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Hackoo
Asked: 2021-10-29 23:24:25 +0800 CST

从批处理文件执行powershell代码时出错“TerminatorExpectedAtEndOfString”？

  • 772

我想使用批处理文件执行 powershell 代码，以从我的桌面截取屏幕截图，我在这里找到它Take a Screenshot of a User's Desktop with PowerShell。

因此，当我将它作为 powershell 脚本执行时，它可以正常工作并且对我 5/5 有效，但是当我将它作为这样的批处理文件时，我遇到了这种类型的错误，我不知道如何正确调试它？

Error "TerminatorExpectedAtEndOfString"

那么，如果有人可以解释我为什么以及这个错误来自哪里？

这是我的批处理代码：

@echo off
Title Get a ScreenShot with Batch and Powershell
Call :ScreenShot
pause
::----------------------------------------------------------------------------------------------------------------------------
:ScreenShot
Powershell ^
$Path = "E:\ScreenCapture\";^
Add-Type -AssemblyName System.Windows.Forms;^
$screen = [System.Windows.Forms.Screen]::PrimaryScreen.Bounds;^
$image = New-Object System.Drawing.Bitmap($screen.Width, $screen.Height);^
$graphic = [System.Drawing.Graphics]::FromImage($image);^
$point = New-Object System.Drawing.Point(0,0);^
$graphic.CopyFromScreen($point, $point, $image.Size);^
$cursorBounds = New-Object System.Drawing.Rectangle([System.Windows.Forms.Cursor]::Position,[System.Windows.Forms.Cursor]::Current.Size);^
[System.Windows.Forms.Cursors]::Default.Draw($graphic, $cursorBounds);^
$FileName = (Get-Date -F dd-MM-yyyy_HH_mm_ss)+".jpg";^
$FilePath = $Path$FileName;^
$FormatJPEG = [System.Drawing.Imaging.ImageFormat]::jpeg;^
$image.Save($FilePath,$FormatJPEG)
Exit /B
::------------------------------------------------------------------------------------------------------------------------
powershell command-line

2 个回答

  1. 错误来自这一行：

    $FilePath = $Path$FileName;^

    您不应该连接变量以形成这样的文件路径。如果你用这样的双引号将它括起来，
    它会$FilePath = "$Path$FileName";^起作用：但是，远更好的是使用Join-Path cmdlet：

    $FilePath = Join-Path -Path $Path -ChildPath $FileName;^

    或者如果您更愿意直接使用 .Net：

    $FilePath = [System.IO.Path]::Combine($Path, $FileName);^
    • 2
  2. Best Answer

    可能你需要：

    1. 用于^_)

    2. 替换"为'_ $var='value'

    3. 在中使用 To.String$((Get-Date).ToString('dd-MM-yyyy_HH_mm_ss')+'.jpg')

    在命令行中测试它：

    Powershell $Path='E:\ScreenCapture\'; Add-Type -AssemblyName System.Windows.Forms;$screen = [System.Windows.Forms.Screen]::PrimaryScreen.Bounds;$image = New-Object System.Drawing.Bitmap($screen.Width, $screen.Height^);$graphic = [System.Drawing.Graphics]::FromImage($image^);$point = New-Object System.Drawing.Point(0,0^);$graphic.CopyFromScreen($point, $point, $image.Size^);$cursorBounds = New-Object System.Drawing.Rectangle([System.Windows.Forms.Cursor]::Position,[System.Windows.Forms.Cursor]::Current.Size^);[System.Windows.Forms.Cursors]::Default.Draw($graphic, $cursorBounds^);$FileName = $((Get-Date^).ToString('dd-MM-yyyy_HH_mm_ss'^)+'.jpg'^);$FilePath = $Path+$FileName;$FormatJPEG = [System.Drawing.Imaging.ImageFormat]::jpeg;$image.Save($FilePath,$FormatJPEG^)

    在你的 bat 文件中测试它：

    @echo off
Title Get a ScreenShot with Batch and Powershell
Call :ScreenShot
pause 
goto :eof
::----------------------------------------------------------------------------------------------------------------------------
:ScreenShot
Powershell $Path='E:\ScreenCapture\';^
Add-Type -AssemblyName System.Windows.Forms;^
$screen = [System.Windows.Forms.Screen]::PrimaryScreen.Bounds;^
$image = New-Object System.Drawing.Bitmap($screen.Width, $screen.Height^);^
$graphic = [System.Drawing.Graphics]::FromImage($image^);^
$point = New-Object System.Drawing.Point(0,0^);^
$graphic.CopyFromScreen($point, $point, $image.Size^);^
$cursorBounds = New-Object System.Drawing.Rectangle([System.Windows.Forms.Cursor]::Position,[System.Windows.Forms.Cursor]::Current.Size^);^
[System.Windows.Forms.Cursors]::Default.Draw($graphic, $cursorBounds^);^
$FileName = $((Get-Date^).ToString('dd-MM-yyyy_HH_mm_ss'^)+'.jpg'^);^
$FilePath = $Path+$FileName;$FormatJPEG = [System.Drawing.Imaging.ImageFormat]::jpeg;^
$image.Save($FilePath,$FormatJPEG^)
    • 2

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