“我的SQL不工作”问题很难回答，因为如果没有OP的数据集，它们几乎是不稳定的，但是。。。

我想你可能把事情弄得太复杂了。

SELECT 
  DISTINCT (p.ID), 
  p.post_title
FROM wp_posts p
JOIN wp_postmeta pm1 ON p.ID = pm1.post_id
JOIN wp_postmeta pm2 ON p.ID = pm2.post_id
WHERE 1 = 1
AND p.post_type =  \'game\'
AND (
  pm1.meta_key =  \'year\' 
  AND pm1.meta_value = \'2014\'
) 
AND (
  pm2.meta_key IN (\'team1\',\'team2\')
  AND pm2.meta_value =  \'yankees\'
)

function swap_key_wpse_144950($where) {
  remove_filter(\'posts_where\',\'swap_key_wpse_144950\');
  $where = str_replace("= \'xxteamsxx\'"," IN (\'team1\',\'team2\')",$where);
  return $where;
}
add_filter(\'posts_where\',\'swap_key_wpse_144950\');

$args = array(
  \'meta_query\' => array(
    array(
      \'key\' => \'year\',
      \'value\' => \'2014\'
    ),
    array(
      \'key\' => \'xxteamsxx\',
      \'value\' => \'yankees\'
    )
  )
);
$q = new WP_Query($args);
var_dump($q->request);

您是否可以尝试将此查询编写为

SELECT DISTINCT (p.ID), p.post_title
FROM wp_posts p
left outer join wp_postmeta pm1 ON p.ID = pm1.post_id AND pm1.meta_key = \'team1\' and pm1.meta_value =  \'yankees\'
left outer join wp_postmeta pm2 ON p.ID = pm2.post_id  and pm2.meta_key =  \'team2\'  and pm2.meta_value =  \'yankees\'
left outer join wp_postmeta pm on p.ID= pm.post_id and pm.meta_key = \'year\' and pm.meta_value = \'2014\'
WHERE p.post_type =  \'game\'

编辑

这个怎么样

SELECT DISTINCT (p.ID), p.post_title
FROM wp_posts p
left outer join wp_postmeta pm1 ON p.ID = pm1.post_id AND pm1.meta_key = \'team1\' 
left outer join wp_postmeta pm2 ON p.ID = pm2.post_id  and pm2.meta_key =  \'team2\'  
left outer join wp_postmeta pm on p.ID= pm.post_id and pm.meta_key = \'year\' 
WHERE p.post_type =  \'game\'  and pm.meta_value = \'2014\' and (pm1.meta_value = \'yankees\' or pm2.meta_value = \'yankees\')