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Old Timestamp
Posted: Sun Oct 09, 2005 7:47 am
by FloSoft
Hi,
i found an old QBasic 3 program of which i have no source. I tried to decode the data-files, but got some problems:
How do I translate an old "timestamp"?
Data in File (Hex) -> Displayed in Program as
A0 94 59 94 -> 10.12.89
E0 95 59 94 -> 30.12.89
60 89 59 94 -> 30.10.89
60 90 43 94 -> 30.10.80
60 F3 71 94 -> 30.10.99
00 53 45 91 -> 30.10.10
Its not a plain integer, (no unix timestamp)
I hope somebody can help me to decode this.
Posted: Sun Oct 09, 2005 8:43 am
by MystikShadows
You could read them as String of 8 characters (if the periods you displayed here are part of the data) or 6 characters (if the periods are just spaces).
Then you could do the following assuming your field is called DateField:
Code: Select all
DIM DayValue AS INTEGER
DIM MonthValue AS INTEGER
DIM YearValue AS INTEGER
DayValue = VAL(LEFT$(DateField, 2))
MonthValue = VAL(MID$(DateField, 3, 2))
YearValue = 1900 + VAL(RIGHT$(DateField, 2))
In the case of the first line of date you should you should then have
DayValue = 10
MonthValue = 12
YearValue = 1989
You can then form a string with this in a function:
Code: Select all
FUNCTION DateString(Day AS INTEGER, Month AS INTEGER, Year AS INTEGER) AS STRING
DIM DayOffset AS STRING
DIM MonthOffset AS STRING
DIM WorkDate AS STRING
IF Day < 10 THEN
DayOffset="0"
ELSE
DayOffset = ""
END IF
IF Month < 10 THEN
MonthOffset="0"
ELSE
MonthOffset = ""
END IF
DateString$ = DayOffset + TRIM$(STR$(Day)) + "-" + MonthOffset + TRIM$(STR$(Month)) + "-" + TRIM$(STR$(Year))
END FUNCTION
And that will return you a date in the form of "DD-MM-YYYY" which you should be able to use.
Hope this helps
Re: Old Timestamp
Posted: Sun Oct 09, 2005 9:02 pm
by moneo
FloSoft wrote:Hi,
i found an old QBasic 3 program of which i have no source. I tried to decode the data-files, but got some problems:
How do I translate an old "timestamp"?
Data in File (Hex) -> Displayed in Program as
A0 94 59 94 -> 10.12.89
E0 95 59 94 -> 30.12.89
60 89 59 94 -> 30.10.89
60 90 43 94 -> 30.10.80
60 F3 71 94 -> 30.10.99
00 53 45 91 -> 30.10.10
Its not a plain integer, (no unix timestamp)
I hope somebody can help me to decode this.
I've been looking at the hex values. They don't make sense.
Maybe the hex values represent the number of offset days from some pivot year. Even thinking this, the values still don't make an sense, because if
A0 94 59 94 means 10.12.89, and
E0 95 59 94 means 30.12.89,
then these supposed dates are only 20 days apart, but the hex values
don't reflect a 20 day difference any way you look at them.
Whoever created these hex values had his own kookie algorithm.
I give up.
*****
Posted: Mon Oct 10, 2005 2:54 am
by FloSoft
@MystikShadows: Hmm the Values are only 4 Bytes.
@mondeo: Yeah that was my problem too. I found following:
Between 30.12.10 und 30.12.11 are exaktly 1280000 "Seconds" in the timestamp (if you interpret the 4 bytes as an 32bit integer) if you divide that value by 3600, you get 355.5555555. But that does not help me much, but perhaps you have another idea?
Posted: Mon Oct 10, 2005 3:03 am
by FloSoft
Sorry, i could not edit my previous post:
I currently found another thing:
2488898720 10.12.89
2488898880 20.12.89
2488899040 30.12.89
the first number is the integer if you interpret those 4 bytes as int (little endian, end to front)
if you subtract them from each other, you get:
30.12.89 -> 20.12.89 : 160
20.12.89 -> 10.12.89 : 160
30.12.89 -> 10.12.89 : 320
so one day is interpreted as a value of 16?
[edit]
ok i found out that all values above the year "00" are strange. if i stay between 70-99 it works:
Day: 16
Month: 1600
Year: 160000
10.08.95:
10*16 + 8*1600 + 95* 160000 + 2474639360 (kind of "base"?) = correct timestamp
the other way round works too.
[/edit]
Posted: Mon Oct 10, 2005 5:19 am
by Antoni
Some Ideas:
1.-
The DOS file timestamp uses bitfields,7 bits for the minute, 5 bits for the hour, 6 bits for day, 4 bits for month and the rest for year since 1980 (iirc) . If you think it could be that I can check my old books for the exact format.
2.-
Microsoft Excel uses a floating point value, the unit is one day.