从发帖的角度来看,我是这个论坛的新手,正如这个问题所显示的那样。我希望数据的显示方式存在一些问题。
所以,我有 3 个表(我只显示我想要的列):
访客:
center_id (where the visitor was) | state_id (where they came from)
中心:
center_id | center
状态:
state_id | state
这是我一直在使用的查询
SELECT states.state, visitors.center_id, visitors.state_id, centers.center, COUNT(visitors.state_id) AS totalCount
FROM visitors
LEFT JOIN states ON states.state_id = visitors.state_id
LEFT JOIN centers ON centers.center_id = visitors.center_id
WHERE visitors.vdate = <some date> AND visitors.state_id <> '0'
GROUP BY centers.center, visitors.state_id
这将产生以下数组:
Array
(
[0] => Array
(
[state] => Connecticut
[location_id] => 1
[state_id] => 8
[center] => Little River
[totalCount] => 1
)
[1] => Array
(
[state] => California
[location_id] => 5
[state_id] => 6
[center] => North Augusta
[totalCount] => 1
)
[2] => Array
(
[state] => Colorado
[location_id] => 5
[state_id] => 7
[center] => North Augusta
[totalCount] => 2
)
[6] => Array
(
[state] => Connecticut
[location_id] => 9
[state_id] => 8
[center] => Santee
[totalCount] => 2
)
[7] => Array
(
[state] => Virginia
[location_id] => 9
[state_id] => 51
[center] => Santee
[totalCount] => 1
)
)
这是我真正想要的:
Array
(
[Little River] => Array
(
[0] => Array
(
[state] => Connecticut
[state_id] => 8
[totalCount] => 1
)
)
[North Augusta] => Array
(
[0] => Array
(
[state] => California
[state_id] => 6
[totalCount] => 1
)
[1] => Array
(
[state] => Colorado
[state_id] => 7
[totalCount] => 2
)
)
[Santee] => Array
(
[0] => Array
(
[state] => Connecticut
[state_id] => 8
[totalCount] => 2
)
[1] => Array
(
[state] => Virginia
[state_id] => 51
[totalCount] => 1
)
)
)
最终我把它放到一个看起来像这样的表中:
__________________
|State | Count|
-------------------
| Santee |
-------------------
| Georgia | 5 |
-------------------
| Alabama | 10 |
-------------------
| North Augusta |
-------------------
| another | 7 |
-------------------
对不起,啰嗦了,但这是我可以描述它的唯一方式。
我也试过用 php 破解它,但我可能在那里也做错了什么。我可以制作一个包含 3 列的表格,其中每个州都列出了中心,但我真的在寻找一行来显示中心,然后是所有州,并计算该中心和下一个中心。
任何援助将不胜感激。