我有点想法。我有一张桌子,其中有以下数据:
ID | timedate | value | ...
1 | 04-16-2017 | 3 | ...
1 | 04-20-2017 | 4 | ...
1 | 04-24-2017 | 10 |...[approx 100k different IDs)
现在,用户搜索(例如)ID = 1.在我的视图中,我想显示不同的cols和Value的差异。但差异应该只是从最后一个时间到最后一个时间日期,如下:
ID | timedate | value | dif.Value
1 | 04-24-2017 | 10 | 6 [10-4]
我的实际mySQL-Query看起来像这样:
SELECT
tab.Id AS `tab__Id`,
tab.timedate AS `tab__timedate`,
tab.value AS `tab__value`,
tab.value - t2.value AS `difValue`,
FROM
tab tab
INNER JOIN tab3 t3 ON t3.Id = tab.Id
INNER JOIN tab t2 ON t2.Id = tab.Id
WHERE
(t3.Id like '1'
AND tab.timedate IN (
SELECT
max(tab.timedate)
FROM
tab
GROUP BY
tab.Id
)
)
GROUP BY
tab.Id
LIMIT
10 OFFSET 0
来自cakePHP3 =
$search= $tab->find('all')
->select(['Id', 'tab.timedate ', 'tab.value','difValue' =>
'tab.value - t2.value'])
->andWhere(['t3.Id LIKE' => $id])
->where(['tab.timedate IN (SELECT max(tab.timedate) FROM tab
GROUP BY tab.Id)'])
->distinct(['tab.Id'])
->join([
't3' => [
'table' => 'tab3',
'conditions' => 't3.Id= tab.Id ',
],
't2' => [
'table' => 'tab',
'conditions' => ' t2.Id= tab.Id
]
]);
如何设置他在t2.value的第二个最后日期加入(t2)?我试图将此添加到我的条件t2,但它需要第一个时间日期
...AND tab.timedate > t2.timedate - INTERVAL 1 DAY'
也许有另一种方法可以解决这个问题。我只想计算差异。最后一个时间:值和倒数第二个时间:值(上一个值 - 倒数第二个值=>“difValue”)
答案 0 :(得分:0)
为了清楚起见,我已经添加,而不是减去 - 而且只是为了与众不同。请注意,外部查询中对“当前”的引用实际上只是对任何恰好是主键的引用。
SELECT * FROM ints;
+---+
| i |
+---+
| 0 |
| 1 |
| 2 |
| 3 |
| 4 |
| 5 |
| 6 |
| 7 |
| 8 |
| 9 |
+---+
SELECT total
FROM
( SELECT @i:=@i+i total
, @i:=i current
FROM ints
, (SELECT @i:=0) vars
ORDER
BY i DESC
LIMIT 2
) x
ORDER
BY current
LIMIT 1;
+-------+
| total |
+-------+
| 17 |
+-------+