如何在现有模式中查找Informix DATETIME字段限定符

Question

我有一张这样的桌子：

create table t (
  t0 datetime year to fraction,
  t1 datetime year to fraction(1),
  t2 datetime year to fraction(2),
  t3 datetime year to fraction(3),
  t4 datetime year to fraction(4)
);

现在，我想对该表的数据类型信息进行反向工程。我对小数秒部分最感兴趣，但是如果我可以找到其他预选赛信息，那就更好了。以下查询无效：

select 
  c.colname::varchar(10) colname,
  informix.schema_coltypename(c.coltype, c.extended_id)::varchar(10) coltypename,
  c.collength,
  informix.schema_precision(c.coltype, c.extended_id, c.collength) precision,
  informix.schema_numscale(c.coltype, c.collength) numscale,
  informix.schema_datetype(c.coltype, c.collength) datetype,
  c.coltype
from syscolumns c
join systables t on c.tabid = t.tabid
where t.tabname = 't'

产生

|colname   |coltypename|collength|precision  |numscale   |datetype   |coltype|
|----------|-----------|---------|-----------|-----------|-----------|-------|
|t0        |DATETIME   |4365     |4365       |           |60         |10     |
|t1        |DATETIME   |3851     |3851       |           |60         |10     |
|t2        |DATETIME   |4108     |4108       |           |60         |10     |
|t3        |DATETIME   |4365     |4365       |           |60         |10     |
|t4        |DATETIME   |4622     |4622       |           |60         |10     |

长度似乎包含相关信息，但是我不能用schema_precision或schema_numscale提取它，否则可能会导致数值精度。另外，schema_datetype不会产生有趣的结果。

如何将coltype信息反向工程回datetime year to fraction(N)？

Answer 1

基于文档Time data types：

对于DATETIME或INTERVAL类型的列，使用以下公式确定长度：

(length * 256) + (first_qualifier * 16) + last_qualifier

长度是DATETIME或INTERVAL字段的物理长度，并且first_qualifier和last_qualifier具有下表所示的值。

+------------------+--------+------------------+-------+
| Field qualifier  | Value  | Field qualifier  | Value |
+------------------+--------+------------------+-------+
| YEAR             |     0  | FRACTION(1)      |    11 |
| MONTH            |     2  | FRACTION(2)      |    12 |
| DAY              |     4  | FRACTION(3)      |    13 |
| HOUR             |     6  | FRACTION(4)      |    14 |
| MINUTE           |     8  | FRACTION(5)      |    15 |
| SECOND           |    10  |                  |       |
+------------------+--------+------------------+-------+

计算：

  t1 datetime year to fraction(1),  15*256 + 0*16+11 = 3851
  t2 datetime year to fraction(2),  16*256 + 0*16+12 = 4108
  t3 datetime year to fraction(3),  17*256 + 0*16+13 = 4365
  t4 datetime year to fraction(4),  18*256 + 0*16+14 = 4622

---

如果长度已知，那么即使使用“蛮力”也可以对其进行反向工程。

查找：

WITH l(v) AS (
  VALUES (12),(13),(14),(15),(16),(17),(18)
), first_q(v, first_qualifier) AS (
  VALUES (0,'YEAR'),(2,'MONTH'),(4,'DAY'),(6,'HOUR'),(8,'MINUTE'),(10, 'SECOND')
), last_q(v, last_qualifier) AS (
  VALUES (11, 'FRACTION(0)'),(12, 'FRACTION(1)'),(13, 'FRACTION(2)'),
         (14, 'FRACTION(3)'),(15, 'FRACTION(4)')
), result AS (
  SELECT  l.v * 256 + (first_q.v * 256) + last_q.v AS collen, *
  FROM l CROSS JOIN first_q CROSS JOIN last_q
)
SELECT *
FROM result
--WHERE collen = 3851

db<>fiddle demo