只看WHERE子句，我们可以看到一些问题：
WHERE wp_term_taxonomy.term_id IN (SELECT term_id FROM wp_terms
WHERE slug='academia' OR slug='law' OR slug='policy')
AND wp_term_taxonomy.term_id IN (SELECT term_id FROM wp_terms
WHERE slug='full-time' OR slug='part-time')
AND wp_term_taxonomy.term_id IN (SELECT term_id FROM wp_terms
WHERE slug='early-career')
这里我们有一个单独的列 (wp_term_taxonomy.term_id)，它必须与 Academia、Law 或 Policy 之一的术语 ID 同时相同（可能是 3 个不同的值），并且还与 Full 之一的术语 ID 相同-Time 或 Part-Time（可能是 2 个不同的值，并且与之前的 3 个值中的每一个都不同）并且也与 Early-Career 的术语 ID 相同（一个值，但与之前的 5 个值中的每一个都不同。所以，单项 ID 必须同时为 3 个不同的值，并且无法管理。
您可能需要使用 3 个不同的别名多次加入 wp_term_taxonomy 表。
WHERE wtt1.term_id IN (SELECT term_id FROM wp_terms
WHERE slug='academia' OR slug='law' OR slug='policy')
AND wtt2.term_id IN (SELECT term_id FROM wp_terms
WHERE slug='full-time' OR slug='part-time')
AND wtt3.term_id IN (SELECT term_id FROM wp_terms
WHERE slug='early-career')
我使用的 3 个别名是 wtt1、wtt2 和 wtt3。它们将列在 JOIN 条件中。
让我们看看选择列表和 FROM 子句
SELECT * FROM $wpdb->posts
LEFT JOIN $wpdb->postmeta ON($wpdb->posts.ID = $wpdb->postmeta.post_id)
LEFT JOIN $wpdb->term_relationships ON($wpdb->posts.ID = $wpdb->term_relationships.object_id)
LEFT JOIN $wpdb->term_taxonomy ON($wpdb->term_relationships.term_taxonomy_id = $wpdb->term_taxonomy.term_taxonomy_id)
LEFT JOIN $wpdb->terms ON($wpdb->terms.term_id = $wpdb->term_taxonomy.term_id)
现在让我们解开一些 PHP 材料，留下常规 SQL：
SELECT *
FROM wp_posts AS p
LEFT JOIN wp_postmeta AS pm ON p.ID = pm.post_id
LEFT JOIN wp_term_relationships AS tr ON p.ID = tr.object_id
LEFT JOIN wp_term_taxonomy AS tt ON tr.term_taxonomy_id = tt.term_taxonomy_id
LEFT JOIN wp_terms AS tm ON tm.term_id = tt.term_id
您可能不希望这里有任何左连接；您不想看到不符合条件的帖子，但使用 LEFT JOIN 将意味着在这部分中选择了许多帖子（尽管所有行随后都被已经讨论过的损坏的 WHERE 条件丢弃）。
一个帖子可能有多个术语关系条目。我们希望帖子至少包含三个任期关系条目：一个用于学术/法律/政策三人组，一个用于全职/兼职二人组，以及早期职业。
SELECT *
FROM wp_posts AS p
JOIN wp_postmeta AS pm ON p.ID = pm.Post_ID
JOIN (SELECT t1.Object_ID
FROM wp_term_relationships AS tr
JOIN wp_term_taxonomy AS tt ON tr.term_taxonomy_id = tt.term_taxonomy_id
JOIN wp_terms AS tm ON tm.term_id = tt.term_id
WHERE tm.slug IN ('academia', 'law', 'policy')
) AS t1 ON p.ID = t1.Object_ID
JOIN (SELECT t1.Object_ID
FROM wp_term_relationships AS tr
JOIN wp_term_taxonomy AS tt ON tr.term_taxonomy_id = tt.term_taxonomy_id
JOIN wp_terms AS tm ON tm.term_id = tt.term_id
WHERE tm.slug IN ('full-time', 'part-time')
) AS t2 ON p.ID = t2.Object_ID
JOIN (SELECT t1.Object_ID
FROM wp_term_relationships AS tr
JOIN wp_term_taxonomy AS tt ON tr.term_taxonomy_id = tt.term_taxonomy_id
JOIN wp_terms AS tm ON tm.term_id = tt.term_id
WHERE tm.slug = 'early-career')
