Question

我有一个带有目录路径的CSV文件，我需要从中提取一些信息来创建一个日志文件名。不知道怎么做？ Select-String？

CSV文件看起来像这样

User,Computer,Directory
john.doe,CAD-12,C:\Cimatron\CimatronE11_64bit\Program\IT\var\post
john.doe,CAD-12,C:\Cimatron\CimatronE11_64bit\Program\IT\var\post2
john.doe,CAD-12,C:\Cimatron\CimatronE11_64bit\Data\templates
john.doe,CAD-12,C:\Program Files\CGTech\VERICUT 7.3.3\library
john.doe,CAD-12,C:\Program Files\CGTech\VERICUT 7.4\library

我知道我必须导入CSV文件，这就是我所拥有的。

$UsersCSV = import-csv C:\test.csv
foreach($Computer in $UsersCSV)
{$Logfile = Select-String "not sure how to search the string for what I want"
$LogFile
}

需要Select-String的帮助。以下是我希望最终得到的内容或最简单的$Logfile唯一名称。

CimE11_post
CimE11_post2
CimE11_templates
VERI_7.3.3_lib
VERI_7.4_lib

感谢您的帮助

Answer 1

TessellatingHeckler在对该问题的评论中建议使用实用解决方案：使用合法字符替换输入路径中的\和:个实例文件名，用于表示作为文件名的每个路径：

Import-Csv C:\test.csv | % {  # % is a built-in alias for ForEach-Object
  $Logfile = ($_.Directory -replace '[\\: ]+', '_') + '.log'
}

这会产生以下文件名：

C_Cimatron_CimatronE11_64bit_Program_IT_var_post.log
C_Cimatron_CimatronE11_64bit_Program_IT_var_post2.log
C_Cimatron_CimatronE11_64bit_Data_templates.log
C_Program_Files_CGTech_VERICUT_7.3.3_library.log
C_Program_Files_CGTech_VERICUT_7.4_library.log

正如评论中所述，您想要的文件名背后没有明确的算法逻辑。

这是一个产生所需文件名的解决方案，尽量保持通用这可能不值得在现实生活中做，但也许它可以启发现实生活中的解决方案：

# Helper function that takes a directory path and converts it to a log filename
# representing that dir.
function pathToLogFilename([string] $path) {

  # Ordered hashtable defining regex-based transformations (replacements).
  # Entries are processed in order, and processing stops as soon as 
  # a match is found and the transformation has been performed.
  # Entry format: <regex> = <replacement>, to be passed to the -replace operator.
  $ohtTansform = [ordered] @{
    '^(.{3}).+?(E\d+).*$' = '$1$2' # 'CimatronE11_64bit' -> 'CimE11', for instance. 
    '^(.{4}).+? (\d.*)$' = '$1_$2' # 'VERICUT 7.3.3' -> 'VER_7.3.3', for instance
    '^(lib)rary$' = '$1'           # 'library' -> 'lib'
    '^(post.*|templates)$' = '$&'  #  preserve, if prefixed with 'post' or equal to 'templates' 
    '.+' = ''                      #  remove all other tokens
  }

  $logFilename = ''; $i = 0
  # Split the .Directory path into tokens (path components) and synthesize
  # the log filename from a subset of the tokens, with transformations applied.
  foreach($token in ($path -split '\\')) {
    # Apply transformation.
    $transformedToken = $token
    foreach($regex in $ohtTansform.Keys) {
      if ($token -match $regex) {
        $transformedToken = $token -replace $regex, $ohtTansform[$regex]
        break
      }
    }
    # Add transformed token to filename, separated with "_".
    if ($transformedToken) {
      $logFilename += $(if ($i++) { '_' } else { '' }) + $transformedToken 
    }
  }
  # Output the synthesized filename.
  $logFilename

}


Import-Csv C:\test.csv | % {
  # Call the helper function to transform the directory path to a log filename.
  $LogFile = pathToLogFilename $_.Directory
}

Answer 2

-replace是你的朋友，下面是我如何从你给出的场景中得到你想要的字符串的一个例子。

希望这会有所帮助，祝你好运。

$ file = import-csv $ filename

foreach（$ file.directory中的$ dir）{ if（$ dir -match＆＃34; VERI＆＃34;）{ （（（（$ dir）.replace（＆＃39; C：\ Program Files \ CGTech \＆＃39;，＆＃39;＆＃39;））。replace（＆＃39; rary＆＃39;，＆＃39;＆＃39;））。替换（＆＃39; CUT＆＃39;，＆＃39; ＆＃39;））。替换（＆＃39; \＆＃39;，＆＃ 39; ＆＃39;） }

if（$ dir -match＆＃34; Cima＆＃34;）{ （（（（$ DIR）.replace（＆＃39; C：\ Cimatron的\＆＃39;＆＃39;＆＃39））代替（＆＃39; ATRON＆＃39;，＆＃39; ＆＃39;。））代替（＆＃39; 64 \数据\＆＃39;＆＃39; ＆＃39;。））代替（＆＃39; 64位\程序\ IT \ VAR \＆＃39;＆＃39;＆＃39;） }

}

Powershell从路径名中选择字符串？

2 个答案: