How do I find the number of hours between two dates?", substituting minutes, seconds, days, or whatever interval you are looking for in the place of 'hours'. Well, in Visual Basic .NET there are two main ways to achieve this result; the DateDiff function or through the TimeSpan structure.

Both of these methods are equally valid, where they overlap in functionality, but each of them has a couple of unique features. Despite their similar functions, the core difference between the two is that DateDiff is a function, so you need to call it every time you need to retrieve a value, whereas TimeSpan is a structure that is created once and then you just work with its various members as needed.

Using DateDiff, you call it with different date interval parameters to retrieve the appropriate value:

Dim D1, D2 As Date
D1 = Date.Now
D2 = #11/9/2004#
'DateDiff
Console.WriteLine("DateDiff")
Console.WriteLine()
Console.WriteLine("{0} Days", _
DateDiff(DateInterval.Day, D1, D2))
Console.WriteLine("{0} Hours", _
DateDiff(DateInterval.Hour, D1, D2))
Console.WriteLine("{0} Minutes", _
DateDiff(DateInterval.Minute, D1, D2))
Console.WriteLine("{0} Seconds", _
DateDiff(DateInterval.Second, D1, D2))
Console.WriteLine()

Alterna tively, a TimeSpan structure can be retrieved as the result of subtracting one date from another, and then querying the various members of that structure.

'TimeSpan
Console.WriteLine("TimeSpan")
Console.WriteLine()
Dim difference As TimeSpan = D2.Subtract(D1)
Console.WriteLine("{0} Days", difference.TotalDays)
Console.WriteLine("{0} Hours", difference.TotalHours)
Console.WriteLine("{0} Minutes", difference.TotalMinutes)
Console.WriteLine("{0} Seconds", difference.TotalSeconds)
Console.WriteLine()

The output of the two different methods is nearly identical, except that the TimeSpan properties are returning Doubles, while DateDiff always returns Longs (Int64).

DateDiff

175 Days
4222 Hours
253345 Minutes
15200730 Seconds

TimeSpan

175.934 383644387 Days
4222.42520746528 Hours
253345.512447917 Minutes
15200730.746875 Seconds

The fractional values returned from the TimeSpan properties are more meaningful that they might seem at first glance; each of these properties is returning the complete difference between the two dates, whereas the whole numbers returned from DateDiff only provide a value to within one interval (day, hour, etc…). Using only one of these properties from TimeSpan, it would be possible to figure out any other time interval (with varying degrees of precision, of course), although that calculation would seldom be required in practice.

The TimeSpan structure provides additional members (.Hours, .Days, etc...) that return values closer to what you see from DateDiff, so you can use those if it is more suitable for your situation. While TimeSpan can appear to have more features and to be more efficient, DateDiff has two benefits that may cause you to continue to use it in your Visual Basic .NET applications:

1. It provides several date intervals that are not covered by TimeSpan, including Weeks and Quarters, and
2. It is compatible with the VB6 function of the same name, allowing you to more easily reuse existing code.

Read the corresponding documentation for more information, but I hope that this has been helpful...

* DateDiff
* TimeSpan

source taken from:
http://blogs.msdn.com/v bfaq/default.aspx

Answered by jaivir , an ibibo Master, at 7:01 PM on September 12, 2008

there a datediff () fucntion in vb and vb.net check that

Answered by Leo G , an ibibo Citizen, at 5:51 PM on September 12, 2008

Suppose you want to know the age of a person given her birth date. Let’s say her birth date is "09-23-1969". Although in this case you might say that we don’t need a program to compute the age, but assuming there are 1000s of records in your database and for every person you have to print a report like:

Sheila is 29 years old
Fatima is 33 years old
Peter is 17 years old
… and so on

In the database we don’t generally store what the person’s age is — we store what the person’s date of birth is. This is because given the date of birth, we can always deduct it from today’s date and calculate the number of years that person has lived so far. Calculating date difference also comes handy if you are running a subscription-based website. You need to know for how many days a person has been using your website and when the subscription is due. So in this tip we learn to write a function that returns us the number of days elapsed between two dates. First the function (I’ll discuss it later on)

function dateDiff($dformat, $endDate, $beginDate)
{
$date_parts1=explode($dformat, $beginDate);
$date_parts2=explode($dformat, $endDate);
$start_date=gregoriantojd($dat e_parts1[0], $date_parts1[1], $date_parts1[2]);
$end_date=gregoriantojd($date_ parts2[0], $date_parts2[1], $date_parts2[2]);
return $end_date - $start_date;
}

Now let us see how we use this function:

$date1="07/11/2003";
$date2="09/04/2004";

print "If we minus " . $date1 . " from " . $date2 . " we get " . dateDiff("/", $date2, $date1) . ".";

which generates If we minus 07/11/2003 from 09/04/2004 we get 421.

The dateDiff() functions uses two PHP functions viz., explode() and gregoriantojd(). The explode() function is used mostly to convert strings into arrays. It takes two arguments: the separator and the string. So if

$string="Trees, plants, bushes";
$stringparts=explode(", ", $string);

then $stringparts[0] holds "Trees", $stringparts[1] holds "plants" and $stringparts[2] holds "bushes". The separator is ", ". Similarly, in the above example, $date_parts1[0] holds the month part, $date_parts1[1] holds the day part and $date_parts1[2] holds the year part and these are the arguments that the function gregoriantojd() needs. The syntax of the gregoriantojd() function is

gregoriantojd($month, $day, $year)

But why do we need this function? Without going into the gory details, it changes the date into a big number from which another number (obtained from another date) can be deducted. Having said that, let us now see a few applications for this function: first the age thing. The following snippet of code gives us the age of a person according to the current date.

$dob="08/12/1975";
echo "If you were born on " . $dob . ", then today your age is approximately " . round(dateDiff("/", date("m/d/Y", time()), $dob)/365, 0) . " years.";

This generates a line appearing below:

If you were born on 08/12/1975, then today your age is approximately 30 years.

Answered by Shobhit , an ibibo Master, at 5:28 PM on September 12, 2008

Calculating the difference between two dates in JavaScript is relatively straightforward, provided you choose the right Date methods to work with. Whichever way you get there, the potential applications on date differences are many, from counting down to a particular event, counting up from a past date, to dynamically indicating what's new on your page. Sounds like fun!

Lets begin this tutorial by getting to the heart of it:
Date.getTime()

For the scripts that follow, the above is our hero people, not some Hollywood actor. Date.getTime() is a prebuilt JS method that returns the time elapsed from January 1st, 1970 to the current Date instance, in milliseconds. Its superpower is not so much its long term memory, as impressive as that may be, but its knack for converting a date to a number (in milliseconds, but nevertheless). And we all know the easiest subjects to perform arithmetic on are numbers.

Answered by Saq , an ibibo Master, at 4:33 PM on September 12, 2008

Using DateDiff, you call it with different date interval parameters to retrieve the appropriate value:

Dim D1, D2 As Date
D1 = Date.Now
D2 = #11/9/2004#
'DateDiff
Console.WriteLine("DateDiff")
Console.WriteLine()
Console.WriteLine("{0} Days", _
DateDiff(DateInterval.Day, D1, D2))
Console.WriteLine("{0} Hours", _
DateDiff(DateInterval.Hour, D1, D2))
Console.WriteLine("{0} Minutes", _
DateDiff(DateInterval.Minute, D1, D2))
Console.WriteLine("{0} Seconds", _
DateDiff(DateInterval.Second, D1, D2))
Console.WriteLine()

Alterna tively, a TimeSpan structure can be retrieved as the result of subtracting one date from another, and then querying the various members of that structure.

'TimeSpan
Console.WriteLine("TimeSpan")
Console.WriteLine()
Dim difference As TimeSpan = D2.Subtract(D1)
Console.WriteLine("{0} Days", difference.TotalDays)
Console.WriteLine("{0} Hours", difference.TotalHours)
Console.WriteLine("{0} Minutes", difference.TotalMinutes)
Console.WriteLine("{0} Seconds", difference.TotalSeconds)
Console.WriteLine()

The output of the two different methods is nearly identical, except that the TimeSpan properties are returning Doubles, while DateDiff always returns Longs (Int64).

DateDiff

175 Days
4222 Hours
253345 Minutes
15200730 Seconds

TimeSpan

175.934 383644387 Days
4222.42520746528 Hours
253345.512447917 Minutes
15200730.746875 Seconds

Answered by Sharad Singh , an ibibo Master, at 2:24 PM on September 12, 2008

hi sidd!

This common question is often phrased as "How do I find the number of hours between two dates?", substituting minutes, seconds, days, or whatever interval you are looking for in the place of 'hours'. Well, in Visual Basic .NET there are two main ways to achieve this result; the DateDiff function or through the TimeSpan structure.

Both of these methods are equally valid, where they overlap in functionality, but each of them has a couple of unique features. Despite their similar functions, the core difference between the two is that DateDiff is a function, so you need to call it every time you need to retrieve a value, whereas TimeSpan is a structure that is created once and then you just work with its various members as needed.

Using DateDiff, you call it with different date interval parameters to retrieve the appropriate value:


Dim D1, D2 As Date D1 = Date.Now D2 = #11/9/2004# 'DateDiff Console.WriteLine("DateDiff") Console.WriteLine() Console.WriteLine("{0} Days", _ DateDiff(DateInterval.Day, D1, D2)) Console.WriteLine("{0} Hours", _ DateDiff(DateInterval.Hour, D1, D2)) Console.WriteLine("{0} Minutes", _ DateDiff(DateInterval.Minute, D1, D2)) Console.WriteLine("{0} Seconds", _ DateDiff(DateInterval.Second, D1, D2)) Console.WriteLine()Alternatively, a TimeSpan structure can be retrieved as the result of subtracting one date from another, and then querying the various members of that structure.
'TimeSpan Console.WriteLine("TimeSpan") Console.WriteLine() Dim difference As TimeSpan = D2.Subtract(D1) Console.WriteLine("{0} Days", difference.TotalDays) Console.WriteLine("{0} Hours", difference.TotalHours) Console.WriteLine("{0} Minutes", difference.TotalMinutes) Console.WriteLine("{0} Seconds", difference.TotalSeconds) Console.WriteLine()The output of the two different methods is nearly identical, except that the TimeSpan properties are returning Doubles, while DateDiff always returns Longs (Int64).

DateDiff

175 Days
4222 Hours
253345 Minutes
15200730 Seconds

TimeSpan

175.93438364 4387 Days
4222.42520746528 Hours
253345.512447917 Minutes
15200730.746875 Seconds

The fractional values returned from the TimeSpan properties are more meaningful that they might seem at first glance; each of these properties is returning the complete difference between the two dates, whereas the whole numbers returned from DateDiff only provide a value to within one interval (day, hour, etc…). Using only one of these properties from TimeSpan, it would be possible to figure out any other time interval (with varying degrees of precision, of course), although that calculation would seldom be required in practice.

The TimeSpan structure provides additional members (.Hours, .Days, etc...) that return values closer to what you see from DateDiff, so you can use those if it is more suitable for your situation. While TimeSpan can appear to have more features and to be more efficient, DateDiff has two benefits that may cause you to continue to use it in your Visual Basic .NET applications:

It provides several date intervals that are not covered by TimeSpan, including Weeks and Quarters, and
It is compatible with the VB6 function of the same name, allowing you to more easily reuse existing code.
for more inf check http://blogs.msdn.com/vbfaq/archive /2004/05/30/144571.aspx

Answered by raj u , an ibibo Master, at 12:03 PM on September 12, 2008

in Visual Basic .NET there are two main ways to achieve this result; the DateDiff function or through the TimeSpan structure.

Both of these methods are equally valid, where they overlap in functionality, but each of them has a couple of unique features. Despite their similar functions, the core difference between the two is that DateDiff is a function, so you need to call it every time you need to retrieve a value, whereas TimeSpan is a structure that is created once and then you just work with its various members as needed.

Using DateDiff, you call it with different date interval parameters to retrieve the appropriate value:


Dim D1, D2 As Date
D1 = Date.Now
D2 = #11/9/2004#
'DateDiff
Console.WriteLine("DateDiff")
Console.WriteLine()
Console.WriteLine("{0} Days", _
DateDiff(DateInterval.Day, D1, D2))
Console.WriteLine("{0} Hours", _
DateDiff(DateInterval.Hour, D1, D2))
Console.WriteLine("{0} Minutes", _
DateDiff(DateInterval.Minute, D1, D2))
Console.WriteLine("{0} Seconds", _
DateDiff(DateInterval.Second, D1, D2))
Console.WriteLine()

a TimeSpan structure can be retrieved as the result of subtracting one date from another, and then querying the various members of that structure.

'TimeSpan
Console.WriteLine("TimeSpan")
Console.WriteLine()
Dim difference As TimeSpan = D2.Subtract(D1)
Console.WriteLine("{0} Days", difference.TotalDays)
Console.WriteLine("{0} Hours", difference.TotalHours)
Console.WriteLine("{0} Minutes", difference.TotalMinutes)
Console.WriteLine("{0} Seconds", difference.TotalSeconds)
Console.WriteLine()

The output of the two different methods is nearly identical, except that the TimeSpan properties are returning Doubles, while DateDiff always returns Longs (Int64).

DateDiff

175 Days
4222 Hours
253345 Minutes
15200730 Seconds

TimeSpan

175.93438364 4387 Days
4222.42520746528 Hours
253345.512447917 Minutes
15200730.746875 Seconds

The fractional values returned from the TimeSpan properties are more meaningful that they might seem at first glance; each of these properties is returning the complete difference between the two dates, whereas the whole numbers returned from DateDiff only provide a value to within one interval (day, hour, etc…). Using only one of these properties from TimeSpan, it would be possible to figure out any other time interval (with varying degrees of precision, of course), although that calculation would seldom be required in practice.

The TimeSpan structure provides additional members (.Hours, .Days, etc...) that return values closer to what you see from DateDiff, so you can use those if it is more suitable for your situation. While TimeSpan can appear to have more features and to be more efficient, DateDiff has two benefits that may cause you to continue to use it in your Visual Basic .NET applications:

It provides several date intervals that are not covered by TimeSpan, including Weeks and Quarters, and
It is compatible with the VB6 function of the same name, allowing you to more easily reuse existing code.
Read the corresponding documentation for more information, but I hope that this has been helpful...

DateDiff
TimeSpan

Answered by Hem@nt .... , an ibibo Master, at 11:54 AM on September 12, 2008

The attached Java file contains three methods: the div() method provides integer division with rounding towards negative infinity, which Java does not provide by itself (the normal division and modulus operations round towards zero). Use of this function makes the other methods usable with dates before 1970 because it will handle negative numbers correctly.

The wdnum() method returns the number of weekdays (excluding weekends) that have passed since Monday, 29 December 1969. It works by calculating the number of days since 1 January, 1970 (getTime() divided by the number of milliseconds in a day), adding 3 and returning the number of week days in full weeks and possibly a partial week that have passed since then.

The diff() method shows one possible use of wdnum(): it calculates the number of weekdays that have passed between two given dates, excluding weekends.

An example use of the diff() function would be to calculate the number of working days in a month. This could be done as follows:

Calendar cal = Calendar.getInstance(TimeZone. getTimeZone("UTC"));
cal.set( year, month, 1);
Date d1 = cal.getTime();
cal.set(year, month + 1, 1);
Date d2 = cal.getTime();
System.out.pri ntln("Month " + month + ", " + year + " has " + diff(d1, d2) + " working days");

Note that the Calendar has to be set to the UTC time zone because the Date objects passed to diff() are expected to be relative to UTC.

For more visit :

http://www.developerfusio n.co.uk/show/5734/

Answered by Ashish Yadav , an ibibo Wizard, at 10:57 AM on September 12, 2008

well;
The dateDiff() functions uses two PHP functions viz., explode() and gregoriantojd(). The explode() function is used mostly to convert strings into arrays. It takes two arguments: the separator and the string. So if

$string="Trees, plants, bushes";
$stringparts=explode(", ", $string);

then $stringparts[0] holds "Trees", $stringparts[1] holds "plants" and $stringparts[2] holds "bushes". The separator is ", ". Similarly, in the above example, $date_parts1[0] holds the month part, $date_parts1[1] holds the day part and $date_parts1[2] holds the year part and these are the arguments that the function gregoriantojd() needs. The syntax of the gregoriantojd() function is

gregoriantojd($month, $day, $year)

But why do we need this function? Without going into the gory details, it changes the date into a big number from which another number (obtained from another date) can be deducted. Having said that, let us now see a few applications for this function: first the age thing. The following snippet of code gives us the age of a person according to the current date.

$dob="08/12/1975";
echo "If you were born on " . $dob . ", then today your age is approximately " . round(dateDiff("/", date("m/d/Y", time()), $dob)/365, 0) . " years.";

This generates a line appearing below:

If you were born on 08/12/1975, then today your age is approximately 30 years.

Answered by Kishore , an ibibo Master, at 8:12 AM on September 12, 2008

Assuming on your previous Qs, that you're asking this to find the difference between two dates in VB.NET. Right? It was an easy job with classic VB to find difference between two dates or times. Even though it is not very difficult, it is a tricky and confusing thing in C# and VB.NET. In Visual Basic .NET there are two main ways to achieve this result; the DateDiff function (you need to call it every time you need to retrieve a value) or through the TimeSpan structure (created once and then you just work with its various members as needed). Both of these methods are equally valid. Let me suggest you to take a look at the below website for a typical examples that employs both these methods:

http://blogs.msdn.com/v bfaq/archive/2004/05/30/144571.aspx

or

http://tinyurl.com/vbdate

Answered by A.R. , an ibibo Master, at 7:40 AM on September 12, 2008