我对wpdb->get_results有一个非常具体的问题我已经实现了自定义搜索请求,它能够搜索我们所有的自定义类型和自定义结构化表。查询运行良好(因此没有数据库连接问题),直到搜索关键字包含与号(&)。真正的奥秘在于,如果我回显请求本身,然后通过phpMyAdminSQL运行它,那么查询会返回预期(和期望)的结果。我试过用mysqli,但没有用。
以下是搜索系统中使用的代码的关键部分:
$request = "SELECT DISTINCT
SQL_CALC_FOUND_ROWS
p.ID,
(SELECT meta_value FROM wp_postmeta WHERE post_id = p.ID AND meta_key = 'latitude') AS latitude,
(SELECT meta_value FROM wp_postmeta WHERE post_id = p.ID AND meta_key = 'longitude') AS longitude
FROM wp_posts p
WHERE p.post_type = 'studia'
AND (SELECT meta_value FROM wp_postmeta WHERE post_id = p.ID AND meta_key = 'latitude') > 0
AND (SELECT meta_value FROM wp_postmeta WHERE post_id = p.ID AND meta_key = 'longitude') > 0
AND ((SELECT meta_value FROM wp_postmeta WHERE post_id = p.ID AND meta_key = 'mesto' LIMIT 1) LIKE '%" . $keyword . "%'
OR p.post_title LIKE '%" . $keyword . "%'
OR (SELECT meta_value FROM wp_postmeta WHERE post_id = p.ID AND meta_key = 'ulice' LIMIT 1) LIKE '%" . $keyword . "%'
OR (SELECT IF(pm.meta_value LIKE 'a:%',(SELECT post_title FROM wp_posts WHERE ID = SUBSTRING_INDEX(SUBSTRING_INDEX(pm.meta_value,'"',2),'"',-1)), pm.meta_value ) FROM wp_postmeta pm WHERE pm.post_id = p.ID AND meta_key = 'studio' ) LIKE '%" . $keyword . "%')";
echo $request;
$studios = $wpdb->get_results( $request);
这是一个呼应的例子(关键字:YOGA&ART)请求,它再次在phpMyAdmin中返回所需的结果,但在网站上没有
SELECT DISTINCT SQL_CALC_FOUND_ROWS p.ID, (SELECT meta_value FROM wp_postmeta WHERE post_id = p.ID AND meta_key = 'latitude') AS latitude, (SELECT meta_value FROM wp_postmeta WHERE post_id = p.ID AND meta_key = 'longitude') AS longitude FROM wp_posts p WHERE p.post_type = 'studia' AND (SELECT meta_value FROM wp_postmeta WHERE post_id = p.ID AND meta_key = 'latitude') > 0 AND (SELECT meta_value FROM wp_postmeta WHERE post_id = p.ID AND meta_key = 'longitude') > 0 AND ((SELECT meta_value FROM wp_postmeta WHERE post_id = p.ID AND meta_key = 'mesto' LIMIT 1) LIKE '%YOGA & ART%' OR p.post_title LIKE '%YOGA & ART%' OR (SELECT meta_value FROM wp_postmeta WHERE post_id = p.ID AND meta_key = 'ulice' LIMIT 1) LIKE '%YOGA & ART%' OR (SELECT IF(pm.meta_value LIKE 'a:%',(SELECT post_title FROM wp_posts WHERE ID = SUBSTRING_INDEX(SUBSTRING_INDEX(pm.meta_value,'"',2),'"',-1)), pm.meta_value ) FROM wp_postmeta pm WHERE pm.post_id = p.ID AND meta_key = 'studio' ) LIKE '%YOGA & ART%')
我找到了答案。简单地说,由于我响应了请求,显示格式与实际发送的格式不相同,事实证明,实际发送的是HTML实体&
,而不是符号&。如此简单的更换解决了我的问题。
$keyword = str_replace("&", "&", $keyword);