) AS t3 ON p.ID = t3.Object_ID
我认为这可能会奏效——但现在已经很晚了，我可能会完全脱离困境。这当然不是一个简单的查询。
假设我的基本 SQL 正确，您只需将表名替换为 PHP 表示法即可：
SELECT *
FROM $wpdb->posts AS p
JOIN $wpdb->postmeta AS pm ON p.ID = pm.Post_ID
JOIN (SELECT t1.Object_ID
FROM $wpdb->term_relationships AS tr
JOIN $wpdb->term_taxonomy AS tt ON tr.term_taxonomy_id = tt.term_taxonomy_id
JOIN $wpdb->terms AS tm ON tm.term_id = tt.term_id
WHERE tm.slug IN ('academia', 'law', 'policy')
) AS t1 ON p.ID = t1.Object_ID
JOIN (SELECT t1.Object_ID
FROM $wpdb->term_relationships AS tr
JOIN $wpdb->term_taxonomy AS tt ON tr.term_taxonomy_id = tt.term_taxonomy_id
JOIN $wpdb->terms AS tm ON tm.term_id = tt.term_id
WHERE tm.slug IN ('full-time', 'part-time')
) AS t2 ON p.ID = t2.Object_ID
JOIN (SELECT t1.Object_ID
FROM $wpdb->term_relationships AS tr
JOIN $wpdb->term_taxonomy AS tt ON tr.term_taxonomy_id = tt.term_taxonomy_id
JOIN $wpdb->terms AS tm ON tm.term_id = tt.term_id
WHERE tm.slug = 'early-career')
) AS t3 ON p.ID = t3.Object_ID
您没有说正在使用哪个 DBMS，但它很可能是 MySQL。如果您使用的是 Oracle，则必须将 AS 排除在表别名之外。标准 SQL 和大多数其他 SQL DBMS 都可以使用 AS 作为表别名。请注意$wpdb-> 符号的使用如何受到表别名使用的限制；它使代码更易于阅读（尽管仍然不容易阅读）。
错误修复和问题解决
未经测试的代码通常有错误；这与任何其他未经测试的代码没有什么不同。
第一个测试步骤是单独运行 FROM 子句中的子查询。这立即表明他们不应该引用t1.Object_ID；在每种情况下都应该是tr.Object_ID。在“早期职业”之后还有一个无关紧要的右括号。一旦我有一个可以运行（子）查询的测试数据库，这些错误就很容易被发现。
SELECT *
FROM wp_posts AS p
JOIN wp_postmeta AS pm ON p.ID = pm.Post_ID
JOIN (SELECT tr.Object_ID
FROM wp_term_relationships AS tr
JOIN wp_term_taxonomy AS tt ON tr.term_taxonomy_id = tt.term_taxonomy_id
JOIN wp_terms AS tm ON tm.term_id = tt.term_id
WHERE tm.slug IN ('academia', 'law', 'policy')
) AS t1 ON p.ID = t1.Object_ID
JOIN (SELECT tr.Object_ID
FROM wp_term_relationships AS tr
JOIN wp_term_taxonomy AS tt ON tr.term_taxonomy_id = tt.term_taxonomy_id
JOIN wp_terms AS tm ON tm.term_id = tt.term_id
WHERE tm.slug IN ('full-time', 'part-time')
) AS t2 ON p.ID = t2.Object_ID
JOIN (SELECT tr.Object_ID
FROM wp_term_relationships AS tr
JOIN wp_term_taxonomy AS tt ON tr.term_taxonomy_id = tt.term_taxonomy_id
JOIN wp_terms AS tm ON tm.term_id = tt.term_id
WHERE tm.slug = 'early-career'
) AS t3 ON p.ID = t3.Object_ID
通过这些修复，查询运行并生成了数据行。您可能会合理地决定您想要结果中的三个子查询中的 slug。您将更改子查询以返回 tr.Object_ID, tm.slug。例如，这个查询变体：
SELECT p.ID, t1.slug_1, t2.slug_2, t3.slug_3, pm.meta_key
FROM wp_posts AS p
JOIN wp_postmeta AS pm ON p.ID = pm.Post_ID
JOIN (SELECT tr.Object_ID, tm.slug AS slug_1
FROM wp_term_relationships AS tr
JOIN wp_term_taxonomy AS tt ON tr.term_taxonomy_id = tt.term_taxonomy_id
JOIN wp_terms AS tm ON tm.term_id = tt.term_id
WHERE tm.slug IN ('academia', 'law', 'policy')
) AS t1 ON p.ID = t1.Object_ID
JOIN (SELECT tr.Object_ID, tm.slug AS slug_2
FROM wp_term_relationships AS tr
JOIN wp_term_taxonomy AS tt ON tr.term_taxonomy_id = tt.term_taxonomy_id
JOIN wp_terms AS tm ON tm.term_id = tt.term_id
WHERE tm.slug IN ('full-time', 'part-time')
) AS t2 ON p.ID = t2.Object_ID
JOIN (SELECT tr.Object_ID, tm.slug AS slug_3
FROM wp_term_relationships AS tr
JOIN wp_term_taxonomy AS tt ON tr.term_taxonomy_id = tt.term_taxonomy_id
JOIN wp_terms AS tm ON tm.term_id = tt.term_id
WHERE tm.slug = 'early-career'
) AS t3 ON p.ID = t3.Object_ID;
在一些测试数据上产生了以下结果：
1575 policy full-time early-career date_legible
1575 policy full-time early-career date_timestamp
1575 policy full-time early-career longitude
1575 policy full-time early-career date_normal
1575 policy full-time early-career url
1575 policy full-time early-career _su_rich_snippet_type
1575 policy full-time early-career _edit_last
1575 policy full-time early-career expiration-date
1575 policy full-time early-career organization
1575 policy full-time early-career latitude
1575 policy full-time early-career location
1575 policy full-time early-career _edit_lock
这表明至少有一篇帖子 (ID = 1575) 具有您需要的三个特征，但这也表明您将不得不更巧妙地处理 PostMeta 数据。结果表明 PostMeta 是一个 EAV (Entity-Attribute-Value) 模型。这将需要仔细处理以提取给定帖子的有用信息（例如纬度和经度）。实际上，您需要为每个要检查的单独元属性建立一个（可能是外部的）连接。
例如，要收集帖子的纬度和经度（如果有），您需要编写：
SELECT p.ID, t1.slug_1, t2.slug_2, t3.slug_3, p1.latitude, p2.longitude
FROM wp_posts AS p
LEFT JOIN
(SELECT Post_ID, Meta_Key AS m1_key, Meta_Value AS latitude
FROM wp_postmeta
WHERE Meta_Key = 'latitude'
) AS p1 ON p.ID = p1.Post_ID
LEFT JOIN
(SELECT Post_ID, Meta_Key AS m2_key, Meta_Value AS longitude
FROM wp_postmeta
WHERE Meta_Key = 'longitude'
) AS p2 ON p.ID = p2.Post_ID
JOIN (SELECT tr.Object_ID, tm.slug AS slug_1
FROM wp_term_relationships AS tr
JOIN wp_term_taxonomy AS tt ON tr.term_taxonomy_id = tt.term_taxonomy_id
JOIN wp_terms AS tm ON tm.term_id = tt.term_id
WHERE tm.slug IN ('academia', 'law', 'policy')
) AS t1 ON p.ID = t1.Object_ID
JOIN (SELECT tr.Object_ID, tm.slug AS slug_2
FROM wp_term_relationships AS tr
JOIN wp_term_taxonomy AS tt ON tr.term_taxonomy_id = tt.term_taxonomy_id
JOIN wp_terms AS tm ON tm.term_id = tt.term_id
WHERE tm.slug IN ('full-time', 'part-time')
) AS t2 ON p.ID = t2.Object_ID
JOIN (SELECT tr.Object_ID, tm.slug AS slug_3
FROM wp_term_relationships AS tr
JOIN wp_term_taxonomy AS tt ON tr.term_taxonomy_id = tt.term_taxonomy_id
JOIN wp_terms AS tm ON tm.term_id = tt.term_id
WHERE tm.slug = 'early-career'
) AS t3 ON p.ID = t3.Object_ID;
产生：
1575 policy full-time early-career -33.8210366 151.1887557
等等。