如何在现有模式中查找Informix DATETIME字段限定符

Question

我有一个这样的表：

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)
);

现在我想对这个表的数据类型信息进行逆向工程。我最感兴趣的是小数秒部分，但如果我能找到其他限定符信息，那就更好了。以下查询不起作用：

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'

它产生

|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     |

collength似乎包含相关信息，但我无法使用schema_precision或schema_numscale提取它，否则可能会导致数字精度。此外，schema_datetype没有产生令人感兴趣的结果。

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

Answer 1

基于文档时间数据类型：

对于DATETIME或INTERVAL类型的列，使用以下公式确定collength：

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

长度是DATETIME或INTERVAL字段的物理长度，first_qualifier和last_qualifiers的值如下表所示。

+------------------+--------+------------------+-------+
| 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(十六进制值，更容易发现图案(：

t1 datetime year to fraction(1),  15*256 + 0*16+11 = 3851   0x0F0B
t2 datetime year to fraction(2),  16*256 + 0*16+12 = 4108   0x100C
t3 datetime year to fraction(3),  17*256 + 0*16+13 = 4365   0x110D
t4 datetime year to fraction(4),  18*256 + 0*16+14 = 4622   0x120E

---

如果长度已知，则即使使用"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<>小提琴演示