Php 是否有任何方法可用于数组内爆（'；或'；，$myArray）的wpdb prepare语句？

我正试图为$wpdb正确准备数据

$region = $wpdb->get_results( $wpdb->prepare( "
      SELECT tr.*, count(*) AS jobs_count FROM {$wpdb->terms} tr
      INNER JOIN {$wpdb->term_taxonomy} tm
        ON ( tm.term_id = tr.term_id )
      INNER JOIN {$wpdb->term_relationships} tmr
        ON ( tmr.term_taxonomy_id = tm.term_taxonomy_id )
      INNER JOIN {$wpdb->posts} p
        ON ( p.ID = tmr.object_id )
      WHERE (tm.parent = '%d'
        AND tm.taxonomy = '%s'
        AND p.post_type = '%s' )
      GROUP BY name HAVING COUNT(name) > '%d'
  ", 0, 'location', 'job', 0 ));

我试过这个来获取地区名称。它完全可以按预期返回数据。但在这里，地区是父分类法，国家是它的子项。因此，我还想为获得国家名单准备数据。但这里的问题是它在父元素下查找（如果存在）。所以我为它创建了一个名为$sql的动态数组。 我想准备的代码在下面以$country\u query的形式给出

在WHERE语句中，我使用php内爆方法构建动态查询。 现在对我来说很有用。但我还想准备数据，就像准备区域一样

foreach ($region as $reg) {
      $sql[] = " $wpdb->term_taxonomy.parent = '$reg->term_id' ";
}
    $country_query = "SELECT $wpdb->terms.*, count(*) AS jobs_count FROM $wpdb->terms
                      INNER JOIN $wpdb->term_taxonomy
                        ON ( $wpdb->term_taxonomy.term_id = $wpdb->terms.term_id )
                      INNER JOIN $wpdb->term_relationships
                        ON ( $wpdb->term_relationships.term_taxonomy_id = $wpdb->term_taxonomy.term_taxonomy_id )
                      INNER JOIN $wpdb->posts
                        ON ( $wpdb->posts.ID = $wpdb->term_relationships.object_id )

                      WHERE (". implode(' OR ', $sql) ." AND $wpdb->term_taxonomy.taxonomy = 'location' ) AND $wpdb->posts.post_type = 'job'
                      GROUP BY name HAVING COUNT(name) > 0";

    $country = $wpdb->get_results($country_query);

当前，我的查询在使用内爆（'OR'，$SQL）this后抛出此SQL语句。如果有人知道怎么做，请告诉我

SELECT wp_terms.*, count(*) AS jobs_count FROM wp_terms
                  INNER JOIN wp_term_taxonomy
                    ON ( wp_term_taxonomy.term_id = wp_terms.term_id )
                  INNER JOIN wp_term_relationships
                    ON ( wp_term_relationships.term_taxonomy_id = wp_term_taxonomy.term_taxonomy_id )
                  INNER JOIN wp_posts
                    ON ( wp_posts.ID = wp_term_relationships.object_id )

                  WHERE ( wp_term_taxonomy.parent = '2'  OR  wp_term_taxonomy.parent = '322'  OR  wp_term_taxonomy.parent = '651'  AND wp_term_taxonomy.taxonomy = 'location' ) AND wp_posts.post_type = 'job'
                  GROUP BY name HAVING COUNT(name) > 0

我也在尽力寻找答案。如果我能找到任何答案，我也会分享我的答案。但问题是我不知道如何准备内爆（'OR'，$sql）。如果这是不可能的，而且任何人都知道，那么我必须忽略这一点，或者尝试另一种方法

提前感谢…

感谢您的提醒。他对我说试试看。因此，最后我从stackoverflow中找到了一个答案：这对我帮助很大

下面给出了我的最终代码，供其他人参考

$country = array();

  $sql = array();
  foreach ($region as $reg) {
    $sql[] = $reg->term_id;
  }

$test = array("location", "job", "0");


$sql_st = "SELECT tr.*, count(*) AS jobs_count FROM {$wpdb->terms} tr
                  INNER JOIN {$wpdb->term_taxonomy} tm
                    ON ( tm.term_id = tr.term_id )
                  INNER JOIN {$wpdb->term_relationships} trm
                    ON ( trm.term_taxonomy_id = tm.term_taxonomy_id )
                  INNER JOIN {$wpdb->posts} p
                    ON ( p.ID = trm.object_id )

                  WHERE tm.parent IN(".implode(', ', array_fill(0, count($sql), '%s')).") AND tm.taxonomy = '%s' AND p.post_type = '%s'
                  GROUP BY name HAVING COUNT(name) > '%s'";

$country_query = call_user_func_array(array($wpdb, 'prepare'), array_merge(array($sql_st), $sql, $test));

$country = $wpdb->get_results($country_query);

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干杯：）

我不明白。你能在中使用mysql吗？我会查一查并让你知道的，谢谢