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See more: VB7.x VB8.0 VB9.0 VB VBScript XML , +
<?xml version="1.0" encoding="UTF-8"?>
-<calls count="31" date="1360309147911"> <call date="1359279658478" photoid="0" name="nish" duration="29" new="1" numbertype="2" modified="1" type="2" number="7998609" iccid="8923001070620096930" modified_time="1360181147995"/>
 
How to decrypt the dates and time in vb.net ?
Posted 21-Feb-13 9:37am
Comments
Sergey Alexandrovich Kryukov at 21-Feb-13 14:41pm
   
How can we know what is it?
—SA
virajdaw at 22-Feb-13 0:34am
   
Really i didn't know as well, but the person who replied this question told the date is in the form on unix format.Well it is true..
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Solution 1

All the times are the same order of magnitude and relatively close to each other, so they aren't actually encrypted.
 
Most likely they are just the total number of some units of time since some origin date.
 
It's either seconds, or milliseconds, since some date.
 
1359279658478 millisecond = aprox 43.07 years (approximating a year at 365.25 days).
 
1 Jan 1970 + 43.07 years = sometime in Jan 2013.
 
Most likely it's just standard unix time (a system of recording time as milliseconds since 1 Jan 1970):
 
http://en.wikipedia.org/wiki/Unix_time[^]
 
Given that there is no notation for the time zone, you have to assume UCT.
 
NOTE: Deriving the exact date requires getting leap year calculations correct and accounting for time zond and daylight savings time differences. Any modern operating system has a function that handles that for you.
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Solution 2

This site UNIX timestamp to System.DateTime[^] help me to decipher the dates which were in UNIX format.
 
On the above site, it uses seconds, for call date="1359279658478" please use milliseconds,It'll work.
 
Foe example replace dateTime = dateTime.AddSeconds(timestamp);
with dateTime = dateTime.AddMilliseconds(timestamp)
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