Yesterday I read about the invoke-expression
from
http://technet.microsoft.com/en-us/library/dd347550.aspx
invoke-expression -command "c:\batfile.bat"
It works very well, but how can I get the results and also , how to add args for .bat file?
Invoke-Expression is to Call up the EXpression,And '.Bat' is not an expression.
So if you want to call .Bat File Justc:\batfile.bat Could Do the trick.
Or if you want to explore on the invoke-expression Copy the commands in Batch file into a variable to see the code being executed.
There was similar instance where I had to do something similar,And I had used it as
$prog="cmd.exe"
$TARGETDIR = 'd:\temp'
$x='C:\Users\v-chetak\Desktop\1.bat'
$params=#("/C";"$x";" >d:\temp\result17.txt")
Start-Process -Verb runas $prog +$params
The $params Will Hold all the arguments you want to pass.
Related
This is my code:
set-location [PATH]
$A = Start-Process -FilePath .\refresh.bat -Wait
set-location C:\
When executed in powershell, the system opens a Command prompt window and executes the bat file without issue. The problem is that the window closes and I cannot see if there was an error if it succeeds.
I want to keep the CMD window open.
I also tried at the end of the bat file:
:END
cmd /k
but no luck.
First, unless you specifically need to run the batch file in a new window, do not use Start-Process - use direct invocation instead, which is implicitly synchronous and allows you to capture or redirect output:
# Invoke the batch file synchronously (wait for it to exit)
# and capture its (standard) output in variable $A
# To print the batch file's output to the console instead, just use:
# .\refresh.bat
$A = .\refresh.bat
See this answer for more information.
Also note Start-Process never allows you to capture the invoked program's output directly (you can only redirect it to files with -RedirectStandardOutput and -RedirectStandardOutput); your specific command captures nothing[1] in $A; adding -PassThru does return something, but not the program's output, but a process-information object (System.Diagnostics.Process).
If you do need to run the batch file in a new window and want to keep that window open:
Start-Process -Wait -FilePath cmd -ArgumentList '/k .\refresh.bat'
Relying on positional parameter binding, the above can be simplified to:
Start-Process -Wait cmd '/k .\refresh.bat'
[1] Strictly speaking, $A is assigned the [System.Management.Automation.Internal.AutomationNull]::Value singleton, which in most contexts behaves like $null.
Thank you mklement0 with your post gave me the solution I wanted. This is how I solved it.
set-location [PATH]
$A = Start-Process -FilePath .\refresh.bat -Wait -NoNewWindow
set-location C:\
-NoNewWindow allowed me to run my batch in the same powershell window getting the feedback of the bat file. That way I have errors if any and success status if no errors.
Thanks!
I have a curious case that I cannot fathom the reason for...
Please know I am a novice to PowerShell.
I am working on a PowerShell menu system to help automate building out new computers in my environment. I have a PS1 file that holds the script for an app install. When I use the script to reference this I am able to run it and have no issue. However, when I try inserting this into a function and referencing it does not.
This works:
4 # Microsoft Office 32-bit
{
Write-Host "`nMicrosoft Office 32-bit..." -ForegroundColor Yellow
# {installMS32Bit}
Invoke-Expression "cmd /c start powershell -NoExit -File '\\**SERVERPATH**\menuItems\ms_office\32-bit\install.ps1'"
Start-Sleep -seconds 2
}
This does not:
function installMS32Bit(){
Invoke-Expression "cmd /c start powershell -NoExit -File '\\**SERVERPATH**\menuItems\ms_office\32-bit\install.ps1'"
}
}
4 # Microsoft Office 32-bit
{
Write-Host "`nMicrosoft Office 32-bit..." -ForegroundColor Yellow
{installMS32Bit}
Start-Sleep -seconds 2}
install.ps1 file:
# Copy MS Office uninstall and setup to local then run and install 32-bit Office
Copy-Item -Path '\\**SERVERPATH**\menuItems\ms_office\setup.exe' -Destination 'C:\temp\' -Force
Copy-Item -Path '\\**SERVERPATH**\menuItems\ms_office\uninstall.xml' -Destination 'C:\temp\' -Force
Copy-Item -Path '\\**SERVERPATH**\menuItems\ms_office\32-bit\Setup.exe' -Destination 'C:\temp' -Force
Invoke-Expression ("cmd /c 'C:\temp\setup.exe' /configure 'C:\temp\uninstall.xml'")
Start-Process -FilePath 'C:\temp\Setup.exe'
Secondary question and a little explanation for Invoke-Expression...
I like to see progress and like to have secondary windows open to monitor the new process being run. I was unable to find a solution with a persistent window that worked for me to do this without Invoke-Expression.
If there is a better way to do this in PowerShell I am all ears!
{installMS32Bit}
As Mathias points out in a comment on the question, this statement doesn't call your function, it wraps it in a script block ({ ... })[1], which is a piece of reusable code (like a function pointer, loosely speaking), for later execution via &, the call (execute) operator.
To call your function, just use its name (by itself here, given that there are no arguments to pass): installMS32Bit
Invoke-Expression should generally be avoided; definitely don't use it to invoke an external program, as in your attempts.
Additionally, there's generally no need to call an external program via cmd.exe (cmd /c ...), just invoke it directly.
For instance, replace the last Invoke-Epression call from your question with:
# If the EXE path weren't quoted, you wouldn't need the &
& 'C:\temp\setup.exe' /configure 'C:\temp\uninstall.xml'
I like to see progress and like to have secondary windows open to monitor the new process being run. I was unable to find a solution with a persistent window that worked for me to do this without Invoke-Expression.
(On Windows), Start-Process by default executes a console application in a new window (unless you specify -NoNewWindow), asynchronously (unless you specify -Wait).
