我正在尝试为$ 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 {$wpdb->posts} p
ON ( p.ID = tmr.object_id )
WHERE (tm.parent = '%d'
AND tm.taxonomy = '%s'
AND p.post_type = '%s' )
GROUP BY name HAVING COUNT(name) > '%d'
", 0, 'location', 'job', 0 ));
我试过这个来获取区域名称。它正好按预期返回数据。但是这里的区域是父分类,而国家是它的子项目。所以我也想准备数据以获得国家名单。但问题是它在父元素下面是否存在。所以我为它命名为$ sql的动态数组。 我想准备的代码在下面给出为$ country_query。
在WHERE语句中,我使用php implode方法进行动态查询构建。 它现在对我有用。但我也想把数据准备好像区域一样。
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);
目前我的查询在使用implode('OR',$ sql)后抛出此SQL语句。如果有人知道该怎么做请帮助我,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
我也尽力找到答案。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,任何人都知道,那么我将忽略这一点或尝试另一种方法。
提前致谢...
答案 0 :(得分:2)
感谢https://stackoverflow.com/users/1704961/mdma提醒。他对我说试试IN。最后我从stackoverflow找到了答案:https://stackoverflow.com/a/10634225/1993427这对我有很大帮助。 :)
所以我的最终代码在下面给出了其他代码。
$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);
干杯:)