我试图用case创建一个搜索查询，但遇到了错误“invalid parameter number”。有人能帮我解决这个问题吗？我该怎么解决？

$query = 'SELECT *,
  CASE WHEN o.title LIKE :keyword THEN 1
       WHEN c.body LIKE :keyword THEN 2
  ELSE 99 END AS priority
  FROM orders AS o INNER JOIN
    comment_relations AS cr ON o.id = cr.target_id INNER JOIN
    comments AS c ON cr.comment_id = c.id
  WHERE cr.type = 2
  ORDER BY priority';

  $stmt = $db->prepare($query);

  $stmt->bindValue(':keyword', '%' . $search_keyword . '%', PDO::PARAM_STR);

table看起来像这样

ORDER:

ID = 95
title = first order

COMMENT RELATIONS:
id = 1241
comment_id = 500
target_id = 95
type = 2

COMMENT: 
id = 500
body = this is the first comment