Terminal command not recognised? - macos

Question

when I type this directly into the terminal - it works as expected and returns 0 if the app is not running, and 1 if it is.
lsappinfo list | grep -v grep | grep bundleID | grep com.test.myapp | wc -l
However, when I use the code below (swift 3 - macOS), it says it is an unrecognised command?
// DECLARE TASK
let task = Process()
// DEFINE THE PATH
task.launchPath = "/usr/bin/lsappinfo"
// DEFINE THE ARGUMENTS
task.arguments = ["list | grep -v grep | grep bundleID | grep com.test.myapp | wc -l"]
// DECLARE outputPipe
let outputPipe = Pipe()
// RUN THE TASK
task.launch()
// DECLARE data
let data = outputPipe.fileHandleForReading.readDataToEndOfFile()
// DECLARE output AS THE UTF-8 STRING OF THE TERMINAL'S OUTPUT
let output = NSString(data: data, encoding: String.Encoding.utf8.rawValue)
print(output!)
if output == "0" {
print("App is not running!")
} else {
print("App is running!")
}
// PAUSE UNTIL COMPLETED
task.waitUntilExit()
Can someone please tell me where I have gone wrong, as I am new to Swift, and still struggling to get my head around the language / syntax.
Thank you all in advance.

Answer 1

Piping operations are not arguments to the process, they are instructions to a shell to connect several separate processes. If the task construct in swift expects a process and a set of command-line argument, then one way to do this here would maybe be to launch the bash binary, and then put -c "lsappinfo list | grep -v grep | grep bundleID | grep com.test.myapp | wc -l" as the full set of arguments. That way, you would let bash sort out what's needed to accomplish the piping.