所以,我有一个我正在创建的数据库。它存储有关家庭和每个家庭成员的信息。然后,它使用这些记录将发票与家庭成员或家庭成员相关联。
我的困境是我需要将所有这些发票列入家庭记录下的页面,即创建与家庭本身或个人家庭成员相关的发票清单。
表格结构
发票
id | date_entered | invoice_date | invoice_number | invoice_amount | client_type | unique_id | supplier_type | supplier_id | category_id | childcare_hours
---+--------------+--------------+----------------+----------------+-------------+-----------+---------------+-------------+-------------+----------------
1 | 1411098397 | 1411048800 | 123 | 0.01 | 0 | 137 | 0 | 139 | 5 | NULL
家庭
id | ufi | last_name | address_1 | address_2 | city_id | phone | mobile | email | f_d_worker_1 | f_d_worker_2 | status_id | trans_date | entry_date | exit_date | eligible_date | active_date | lga_loc_id | facs_loc_id | ind_status_id | referral_id | active_status | comm_org_id | notes
---+----------+-----------------+-------------+-----------+---------+-------+--------+-------+--------------+--------------+-----------+------------+------------+-----------+---------------+-------------+------------+-------------+---------------+-------------+---------------+-------------+-------
1 | 1-XEWUDZ | Forsyth - Ennis | Skinner St. | NULL | NULL | NULL | NULL | NULL | 13 | NULL | 1 | NULL | 1341324000 | NULL | 1341842400 | 1342620000 | 7 | 1 | 3 | NULL | 1 | 1 | NULL
客户(家庭成员)
id | upi | last_name | first_name | birthdate | sex | phone | mobile | email | indig_status_id | referral_id | relationship_id | preschool_id | family_id | notes
---+----------+-----------+------------+------------+-----+-------+--------+-------+-----------------+-------------+-----------------+--------------+-----------+------
1 | 1-XFCBBP | Ennis | Jason | 20/09/1996 | 1 | NULL | NULL | NULL | 3 | NULL | NULL | NULL | 1 | NULL
我当前的SQL看起来像:
SELECT `invoices`.`id`, `invoices`.`date_entered`, `invoices`.`invoice_date`, `invoices`.`invoice_number`, `invoices`.`invoice_amount`, `invoices`.`client_type`, `invoices`.`unique_id`, `unique1`.`ufi`, `unique2`.`upi`, `unique1`.`last_name`, `invoices`.`supplier_type`, `invoices`.`supplier_id`, `suppliers`.`name`, `invoices`.`category_id`, `cat1`.`name`, `cat2`.`name`, `invoices`.`childcare_hours`
FROM `invoices`
LEFT OUTER JOIN `suppliers` ON `suppliers`.`id` = `invoices`.`supplier_id`
LEFT OUTER JOIN `categories` cat1 ON `cat1`.`id` = `invoices`.`category_id`
LEFT OUTER JOIN `preschool_types` cat2 ON `cat2`.`id` = `invoices`.`category_id`
LEFT OUTER JOIN `families` unique1 ON `unique1`.`id` = `invoices`.`unique_id`
LEFT OUTER JOIN `clients` unique2 ON `unique2`.`id` = `invoices`.`unique_id`
WHERE (`invoices`.`unique_id` = ? AND `unique1`.`ufi` = ?) LIMIT 0, 10
但我需要一个检查client_type
列的查询,如果它等于1,则需要查看clients
表,但是需要查找同一系列的成员,由families
表中的id行
好的,经过多次,很多(很多)搞砸了一点研究。似乎@cupid是正确的(尽管他的回答非常简短)。
我会更好地解释解决方案(希望以后能帮助别人)。
MySQL中的UNION选项(很可能是其他SQL)允许您将两个(或更多)SELECT查询的结果集合并到一个结果集中。如果您有相似的数据,这可能非常有用,您可以在单独的表中轻松选择并作为一个请求进行处理。通过允许您使用SQL的LIMIT选项,对于分页也很有用(在我的情况下)。
要考虑的一点是,UNION语法使用第一个SELECT语句中的列作为所有后续查询的列名,还需要确保在所有查询中选择了相同数量的列为了这个工作。
(
SELECT
`invoices`.`id`,
`invoices`.`date_entered`,
`invoices`.`invoice_date`,
`invoices`.`invoice_number`,
`invoices`.`invoice_amount`,
`invoices`.`client_type`,
`invoices`.`unique_id`,
`clients`.`upi`,
`clients`.`last_name`,
`clients`.`family_id`,
`invoices`.`supplier_type`,
`invoices`.`supplier_id`,
`suppliers`.`name`,
`invoices`.`category_id`,
`cat1`.`name`,
`cat2`.`name`,
`invoices`.`childcare_hours`
FROM
(
`invoices`
LEFT OUTER JOIN `suppliers` ON `suppliers`.`id` = `invoices`.`supplier_id`
LEFT OUTER JOIN `categories` cat1 ON `cat1`.`id` = `invoices`.`category_id`
LEFT OUTER JOIN `preschool_types` cat2 ON `cat2`.`id` = `invoices`.`category_id`
LEFT OUTER JOIN `clients` ON `clients`.`id` = `invoices`.`unique_id`)
WHERE
`clients`.`family_id` = 47 AND `invoices`.`client_type` = 1
)
UNION
(
SELECT
`invoices`.`id`,
`invoices`.`date_entered`,
`invoices`.`invoice_date`,
`invoices`.`invoice_number`,
`invoices`.`invoice_amount`,
`invoices`.`client_type`,
`invoices`.`unique_id`,
`families`.`ufi`,
`families`.`last_name`,
`families`.`id`,
`invoices`.`supplier_type`,
`invoices`.`supplier_id`,
`suppliers`.`name`,
`invoices`.`category_id`,
`cat1`.`name`,
`cat2`.`name`,
`invoices`.`childcare_hours`
FROM `invoices`
LEFT OUTER JOIN `suppliers` ON `suppliers`.`id` = `invoices`.`supplier_id`
LEFT OUTER JOIN `categories` cat1 ON `cat1`.`id` = `invoices`.`category_id`
LEFT OUTER JOIN `preschool_types` cat2 ON `cat2`.`id` = `invoices`.`category_id`
LEFT OUTER JOIN `families` ON `families`.`id` = `invoices`.`unique_id`
WHERE
`invoices`.`unique_id` = 47 AND `invoices`.`client_type` = 0
)
答案 0 :(得分:1)
您是否考虑过使用UNION?