Question

我需要编写NuGet包install.ps1脚本，它将原生dll文件复制到输出目录，但我找不到获取输出文件夹路径的方法。

我认为解决方案是使用以下内容：

$solutionDir = [System.IO.Path]::GetDirectoryName($dte.Solution.FullName) + "\"

如何通过PowerShell获取输出目录路径？

Answer 1

“输出目录”是指项目的输出目录吗？如果是这样，您可以遍历项目以按索引查找，然后像这样获取项目的基础目录（假设您已确定您感兴趣的项目位于索引3：

 $project = $dte.Solution.Projects.Item(3)
 ($project.Properties | Where Name -match FullPath).Value

然后，要获取构建路径（bin \ debug或bin \ release），请执行以下操作：

($project.ConfigurationManager.ActiveConfiguration.Properties | Where Name -match OutputPath).Value

您还可以像这样访问解决方案中的活动项目：

 $project = $dte.ActiveSolutionProjects

Answer 2

我开始“走上去”直到找到它：

param($installPath, $toolsPath, $package, $project)
if ($project -eq $null) {
$project = Get-Project
}


Write-Host "installPath:" "${installPath}"
Write-Host "toolsPath:" "${toolsPath}"
Write-Host "package:" "${package}"
<# Write-Host "project:" "${project}" #>
Write-Host " "

<# Recursively look for a .sln file starting with the installPath #>
$parentFolder = (get-item $installPath)
do {
        $parentFolderFullName = $parentFolder.FullName

        $latest = Get-ChildItem -Path $parentFolderFullName -File -Filter *.sln | Select-Object -First 1
        if ($latest -ne $null) {
            $latestName = $latest.name
            Write-Host "${latestName}"
        }

        if ($latest -eq $null) {
            $parentFolder = $parentFolder.parent    
        }
}
while ($parentFolder -ne $null -and $latest -eq $null)
<# End recursive search for .sln file #>


if ( $parentFolder -ne $null -and $latest -ne $null )
{
    <# Create a base directory to store Solution-Level items #>
    $myFolderFullName = $parentFolder.FullName
}

如何通过PowerShell获取NuGet输出目录？

2 个答案: