Question

我有两个不同的表值，一个有15分钟的间隔，一个有小时值。我想创建一个包含PHF列的新表其中PHF =每小时值/ 4 *（该小时最长15分钟间隔）如何为此编写查询？

TABLE1-15 min interval

ID    PHASE       STARTIME               COUNT
1201; 1;     "2015-01-05 12:15:00-06";    234
1201; 1;     "2015-01-05 12:30:00-06";     344
1201; 1;     "2015-01-05 12:45:00-06";     755
1202; 2;     "2015-01-05 01:00:00-06";     234
1202; 2;     "2015-01-05 01:15:00-06";     244
1202; 2;     "2015-01-05 01:30:00-06";     245
1202; 2;     "2015-01-05 01:45:00-06";     246

TABLE2-上表每小时间隔

ID   PHASE   STARTIME                      COUNT
1201; 1;     "2015-01-05 12:00:00";        1333
1202; 2;     "2015-01-05 01:00:00";        969

输出

ID   PHASE   STARTIME                      COUNT     PHF
1201; 1;     "2015-01-05 12:00:00";        1333;     0.44
1202; 2;     "2015-01-05 01:00:00";        969;       0.9847

PHF定义为：每小时值/（4 *最大值（15分钟值）

所以从上面的例子来看，phf1 = 1333 /（4 * 755）=。44

和phf2 =（969 /（4 * 246）=。9847

Answer 1

<强> SQL Fiddle Example

SELECT a.id,
  a.phase,
  b.starttime,
  b.count/cast((4 * a.count) as numeric) AS "PHF"
FROM TableA a
JOIN TableB b on a.id = b.id
JOIN (SELECT id, phase, max(starttime) as "starttime" 
      FROM TableA GROUP BY id, phase) c on a.id = c.id and a.starttime = c.starttime

如何在postgres中进行计算

1 个答案: