PowerShell：使用正文作为变量发送电子邮件

Question

我有一个脚本，用于检查用户是否在两个报告之间匹配。如果有任何用户匹配，则应发送一条消息。如果用户不匹配，我希望它发送一条消息，上面写着＆＃34;没有用户移动。＆＃34;我的问题是，无论我做什么，它总是发送相同的信息。

$TermRep = Import-Csv TestFileTerm.csv
$Donna = Import-Csv TestFileDonna.csv

#Job to match users between CSVs
$Job = ForEach($i in $TermRep){
$TID = $($i.UserID)
ForEach($u in $Donna){
$DID = $($u.UserID)
If($TID -eq $DID){
"Move-ADObject -Identity $TID -TargetPath 'NEWPATH' `r`n"
            }
    }
}

$smtpServer = "server.stuff.com"
$msg = new-object Net.Mail.MailMessage
$smtp = new-object Net.Mail.SmtpClient($smtpServer)
$msg.From = "PowerShell.Reports@work.com"
$msg.To.Add("me@work.com")
#$msg.To.Add("administrator2@test.com")
$msg.Priority = "High"
$msg.Subject = "Users To Move"
$msg.Body = "The following users exist in Termination Report and need to be moved.  Please run the following commands: `r`n $Job"
$smtp.Send($msg)

我已尝试将$msg.body变成

If($TID -eq $DID){
$msg.Body = "The following users exist in Termination Report and need to be moved.  Please run the following commands: `r`n $Job"}
Else{$msg.Body = "No users to move."}

无论If是否为真，它仍然只会推出第一个选项。我还尝试直接在ForEach

之下/之外添加

If($TID -ne $DID){
$Job = "No users to move."
}

如果我运行$Job，其值为"No users to move."，但它仍会在电子邮件中运行另一行。有人可以帮忙吗？ PowerShell v2.0。我尝试将$Job = @()放在If中以便先清除它，但不做任何更改。

额外尝试

我也尝试过这种方式格式化：

$msg.Body = "The following users exist in Termination Report and need to be moved.  Please run the following commands:`r`n"
If($Job -like "*"){
$msg.Body += "$Job"
}
Else{
$msg.Body = "No user to move for $Date."
}

如果$Job有任何价值，我仍会得到正确的结果，但Else声明再次在此处无效。我也试过了$msg.Body += "No user to move for $Date."这也行不通。

解决了它。

我将最后一个从-like更改为-notlike，这似乎修复了它：

If($Job -notlike ""){
$msg.Body += "$Job"
}
Else{
$msg.Body += "No users to move for $Date."
}
$smtp.Send($msg)