CODEWITHSUNDEEP
powershellscripting
user3494110
user3494110

Reputation: 427

Powershell interrogate file name for date

I am just starting to get into powershell as a solution for a file collection process. Basically the gist of it is I want to loop through a directory of files that have the beginning wildcard, interrogate the julian date, and then rename the file based on that date (in Gregorian).

Right now I only have code (shown below) to rename the file based on today's date, but that is just because I am having trouble writing something advanced enough to do what I want it to do. The shortcut won't work 100% of the time. Can anyone offer any help on how to interrogate the julian date and rename?

Here is an example of how the file looks before I rename it, and how I want it to look after I rename it.

Before rename - COMPANY.DATA.J2016183.Y98173987 (the first 3 parts are constant aside from the date changing. Only the last delimited section after the dot is a random key that I don't need. The "J" is a constant as it just refers to julian. What I need to interrogate is the date after the "J").

Desired file name after renaming - DATA.2016-07-04.txt

$Today = (Get-Date).Date
$Day = $Today.AddDays(-2)
$newday = $Day.ToString("yyyy-MM-dd")
Get-ChildItem "\\fileserv\folder\COMPANY.DATA.*" -Recurse| 
Rename-Item -NewName {'DATA.' + $newday + '.txt' }

(the -2 just has to do with the file name/data difference. File name is 2 days ahead of actual data).

Any help would be appreciated. Thank you,

Upvotes: 1

Views: 386

Answers (2)

morgb
morgb

Reputation: 2312

Here is a function to convert Julian to Gregorian (source: http://powershell.com/cs/media/p/18723.aspx):

<# 
ALZDBA 20120831 

#> 
function Convert-JulianDateFormat2Date ( [int]$YYYYddd ) { 
####################### 
<# 
.AUTHOR 
    ALZDBA 20120831 
.SYNOPSIS 
    Convert YYYYddd to [date]. 
.DESCRIPTION 
    convert a numeric Julian date format to a regular date ( date datatype ) 
.INPUTS 
    $YYYYddd e.g. 2012245 
.OUTPUTS 
    [date] 
.EXAMPLE 
Clear-Host  
    $mydate = Convert-JulianDateFormat2Date -YYYYddd 2012263  
    $mydate  
    $mydate.dayofyear  

#> 
    $year = [int]( $YYYYddd / 1000 )*1000 
    $DayOfYear = $YYYYddd - $year 
    $date = Get-Date -Year ( $year/1000 ) -Month 1 -Day 1 -Hour 0 -Minute 0 -Second 0 
    #$date.DayOfYear  
    $date = $date.AddDays( $DayOfYear - 1 ) 
    $date 
    }

Using this, you can rename the files as follows:

$delim = "."
Get-ChildItem "\\fileserv\folder\COMPANY.DATA.*" -Recurse | 
foreach { 
    $nameArray = $_.Name.Split($delim)

    # CONVERT DATE:
    $jDate = $nameArray[2].Replace("J","")
    $mydate = Convert-JulianDateFormat2Date -YYYYddd $jDate
    $newDate = $mydate.ToString("yyyy-MM-dd")

    #RENAME FILES:
    $newName = $nameArray[1]+$delim+$newDate+$delim+$nameArray[4]
    Rename-Item $_ -NewName $newName 
}

Upvotes: 2

GeoGeoGeometry
GeoGeoGeometry

Reputation: 164

Ryan Drane has a great post on this problem here: http://www.ryandrane.com/2015/04/convert-julian-date-to-gregorian-date-in-powershell/

His approach, modified for your application, would look something like this:

$filename = "COMPANY.DATA.J2016183.Y98173987"
$JulianYear = $filename.Substring(14,4)
$JulianDay = $filename.Substring(18,3)
$GregorianDate = (Get-Date -Year $JulianYear -Month 01 -Day 01).AddDays($JulianDay - 1)

Although that code does return the gregorian date with the time attached, it shouldn't be much work to remove that, and add in your file handling method.

Upvotes: 0

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