从二进制输出中选择列和精确的字符串匹配

Question

我在PowerShell中寻找一个等价的命令：

Linux/macOS：

cat mmls.txt | awk '{print $3}' | egrep -o '[0,9]{10}' | awk 'NR==1' | sed -e 's/^[0]*//'

源(mmls.exe的输出)：

GUID Partition Table (EFI)
Offset Sector: 0
Units are in 512-byte sectors

Slot            Start        End          Length       Description
000:  Meta      0000000000   0000000000   0000000001   Safety Table
001:  -------   0000000000   0000002047   0000002048   Unallocated
002:  Meta      0000000001   0000000001   0000000001   GPT Header
003:  Meta      0000000002   0000000033   0000000032   Partition Table
004:  000       0000002048   0000616447   0000614400   Basic data partition
005:  001       0000616448   0001665023   0001048576   EFI system partition
006:  002       0001665024   0001927167   0000262144   Microsoft reserved partition
007:  003       0001927168   0300832767   0298905600   Basic data partition
008:  004       0300832768   0500117503   0199284736   Basic data partition
009:  -------   0500117504   0500118191   0000000688   Unallocated

我只想提取开始偏移没有前导零。使用awk 'NR==x‘，我选择行号。列用2-3个空格隔开.

提前感谢！

Answer 1

下面按照您的示例选择第一个数字

Select-String -Pattern '\d{10}' -Path mmls.txt | 
    ForEach-Object { $PSItem.Matches[0].Value -replace '^0+', '' } | 
    Where-Object { $PSItem -ne '' } | 
    Select -First 1

或者更简洁一点

sls '\d{10}' mmls.txt | 
   %{$_.Matches[0].Value -replace '^0+', ''} | 
   ? {$_ -ne ''} | 
   select -f 1

替代方案

gc mmls.txt|%{($_ -split '\s+')[2]}|?{$_.Length -eq 10}|%{$_ -replace '^0+'}|?{$_ -ne ''}|select -first 1