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How to get file version

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ur

How to get the thrid number of a file version.

Example: Let's say I have a file with below properties.

I want only the third number like here it is 1941 as highlighted.

Is there any built in function to get it??

asdsd.png

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AutoBert

You can use StringSplit for splitting the return of FileGetVersion:

#include <MsgBoxConstants.au3>

Example()

Func Example()
    ; Retrieve the file version of the AutoIt executable.
    Local $sFileVersion = FileGetVersion(@AutoItExe)
    Local $aSplit=StringSplit($sFileVersion,'.')
    If IsArray($aSplit) Then MsgBox($MB_SYSTEMMODAL, "", $aSplit[3])
EndFunc   ;==>Example

 

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ur

yeah I tried the below one and working.Used the inputs from Anoop.

 

#include <MsgBoxConstants.au3>
#include <Array.au3>

$sFilePath = FileOpenDialog("Select a file to read attributes",@ScriptDir,"All (*.*)")
MsgBox(0,$sFilePath,getFileVersionBuildNumber($sFilePath))

Func getFileVersionBuildNumber($sFilePath)

$sStringName = "FileVersion"
Local $sStringValue = FileGetVersion($sFilePath, $sStringName)

If Not @error Then
    ;Display the property value.
    $b = StringSplit ( $sStringValue, '.' )[3]
    return $b
Else
    ;If error, display an error message.
    MsgBox($MB_ICONERROR, "", "Error when getting " & '"' & $sStringName  & '"')
EndIf

EndFunc

 

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Anoop
10 minutes ago, ur said:

yeah I tried the below one and working.Used the inputs from Anoop.

 

#include <MsgBoxConstants.au3>
#include <Array.au3>

$sFilePath = FileOpenDialog("Select a file to read attributes",@ScriptDir,"All (*.*)")
MsgBox(0,$sFilePath,getFileVersionBuildNumber($sFilePath))

Func getFileVersionBuildNumber($sFilePath)

$sStringName = "FileVersion"
Local $sStringValue = FileGetVersion($sFilePath, $sStringName)

If Not @error Then
    ;Display the property value.
    $b = StringSplit ( $sStringValue, '.' )[3]
    return $b
Else
    ;If error, display an error message.
    MsgBox($MB_ICONERROR, "", "Error when getting " & '"' & $sStringName  & '"')
EndIf

EndFunc

 

Great to know that!!! :)

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