Question

我有一个脚本使用get-Content监视日志文件，它等待写入新行，以便检测更改。记录器每天更改日志文件的名称以使其成为System_ $ date.log。

我正在尝试让我的Get-Content等待新行，并在日期更改时中断，然后使用文件名中的新日期重新执行。有人知道如何做到这一点吗？

由于

修改

脚本如下所示：

Get-Content System_201371.log -wait | where {$_ -match "some regex"} | 
foreach {   
   send_email($_)
}

System_201371的文件名每天都在变化，它将是System_201372等等，此脚本也作为服务运行，我需要它以新的文件名来打破并重新执行自己

Answer 1

您可以使用Jobs来实现此目的。在某个后台作业中读取文件，让“前台”脚本等到日期发生变化。一天结束后，请删除旧作业并开始查看新文件。

while($true)
{
    $now = Get-Date
    $fileName = 'System_{0}{1}{2}.log' -f $now.Year, $now.Month, $now.Day
    $fullPath = "some directory\$fileName"

    Write-Host "[$(Get-Date)] Starting job for file $fullPath"
    $latest = Start-Job -Arg $fullPath -ScriptBlock {
        param($file)

        # wait until the file exists, just in case
        while(-not (Test-Path $file)){ sleep -sec 10 }

        Get-Content $file -wait | where {$_ -match "some regex"} | 
          foreach { send_email($_) }
    }

    # wait until day changes, or whatever would cause new log file to be created
    while($now.Date -eq (Get-Date).Date){ sleep -Sec 10 }

    # kill the job and start over
    Write-Host "[$(Get-Date)] Stopping job for file $fullPath"
    $latest | Stop-Job
}

Answer 2

这应该始终监控今天的日志并检测日期变化：

do{
$CurrentDate = get-date -uformat %Y%j
$CurrentLog = ("C:\somepath\System_" + $CurrentDate + ".log")
Start-Job -name Tail -Arg $CurrentLog -Scriptblock{
    While (! (Test-Path $CurrentLog)){ sleep -sec 10 }
    write-host ("Monitoring " + $CurrentLog)
    Get-Content $CurrentLog -wait | where {$_ -match "some regex"} | 
    foreach { send_email($_) }
}
while ($CurrentDate -eq (get-date -uformat %Y%j)){ sleep -sec 10}

Stop-Job -name Tail
} while ($true)

使用Get-Content -wait监视日志文件，直到当天结束

2 个答案: