta_键'Location'

谢谢，这正是我想要的。

if(isset($location)) {
        $query .= " AND 
        CASE 
         WHEN wp_postmeta.meta_key='location' THEN wp_postmeta.meta_key 
         LIKE '".mysql_real_escape_string ($location)."'   
         ELSE 1=1
        END ";
      }