PowerShell 和全局函数

为什么下面的代码不起作用？根据这篇文章，全局的用法应该是正确的：http://technet.microsoft.com/en-us/library/ff730957.aspx http://technet.microsoft.com/en-us/library/ff730957.aspx

Function global:writeLog {
    param($logType, $logString, $logFile)

    $fileStream = New-Object IO.FileStream $logFile ,'Append','Write','Read'
    $streamWriter = New-Object System.IO.StreamWriter $fileStream

    $time = get-date -Format "hh:mm:ss"
    $streamWriter.writeLine("[${time}][$logType] ${logString}")

    $streamWriter.close()

}

$temp = {
    writeLog -logType "INFO" -logString "Test" -logFile "d:\scripts\powershell\logtest.txt"
}

Start-Job -ScriptBlock $temp 
get-job | receive-job -AutoRemoveJob -Wait

这是powershell抛出的异常

The term 'writeLog' is not recognized as the name of a cmdlet, function, script file, or operable program. Check the spelling of the name, or if a path was included, 
verify that the path is correct and try again.
    + CategoryInfo          : ObjectNotFound: (writeLog:String) [], CommandNotFoundException
    + FullyQualifiedErrorId : CommandNotFoundException
    + PSComputerName        : localhost

从文档Start-Job:

Windows PowerShell 后台作业“在后台”运行命令，而不与当前会话交互。

因此，当前会话范围被忽略。

简单的解决方案：在脚本块内定义函数。

$JobScript = { 
    function write-log {
        ....
    }
    write-log <parameters>
}

或者，检查以下相关问题：

Powershell：将参数传递给作业 https://stackoverflow.com/questions/4472284/powershell-passing-parameters-to-a-job

开始作业中的变量 https://stackoverflow.com/questions/10196489/variables-in-start-job