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Can a script determine where "show-output" is going?

Hi

I am running a script, sometimes using "out-file" to catch messages, sometimes not.
In my script is a "write-host" cmdlt so I can see the messages and then I do a write-output cmdlt to ensure it is also in the out-file.

If I have not used an "out-file", I get the information twice on the screen.
To avoid this, can I detect if a running script is using  "out-file"?
Then I could supress "write-output" when there is no out-file.
Thanks.
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dreadman2k
Asked:
dreadman2k
1 Solution
 
footechCommented:
Why not just use Write-Output and skip the Write-Host commands?  Then if you if you have the Out-File it will go to a file, and if not then it will display on screen.  Then you're just choosing which you want.

Another option is to create your script using a parameter that you set as desired for whether you want to display output or not.  For example, setting a variable like $display=$false, and checking throughout the script if $display -eq $false then skipping the Write-Host commands.

Third option would be instead of piping to Out-File, pipe to Out-Null instead.

In short, I haven't heard of nor can I think of any way to detect the use of Out-File (and what about any other cmdlets that produce files like Set-Content, Export-CSV, etc.).  A script is processed in a synchronous/serial (i.e. one thing after another) fashion.  A pipeline isn't going to know what is at the end of it until it gets to it.
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dreadman2kAuthor Commented:
Thanks footech,

I will go with the parameter & send write-output to out-null if no parameter is specified.
Your last paragraph makes sense - I guess I was approaching this from the wrong direction.
Cheers!
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