Question

我需要从下面的列表中获取最新的文件。每天创建3种类型的文件，我只需要3种类型的文件。

示例：

my_file_dob_6545d.txt   12/3/2017 5:00 PM
my_file_csm_6545d.txt   12/3/2017 5:00 PM
my_file_6545d.txt       12/3/2017 5:00 PM  <--- Need to get this.
my_file_dob_6544d.txt   12/2/2017 5:00 PM
my_file_csm_6544d.txt   12/2/2017 5:00 PM
my_file_6544d.txt       12/2/2017 5:00 PM
my_file_dob_6543d.txt   12/1/2017 5:00 PM
my_file_csm_6543d.txt   12/1/2017 5:00 PM
my_file_6543d.txt       12/1/2017 5:00 PM

我有来自here

的以下脚本

$dir = "c:\my_folder\"
$filter="my_file_*d.txt" <--How can modify this to get only  my_file_XXXXd.txt files? XXXX keeps incrementing everyday
$latest = Get-ChildItem -Path $dir -Filter $filter | Sort-Object LastAccessTime -Descending | Select-Object -First 1
$latest.name

Answer 1

语法稍微简单：

$latest = Get-ChildItem C:\my_folder\my_file_????d.txt | Sort-Object LastAccessTime -Descending | Select -First 1

Answer 2

没关系，从示例here

中弄清楚自己

$dir = "c:\my_folder\"
$latest = Get-ChildItem -Path $dir | Sort-Object LastAccessTime -Descending | Where-object { $_.Name -match "my_file_[0-9][0-9][0-9][0-9]d.txt"} | Select-Object -First 1

从目录中的一堆类似命名文件中获取最新文件 - powershell

2 个答案: