Powershell 1 - 列出一個路徑下所有文件； 定義常量變量和數組

列出一個路徑下的所有文件名，或者文件夾名，或者所有名字

$path = "D:\"
Write-Host ("Get file names (not including folder names) only");

Get-ChildItem $path | ForEach-Object -Process{# 注意: { 必須緊跟着 Process if ($_ -is [System.IO.FileInfo]) #如果想要得到文件就用 System.IO.FileInfo
    {
        Write-Host ($_.name);
    }

}

Write-Host ("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~");
Write-Host ("Get all names whatever files or folders.");
Get-ChildItem $path | ForEach-Object -Process{
    Write-Host ($_.name);
}

Write-Host ("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~");
Write-Host ("Get all folders' names.");
Get-ChildItem $path | ForEach-Object -Process{
    if ($_ -is [System.IO.DirectoryInfo]) #如果想要得到文件夾，就用 System.IO.DirectoryInfo
    {
        Write-Host ("Folder: " + $_.name);
    }
}

如何列出文件的創建時間有關的信息？

$path = "C:\Program Files (x86)\Jenkins\jobs";
Get-ChildItem $path | ForEach-Object -Process{
    if($_ -is [System.IO.FileInfo] -and ($_.CreationTime -ge [System.DateTime]::Today)) #如果文件是在今天創建
    {
        Write-Host("File " + $_.name + " is created at " + $_.CreationTime);
    }
    
    if ($_ -is [System.IO.DirectoryInfo] -and ($_.CreationTime -ge (Get-Date).Date.AddDays(-1)))#如果文件夾是在過去一天內創建
    {
        Write-Host ("Folder " + $_.name + " is created at " + $_.CreationTime)
    }
}

也看到用下面方法找到文件或者文件夾的：

#找出D盤根目錄下的所有文件
Get-ChildItem d:\ | ?{$_.psiscontainer -eq $false} #如果是 true 則為文件夾

 

 另外，如果想遍歷一個文件夾的所有子文件夾和文件，可以用  --recurse 

　　　　Get-ChildItem -Recurse

 

如何定義一個常量？

New-Variable var1 -Value "apple" -Option Constant
$var1

 

如何定義一個變量？

很簡單， 

$var = "2"
$var
$var = "3"
$var

輸出：

2

3

其實變量有很多作用域，根據http://www.pstips.net/powershell-scope-of-variables.html，可以有下面的修飾符：$global, $private, $local, $script

 

定義數組？

$family = "0allen", "1jenny", "2alex", "3jessica"
$family[0]
$family[2]
Write-Host ("----")
$family[0, 1,3]
$family.Count
$family[$family.Count - 2]
Write-Host ("----")
#將數組逆序輸出
$family[($family.Count)..0]
Write-Host ("----")
$family[2..0]
Write-Host ("----")
#添加數組元素
$family+="4song";
$family;
Write-Host ("----")
#刪除數組元素
$family= $family[0..2]+$family[4]
$family

 

結果：

0allen
2alex
----
0allen
1jenny
3jessica
4
2alex
----
3jessica
2alex
1jenny
0allen
----
2alex
1jenny
0allen
----
0allen
1jenny
2alex
3jessica
4song
----
0allen
1jenny
2alex
4song