我正在编写一个脚本，用于重命名报告并将其放在文件夹中。
报告名称必须更改为正确的日期(日期必须在文件名中)。
例如，我有一个名为“Report_2015_02”的报告/我必须从文件名中减去一个月而不是系统时间。
在这种情况下，新的报告名称必须为“Report_2015_01”
但目前我正在使用systemtime，因此报表名称将为“Report_2017_02”，因为Systemtime(2017_03)减去一个月就是2017_02 ...
问:如何使用文件名的日期而不是系统时间？
这是我的代码:
#today's date (year-month-day)
$todaydate = Get-Date -Format yyyy-MM-dd
#arrays (today => systemtime)
$todaydate = $todaydate.Split('-')
$todaydate[0] #year
$todaydate[1] #month
$todaydate[2] #day
#arrays yesterday (systemtime - one day)
$yesterdaysdate = Get-Date((Get-Date).AddDays(-1)) -Format yyyy-MM-dd
$yesterdaysdate = $yesterdaysdate.Split('-')
$yesterdaysdate[0] #year
$yesterdaysdate[1] #month
$yesterdaysdate[2] #day
#arrays yesterday (systemtime - one day)
$lastmonth = Get-Date((Get-Date).AddMonths(-1)) -Format yyyy-MM-dd
$lastmonth = $lastmonth.Split('-')
$lastmonth[0] #year
$lastmonth[1] #month
$lastmonth[2] #day
#Example 1: Filename "Report_Telephone_yyyy-mm" => in this case "Report_Telephone_2016-12"
#it renames the file -> minus one month, so the name must be "Report_Telephone_2016-11"
$filename='Report_Telephone_'+ $lastmonth[0]+'-'+ $lastmonth[1] + '.xlsx'
write-host $filename
rename-item 'c:\Reporting\Report_Telephone_' + $todaydate[0] +'-' + $todaydate[1] + '.xlsx' -NewName $filename
$sourcepath='C:\Reporting\'+ $filename
write-host $sourcepath
$destinationpath='C:\Reporting\'+ $lastmonth[0]+'\'+ $lastmonth[1]
write-host $destinationpath
if(test-path $destinationpath)
{
}
else
{
mkdir $destinationpath
}
move-item -path $sourcepath -destination $destinationpath
#Example 2: Filename "Report_Outlook_yyyy-mm-dd" => in this case "Report_Outlook_2016-12-14"
#it renames the file -> minus one day, so the name must be "Report_Outlook_2016-12-13"
$filename='Report_Outlook_'+ $lastmonth[0]+'-'+ $lastmonth[1] + '.xlsx'
write-host $filename
rename-item 'c:\Reporting\Report_Outlook_' + $todaydate[0] +'-' + $todaydate[1] + '.xlsx' -NewName $filename
$sourcepath='C:\Reporting\'+ $filename
write-host $sourcepath
$destinationpath='C:\Reporting\'+ $yesterdaysdate[0]+'\'+ $yesterdaysdate[1]
write-host $destinationpath
if(test-path $destinationpath)
{
}
else
{
mkdir $destinationpath
}
move-item -path $sourcepath -destination $destinationpath
#that is just for generating log-files
[System.IO.File]::WriteAllText("$sourcepath\Protokoll_$todaydate.xls", $output)
最佳答案
您可以使用简单的正则表达式来获取月份，将其转换为整数，减去1，最后将其格式化为带有两位小数的字符串:
$fileName = 'Report_2015_02' # Probably want to use Get-ChildItem here...
$fileMonth = [int][regex]::Match($fileName, '\w+_\d+_(\d+)').Groups[1].Value
$fileMonth = "{0:D2}" -f ($fileMonth - 1)
关于powershell - 按文件名重命名报告，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/42620238/