You cannot pass a .ps1 script directly to Start-Process (it will be treated like a document to open rather than an executable to call), but you can pass it to PowerShell's CLI via the -File parameter:
Start-Process powershell.exe '-File install.ps1'
The above is short for:
Start-Process -FilePath powershell.exe -ArgumentList '-File install.ps1'
That is, PowerShell will execute the following in a new window:
powershell.exe -File install.ps1
[1] Since you're not assigning the script block being created to a variable, it is implicitly output (printed to the display, in the absence of a redirection); a script block stringifies by its literal contents, excluding the enclosing { and }, so string installMS32Bit will print to the display.
I'm trying to capture the Verbose, Error and other streams of a PowerShell script in a file. This to monitor the output of my script.
The following code works fine:
$LogFile = 'S:\ScriptLog.log'
$ScriptFile = 'S:\TestieScript.ps1'
powershell -Command $ScriptFile *>&1 > $LogFile
However, the moment I try to put a space in one of the file paths, it's no longer working. I tried a lot of things, like double quotes, single quotes, .. but no luck.
To illustrate, the following code doesn't work:
$LogFile = 'S:\ScriptLog.log'
$ScriptFile = 'S:\Testie Script.ps1'
powershell -Command $ScriptFile *>&1 > $LogFile
One person in this thread has the same issue.
Thank you for your help.
You're trying to run a file whose name contains a space as a command without proper quoting, so you're most likely getting an error like this in your log:
The term 'S:\Testie' is not recognized as the name of a cmdlet, function, script file, or operable program.
Either add proper quoting (and the call operator &, because your path is now a string):
powershell -Command "& '$ScriptFile'" *>&1 > $LogFile
or (better) use the -File parameter, as #CB. already suggested:
powershell -File $ScriptFile *>&1 > $LogFile
which has the additional advantage that the call will return the actual exit code of the script.
Edit: If you want to run the command as a scheduled task you'll need to use something like this:
powershell -Command "& 'S:\Testie Script.ps1' *>&1 > 'S:\ScriptLog.log'; exit $LASTEXITCODE"
because the redirection operators only work inside a PowerShell process.
try using -file parameter:
powershell -file $ScriptFile *>&1 > $LogFile
This is my powershell script test.ps1:
Write-Output $args;
Now suppose I have a batch script that calls this powershell script with all kinds of paths. One of those is c:\:
powershell -executionpolicy Bypass -file test.ps1 "c:\"
The output is:
c:"
Is there any way to quote my arguments such that c:\ would actually be taken and stored as is in the $args[0] variable? I know I can solve this quick'dirty by passing "c:\\", but that's not a real solution.
EDIT: using named parameters in test.ps1 doesn't make any difference:
[CmdletBinding()]
param(
[string]$argument
)
Write-Output $argument;
EDIT2: using a batch file instead works fine.
My test.bat script:
echo %~1
I run it:
test.bat "c:\"
Returns nicely:
c:\
Are you sure this comes form powershell and not from the program which invokes your statement? The backslash is no escape code in powershell.
my test.ps1 is working, when run from ise.
this works for me:
powershell -executionpolicy Bypass -command "test.ps1 -argument 'C:\'"
(end with quote double-quote)
Help file for PowerShell.exe says:
File must be the last parameter in the command, because 'all characters' typed after the file parameter name are "interpreted" as the script file path followed by the script parameters.
You are against Powershell.exe's command line parser, which uses "\" to escape quotes. Do you need quotes? Not in your case:
powershell -file test.ps1 c:\
prints
c:\
Similarly, this works too
powershell -file test.ps1 "c:\ "
c:\
but then your arg has that extra space which you would want to trim. BTW, Single quotes do not help here:
powershell -file test.ps1 'c:\'
'c:\'
If you need the final backlash to be passed to the command, you can use
$ArgWithABackslashTemp = $ArgWithABackslash -replace '\\$','\\'
&$ExePath $ArgWithABackslashTemp
Or, if the exe is smart enough to handle it without the trailing backslash
&$ExePath $ArgWithABackslash.trim('\')
I'm new to these awesome Power shell world. I have a problem with script and really apreciate your help.
I have a script "cmd4.ps1" that needs to run another script "Transfer.ps1" that needs to receive 3 strings params and it needs to be run as other process thead different to "cmd4.ps1".
cmd4.ps1:
$Script="-File """+$LocalDir+"\Remote\Transfer.ps1"" http://"+$ServerIP+"/Upload/"+$FileName+" "+$ServerIP+" "+$LocalIP
Start-Process powershell.exe -ArgumentList $Script
After ejecution, the $Script cointain a value similar to
-File "c:\temp re\Remote\Transfer.ps1" http://10.1.1.1/Upload/file.txt 10.1.1.1 10.1.1.10
containing the syntax to use -File parameter to run a script of Powershell.exe, and three parameters that Transfer.ps1 needs ("http://10.1.1.1/Upload/file.txt", 10.1.1.1, 10.1.1.10).
When I write these instructions in PowerShell ISE I can see every values are right and PowerShell.exe is executed with right values, everything work fine!, but if I put these instructions inside "cmd4.ps1" script it doesn't work, I mean something is not right with parameters because I can see it start powershell but it never ends.
-ArgumentList is expecting an array of string arguments instead of a single string. Try this instead:
$ScriptArgs = #(
'-File'
"$LocalDir\Remote\Transfer.ps1"
"http://$ServerIP/Upload/$FileName $ServerIP $LocalIP"
)
Start-Process powershell.exe -ArgumentList $ScriptArgs
Note that you can simplify the string construction of the args as shown above.
Why don't you put this in cmd4.ps1?
& "c:\temp re\Remote\Transfer.ps1" "http://10.1.1.1/Upload/file.txt" "10.1.1.1" "10.1.1.10"