我得到表单db字段（organisations.paths）以下字符串：
/ 1/2/3/4
每个数字都是此数据库表中组织名称的ID：
+----+-------------+----------+----------+----------------------+
| id | frameworkid | path | parentid | fullname |
+----+-------------+----------+----------+----------------------+
| 1 | 1 | /1 | 0 | NYC University |
| 2 | 1 | /2 | 0 | Board of directors |
| 3 | 1 | /1/2/3 | 1 | Math faculty |
| 4 | 1 | /1/2/3/4 | 3 | Statistic department |
| 5 | 1 | /1/2/3/5 | 2 | Linguist department |
+----+-------------+----------+----------+----------------------+
然后我为每个组织提供description table：
+----+----------+---------+----------------+
| id | data | fieldid | organisationid |
+----+----------+---------+----------------+
| 1 | HQ | 1 | 1 |
| 2 | advisory | 1 | 2 |
| 3 | advisory | 1 | 3 |
| 4 | bottom | 1 | 4 |
| 5 | advisory | 1 | 5 |
+----+----------+---------+----------------+
如何join两个说明表和主表并仅通过其说明中包含HQ或advisory的组织进行循环播放？所以它变成了：
纽约大学，董事会，数学系（统计部门 - 不会显示，因为它与描述
bottom）
答案 0 :(得分：1)
您需要使用PHP和mysql的# assuming that @user.accounts returns all accounts
user_accounts = @user.accounts.group_by(&:name)
# querying the hash. use `first` to receive the first matching
# account (even if there is only one)
user_accounts['nameofproduct'].first
，explode和IN函数。
join制作循环以获取名称并根据需要显示它们。