我有几个表,我在一个应用程序中组合,我在PHP中创建基本上创建一个检查列表。我意识到我可以使用PHP中的条件来解决这个问题,但我很好奇MySQL是否能够实现这一点,如果是这样,怎么样?具体来说,我有四个表,使用以下语句查询:
SELECT
cl_status.status,
users.user_first,
cl_status.date AS status_date,
cl_status.id AS status_id,
cl_status.criteria_id,
cl_criteria.id AS cid,
cl_criteria.description AS description
FROM cl_criteria
LEFT JOIN cl_lists
ON cl_criteria.cl_id = cl_lists.id
RIGHT JOIN cl_status
ON cl_criteria.id = cl_status.criteria_id
LEFT JOIN users
ON cl_status.user_id = users.user_id
WHERE cl_lists.id = '1'
Table one - cl_lists:
+----+------------------+------------+------------+-------+
| id | title | date | comp_level | owner |
+----+------------------+------------+------------+-------+
| 1 | Newcomer's guide | 1452473606 | 1 | 1 |
+----+------------------+------------+------------+-------+
Table two - cl_assign:
+----+-------+-------+------------+
| id | cl_id | owner | date |
+----+-------+-------+------------+
| 1 | 1 | 1 | 1455843514 |
+----+-------+-------+------------+
Table three - cl_status:
+----+-------------+---------+-------------+------------+--------+----------+
| id | criteria_id | user_id | description | date | status | comments |
+----+-------------+---------+-------------+------------+--------+----------+
| 2 | 66 | 1 | | NULL | 1 | NULL |
| 15 | 65 | 1 | | 1455842197 | 5 | NULL |
| 16 | 67 | 1 | | 1455842201 | 5 | NULL |
| 17 | 68 | 1 | | 1455842203 | 5 | NULL |
| 18 | 69 | 1 | | 1455842217 | 0 | NULL |
| 19 | 70 | 1 | | 1455842222 | 5 | NULL |
| 20 | 72 | 1 | | 1455842237 | 1 | NULL |
| 21 | 71 | 1 | | 1455842234 | 0 | NULL |
| 22 | 73 | 1 | | 1455842246 | 5 | NULL |
| 23 | 76 | 1 | | 1455842249 | 5 | NULL |
| 24 | 77 | 1 | | 1455842268 | 5 | NULL |
| 25 | 78 | 152 | | 1455854420 | 3 | NULL |
| 26 | 81 | 1 | | 1455843660 | 5 | NULL |
+----+-------------+---------+-------------+------------+--------+----------+
Table four - users:
+---------+------------+
| user_id | user_first |
+---------+------------+
| 1 | Mark |
| 2 | Test |
+---------+------------+
理想情况下,我希望联接看起来像这样:
+--------+------------+-------------+-----------+-------------+------+-----------------------------+
| status | user_first | status_date | status_id | criteria_id | cid | description |
+--------+------------+-------------+-----------+-------------+------+-----------------------------+
| 5 | Mark | 1455842197 | 15 | 65 | 65 | Tour of facility |
| 5 | Mark | 1455842201 | 16 | 67 | 67 | Tax forms |
| 5 | Mark | 1455842203 | 17 | 68 | 68 | 2 forms of ID |
| 0 | Mark | 1455842217 | 18 | 69 | 69 | Benefits | |
| 5 | Mark | 1455842246 | 22 | 73 | 73 | Intro to policies |
| 5 | Mark | 1455842249 | 23 | 76 | 76 | Setup email account |
| NULL | NULL | NULL | NULL | 78 | 78 | Setup Computer account |
+--------+------------+-------------+-----------+-------------+------+-----------------------------+
然而,它看起来像这样:
+--------+------------+-------------+-----------+-------------+------+-----------------------------+
| status | user_first | status_date | status_id | criteria_id | cid | description |
+--------+------------+-------------+-----------+-------------+------+-----------------------------+
| 5 | Mark | 1455842197 | 15 | 65 | 65 | Tour of facility |
| 5 | Mark | 1455842201 | 16 | 67 | 67 | Tax forms |
| 5 | Mark | 1455842203 | 17 | 68 | 68 | 2 forms of ID |
| 0 | Mark | 1455842217 | 18 | 69 | 69 | Benefits |
| 5 | Mark | 1455842246 | 22 | 73 | 73 | Intro to policies |
| 5 | Mark | 1455842249 | 23 | 76 | 76 | Setup email account |
| 3 | Temp | 1455854420 | 25 | 78 | 78 | Setup Computer account |
+--------+------------+-------------+-----------+-------------+------+-----------------------------+
有没有办法在加入之前应用条件?或者另一种方法来完成我想要的结果集?
标准表将包括我拥有的每个清单的步骤。列表是各种清单的列表。状态表允许每个用户(例如Mark或Test)查看相同的清单并完成它,就像它是一个单独的文档一样。它还会填充该用户更新项目的日期/时间。
答案 0 :(得分:1)
我怀疑您查询中的RIGHT JOIN
会导致您要显示的记录被过滤掉。请注意,t1 RIGHT JOIN t2
与t2 LEFT JOIN t1
相同,这意味着t1
会丢失任何未出现在t2
中的记录,t2
会保留其所有记录。试试这个:
SELECT cl_status.status, users.user_first, cl_status.date AS status_date,
cl_status.id AS status_id, cl_status.criteria_id, cl_criteria.id AS cid,
cl_criteria.description AS description
FROM cl_criteria LEFT JOIN cl_lists
ON cl_criteria.cl_id = cl_lists.id
LEFT JOIN cl_status
ON cl_criteria.id = cl_status.criteria_id
LEFT JOIN users
ON cl_status.user_id = users.user_id
WHERE cl_lists.id = '1'