我的想法是我只捕获脚本的内容，看看它是否只包含一个或多个：函数X { * }，除此之外什么都没有。
真是个好计划！
[...]可能使用正则表达式来匹配该模式。
这计划太烂了！
使用PowerShell自己的解析器来解析现有脚本，而不是regex：

using namespace System.Management.Automation.Language

# start by enumerating all the script files in scope
$results = Get-ChildItem folder\with\script\files -Filter *.ps1 -Recurse |ForEach-Object {
  # construct an object that will hold the output for each file analyzed
  $result = [pscustomobject]@{
    Name          = $_.Name
    Path          = $_.FullName
    FunctionCount = 0
    PipelineCount = 0
    OtherCount    = 0
    HadErrors     = $false
  }

  try {
    # attempt to parse the script file
    $scriptAst = [Parser]::ParseFile($_.FullName, [ref]$null, [ref]$null)

    # inspect the top-level statements in the resulting AST, capture the counts of particularly interesting syntax elements
    $result.FunctionCount = $scriptAst.EndBlock.Statements.Where({$_ -is [FunctionDefinitionAst]}).Count
    $result.PipelineCount = $scriptAst.EndBlock.Statements.Where({$_ -is [PipelineAst]}).Count
    $result.OtherCount    = $scriptAst.EndBlock.Statements.Where({$_ -isnot [PipelineAst] -and $_ -isnot [FunctionDefinitionAst]}).Count
  }
  catch {
    # in case of failure we want to report that to the caller
    $result.HadErrors = $true
  }

  # output the details captured
  $result
}

现在我们已经分析了所有脚本中的顶级语句，我们可以开始根据大多数可能是“模块式”的期望来过滤它们：

$results |Where-Object { $_.FunctionCount -gt 0 -and $_.PipelineCount -eq 0 } |ForEach-Object {
    Write-Host "Script at '$($_.Path)' looks like a module! No pipelines, just functions!"
}