从路径名中选择字符串？
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文件如下所示
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"
$LogFile
}需要Select-String的帮助。这里是我想要结束的，或者任何最简单的，给$Logfile一个唯一的名字。
CimE11_post
CimE11_post2
CimE11_templates
VERI_7.3.3_lib
VERI_7.4_lib谢谢你的帮助
回答 2
Stack Overflow用户
发布于 2016-11-04 02:04:34
TessellatingHeckler在关于以下问题的评论中提出了一个实用解决方案：将输入路径中的\和:实例替换为文件名中合法的字符，以获得作为文件名工作的每个路径的表示形式。
Import-Csv C:\test.csv | % { # % is a built-in alias for ForEach-Object
$Logfile = ($_.Directory -replace '[\\: ]+', '_') + '.log'
}这将产生以下文件名：
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正如注释中所述，在您说要的文件名后面没有明确的算法逻辑。
下面是一个生成所需文件名的解决方案，试图保持尽可能的泛型。
这在现实生活中可能不值得去做，但也许它能激发出一种真实的解决方案：
# 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
}Stack Overflow用户
发布于 2016-11-02 23:37:54
-replace是您的朋友，下面是我如何从您提供的场景中获取您想要的字符串的示例。
希望这能有所帮助，祝你好运。
$file =导入-csv $filename
foreach($dir in $file.directory){ if($dir -match "VERI"){ ($dir).replace(‘C：\Program\CGTech\ '')).replace('rary'，'') .replace(’$dir '，‘’).replace(‘\’，‘’)}
($dir -match "Cima"){ ($dir).replace(‘C：\Cimatron\，'')).replace('atron'，'') ).replace(’64位\Data‘，’‘).replace(’64位\Program\IT\var‘，’‘)}
}
https://stackoverflow.com/questions/40388371
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