Powershell 以字符串模式强制 5 位数字格式

问题描述

所以我现在所拥有的(下面)将搜索 XX-##-# 并将其强制为 XX-##-#0000.

So what I have now will (below) will search for XX-##-# and force it to XX-##-#0000.

如何才能返回XX-##-0000#?

有没有办法在末尾强制 5 位数字，填充前面的 0，以覆盖其他可能性(XX-##-##, XX-##-###, XX-##-####)?相对于复制 4 次，每次稍微调整.

Is there a way to force 5 digits at the end, filling preceding 0s, to cover the other possibilities (XX-##-##, XX-##-###, XX-##-####)? As opposed to copying this 4 times, slightly adjusting for each.

$Pattern1 = '[a-zA-Z][a-zA-Z]-[0-9][0-9]-[0-9]'
Get-ChildItem 'C:\path\to\file\*.txt' -Recurse | ForEach {
     (Get-Content $_ | 
     ForEach  { $_ -replace $Pattern1, ('$1'+'0000')}) | 
     Set-Content $_
}

谢谢.

我想做以下事情

Search           Replacement
XX-##-#          XX-##-0000#
XX-##-##         XX-##-000##
XX-##-###        XX-##-00###
XX-##-####       XX-##-0####

推荐答案

遗憾的是，PowerShell 的 -replace 运算符不支持将 表达式(脚本块)作为替换字符串，这里需要一个简洁的解决方案.

Unfortunately, PowerShell's -replace operator doesn't support passing an expression (script block) as the replacement string, which a succinct solution would require here.

但是，您可以使用适当的[regex] .NET 类型的 .Replace() 方法重载:

However, you can use the appropriate [regex] .NET type's .Replace() method overload:

# Define sample array.
$lines = @'
Line 1 AB-00-0 and also AB-01-1
Line 2 CD-02-22 after
Line 3 EF-03-333 it
Line 4 GH-04-4444 goes 
Line 5 IJ-05-55555 on
'@ -split "`n"

# Loop over lines...
$lines | ForEach-Object {
  # ... and use a regex with 2 capture groups to capture the substrings of interest
  #     and use a script block to piece them together with number padding
  #     applied to the 2nd group
  ([regex] '\b([a-zA-Z]{2}-[0-9]{2}-)([0-9]+)').Replace($_, { 
    param($match)
    $match.Groups[1].Value + '{0:D5}' -f [int] $match.Groups[2].Value
  })
}

以上产生:

Line 1 AB-00-00000 and also AB-01-00001
Line 2 CD-02-00022 after
Line 3 EF-03-00333 it
Line 4 GH-04-04444 goes 
Line 5 IJ-05-55555 on

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