Question

当我只让一个WHERE并删除另一个时，它会返回正确的结果。当我拥有它时，它不显示任何东西。我怎样才能正确编写下面的逻辑？

我的目标是只显示满足“extra1 = $ value1和extra2 = $ value2 and extra3 = $ value3”的结果。

我在某处读到了我必须添加 addslashes 。有什么用？

$value1 = addslashes($_GET['extra1']);
$value2 = addslashes($_GET['extra2']);
$value3 = addslashes($_GET['extra3']);

$theposts = $wpdb->get_results("SELECT post_title FROM {$wpdb->posts}
WHERE ID IN
(
   SELECT DISTINCT post_id
   FROM {$wpdb->postmeta}
   WHERE meta_key = 'extra1' AND meta_value = '$value1'
   WHERE meta_key = 'extra2' AND meta_value = '$value2'
   WHERE meta_key = 'extra3' AND meta_value = '$value3'

)");

Answer 1

每个（子）查询只能有一个WHERE。用AND替换后续的WHERE。顺便问一下：错误信息是什么？

Answer 2

如果您的数据库支持交叉点，那么您可以这样做：

SELECT post_id
FROM {$wpdb->postmeta}
WHERE meta_key = 'extra1' AND meta_value = '$value1' 
    INTERSECT
SELECT post_id
FROM {$wpdb->postmeta}
WHERE meta_key = 'extra2' AND meta_value = '$value2'
    INTERSECT
SELECT post_id
FROM {$wpdb->postmeta}
WHERE meta_key = 'extra3' AND meta_value = '$value3'

这将为您提供具有extra1 / value1，extra2 / value2和extra3 / value3的所有post_id。

如果您没有INTERSECT（例如，AFAIK，MySQL），您可以尝试HAVING：

SELECT post_id
FROM {$wpdb->postmeta}
WHERE meta_key = 'extra1' AND meta_value = '$value1'
   OR meta_key = 'extra2' AND meta_value = '$value2'
   OR meta_key = 'extra3' AND meta_value = '$value3'
GROUP BY post_id
HAVING COUNT(*) = 3

Answer 3

将它们与OR关键字

结合使用

 WHERE meta_key = 'extra1' AND meta_value = '$value1' 
 OR 
 meta_key = 'extra2' AND meta_value = '$value2' 
 OR
 meta_key = 'extra3' AND meta_value = '$value3'

Answer 4

你也可以使用它：

SELECT p.post_title FROM {$wpdb->posts} AS p
WHERE EXISTS
  ( SELECT *
    FROM {$wpdb->postmeta} AS pm
    WHERE pm.meta_key = 'extra1' AND pm.meta_value = '$value1' 
      AND pm.post_id = p.ID
  )
  AND EXISTS
  ( SELECT * 
    FROM {$wpdb->postmeta} AS pm
    WHERE pm.meta_key = 'extra2' AND pm.meta_value = '$value2'
      AND pm.post_id = p.ID
  )
  AND EXISTS
  ( SELECT *
    FROM {$wpdb->postmeta} AS pm
    WHERE pm.meta_key = 'extra3' AND pm.meta_value = '$value3'
      AND pm.post_id = p.ID
  )

我在错误的多个查询中遇到了一些困难

4 个答案: