#196767 - 2009-11-19 02:07 PM
(Not) Fun with the ScriptControl object
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Skatterbrainz
Starting to like KiXtart
Registered: 2002-10-17
Posts: 172
Loc: Virginia, USA
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I'm curious what is causing this to not work in 4.61. The DateDiff function works, but DateAdd and DatePart don't return any result. Maybe I'm doing something wrong?
Function DateDiff($unit, $date1, $date2)
Dim $sc, $result
$sc = CreateObject("ScriptControl")
$sc.Language = "vbscript"
$result = $sc.Eval("DateDiff("+Chr(34)+$unit+Chr(34)+", "+Chr(34)+$date1+Chr(34)+", "+Chr(34)+$date2+Chr(34)+")")
$sc = 0
$DateDiff = $result
EndFunction
Function DateAdd($unit, $incr, $date)
Dim $sc, $result
$sc = CreateObject("ScriptControl")
$sc.Language = "vbscript"
$result = $sc.Eval("DateAdd("+Chr(34)+$unit+Chr(34) + ", " + $incr + ", " + Chr(34)+$date+Chr(34)+")")
$sc = 0
$DateAdd = $result
EndFunction
Function DatePart($unit, $date)
Dim $sc, $result
$sc = CreateObject("ScriptControl")
$sc.Language = "vbscript"
$result = $sc.Eval("DatePart("+Chr(34)+$unit+Chr(34)+", "+Chr(34)+$date+Chr(34)+")")
$sc = 0
$DatePart = $result
EndFunction
$testvalue = '2009/11/03 12:34:56'
$test1 = FormatDateTime($testvalue, 'vbShortDate')
$test2 = FormatDateTime($testvalue, 'vbLongDate')
$test3 = FormatDateTime($testvalue, 'vbLongTime')
$test4 = DateDiff('d', '7/19/2009', $testvalue)
$test5 = DateAdd('m', 1, $testvalue)
$test6 = DatePart('m', $testvalue)
? "shortdate...: $test1"
? "longdate....: $test2"
? "longtime....: $test3"
? "datediff....: $test4"
? "dateadd.....: $test5"
? "datepart....: $test6"
when I run this, variables $test5 and $test6 contain no value. My computer is running Windows XP Pro SP3, using KiXtart 4.61 final release.
Edited by Skatterbrainz (2009-11-19 02:08 PM)
_________________________
silence is golden, but duct tape is silver
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#196772 - 2009-11-19 05:14 PM
Re: (Not) Fun with the ScriptControl object
[Re: Skatterbrainz]
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Richard H.
Administrator
Registered: 2000-01-24
Posts: 4946
Loc: Leatherhead, Surrey, UK
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It's because the date returned is a complex data type.
If you cast it to a simple data type (using CStr() in this example) then it'll come back in a usable form.
I've added a simple FormatDateTime(), as you missed it in your post. I've also collapsed the Chr()s, as they were giving me a headache 
Function FormatDateTime($v, $f)
Dim $sc, $result
$sc = CreateObject("ScriptControl")
$sc.Language = "vbscript"
$result = $sc.Eval('FormatDateTime("'+$v+'",'+$f+')')
$sc = 0
$FormatDateTime = $result
EndFunction
Function DateDiff($unit, $date1, $date2)
Dim $sc, $result
$sc = CreateObject("ScriptControl")
$sc.Language = "vbscript"
$result = $sc.Eval('DateDiff("'+$unit+'", "'+$date1+'", "'+$date2+'")')
$sc = 0
$DateDiff = $result
EndFunction
Function DateAdd($unit, $incr, $date)
Dim $sc, $result
$sc = CreateObject("ScriptControl")
$sc.Language = "vbscript"
$result = $sc.Eval('CStr(DateAdd("'+$unit+'", ' + $incr + ', "'+$date+'"))')
$sc = 0
$DateAdd = $result
EndFunction
Function DatePart($unit, $date)
Dim $sc, $result
$sc = CreateObject("ScriptControl")
$sc.Language = "vbscript"
$result = $sc.Eval('CStr(DatePart("'+$unit+'", "'+$date+'"))')
$sc = 0
$DatePart = $result
EndFunction
$testvalue = '2009/11/03 12:34:56'
$test1 = FormatDateTime($testvalue, 'vbShortDate')
$test2 = FormatDateTime($testvalue, 'vbLongDate')
$test3 = FormatDateTime($testvalue, 'vbLongTime')
$test4 = DateDiff('d', '7/19/2009', $testvalue)
$test5 = DateAdd('m', 1, $testvalue)
$test6 = DatePart('m', $testvalue)
? "shortdate...: $test1"
? "longdate....: $test2"
? "longtime....: $test3"
? "datediff....: $test4"
? "dateadd.....: $test5"
? "datepart....: $test6"
This gives me:
shortdate...: 11/3/2009
longdate....: Tuesday, November 03, 2009
longtime....: 12:34:56 PM
datediff....: 107
dateadd.....: 12/3/2009 12:34:56 PM
datepart....: 11
Which is what you want, I think.
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