CODEWITHSUNDEEP
powershellcmd
KoLa
KoLa

Reputation: 33

Convert cmd to powershell

How can I convert this cmd to powershell

cd C:\Program Files (x86)\Image Instruments\abcClient
“XYZ.exe” -PatId=”11111” -Firstname=“John” -Lastname=“Do” -Birthday=”19971101” -Sex=“M” -City=“Frankfurt”

I tried something like this, but does not work. It opens the .exe without arguments:

$pathExe = Set-Location “C:\Program Files (x86)\Image Instruments\abcClient”
$openExe = Start-Process “XYZ.exe”
$PatId = ”11111”
$Firstname = “John”
$Lastname = “Doe”
$Birthday = ”19971101”
$Sex = “M”
$City = “Frankfurt”
$pathExe; & $openExe@('-PatID='+$PatId, '-Firstname=’+$Firstname, '-Lastname='+$Lastname, ’-Birthday=’+$Birthday, ‘-Sex=’+$Sex, '-City='+$City)

As I have no experience with programming, how I could bring this to work?

Thank you for any advice

Upvotes: 3

Views: 10831

Answers (3)

ThePennyDrops
ThePennyDrops

Reputation: 171

Get / Convert / Access / Translate Cmd COMMAND Variables in PowerShell:

You may want to try the following which I have 'found' works in a single PC environment:

#- Set up Cmd Variable -----
$zCmdVariable = "%UserProfile%"
#- Get Contents from Cmd -----
$zConvertedVariable = (cmd /c "echo $zCmdVariable")
#- Use contents in Powershell -----
Get-Item -Path $zConvertedVariable

I hope this helps; Pass It forward.

Upvotes: 0

mklement0
mklement0

Reputation: 437062

Just like cmd, PowerShell is a shell that supports invoking programs directly, with arguments.

Therefore, your command looks very similar in PowerShell:

Set-Location "C:\Program Files (x86)\Image Instruments\abcClient"
XYZ.exe -PatId="11111" -Firstname="John" -Lastname="Do" -Birthday="19971101" -Sex="M" -City="Frankfurt"
  • The equivalent of cd in PowerShell is Set-Location (although cd works too, because it is defined as an alias for it).

  • Note the need to quote, the target directory passed to Set-Location, given that its path contains spaces.

  • Unquoted use of XYZ.exe allows its direct invocation; if you quote the program name (e.g., as "XYZ.exe"), you would have to prepend & , the call operator.

  • If you want to use variables in your arguments, simply use them in lieu of the literal strings; e.g. -PatId="11111" can be replaced with -PatId="$PatId", assuming that you've defined $PatId = "11111" beforehand.

  • Though they do not come into play here, there are differences between cmd and PowerShell with respect to what characters need quoting when you pass arguments and how special characters are escaped - see Get-Help about_parsing

  • If XYZ.exe is a console application (which appears to be the case), PowerShell will execute it synchronously - that is, PowerShell will wait until the application has exited and will reflect its exit code in automatic variable $LASTEXITCODE.

Upvotes: 2

Ctznkane525
Ctznkane525

Reputation: 7465

So, the ProcessStartInfo object defines parameters to pass into a process:

FileName - This is what you'll be executing

WorkingDirectory - The execution path that it will be running from (not always the same as the file's directory but often is). Important if you have things like relative paths

Arguments - This is what is getting passed as the process arguments.

$startInfo = New-Object System.Diagnostics.ProcessStartInfo
$startInfo.FileName = "C:\Program Files (x86)\Image Instruments\abcClient\XYZ.exe"
$startInfo.WorkingDirectory = "C:\Program Files (x86)\Image Instruments\abcClient\"
$startInfo.Arguments = "-PatId=11111 -Firstname=John -Lastname=Do -Birthday=19971101 -Sex=M -City=Frankfurt"

Then you start a process object, and wait for it to complete.

$process = New-Object System.Diagnostics.Process
$process.StartInfo = $startInfo
$process.Start()
$process.WaitForExit()

Then get it's return code (usually EXEs with a 0 mean success).

$exitCode = $process.ExitCode

Upvotes: -1

Related Questions