Command line "launch path not accessible" - swift

Question

I recently found out that I can create Swift command line scripts.
I decided to see if I could build my Xamarin project using it.
Unfortunately I am getting the following error and I don't know how to fix it.
*** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: 'launch path not accessible'
Here is my script:
#!/usr/bin/env swift
import Foundation
print("Building Script")
let fileManager = NSFileManager.defaultManager()
let path = fileManager.currentDirectoryPath
func shell(launchPath: String, arguments: [String] = []) -> NSString? {
let task = NSTask()
task.launchPath = launchPath
task.arguments = arguments
let pipe = NSPipe()
task.standardOutput = pipe
task.launch()
let data = pipe.fileHandleForReading.readDataToEndOfFile()
let output = NSString(data: data, encoding: NSUTF8StringEncoding)
return output
}
if let output = shell("/Applications/Xamarin\\ Studio.app/Contents/MacOS/mdtool", arguments: ["-v build", "\"--configuration:Beta|iPhone\"", "MyApp.iOS.sln"]) {
print(output)
}
Any thoughts?

Answer 1

I think the problem is that you actually want to execute the shell and have it execute the mdtool, rather than directly execute mdtool
Try passing "/bin/bash" as the launchpath, and then include the path to mdtool as part of the argument string.

Answer 2

Recently I encounter a similar problem, but my solution is different.
I fix the problem by changing the permission mode of the script. Specifically, chmod 777 xxx. The key is to give the execution permission.
Let's conduct a controlled experiment to verify it:
prepare a script with path /tmp/a.sh and permission 666
The content of /tmp/a.sh:
#!/bin/sh
echo aaa
prepare a swift script to launch the script:
import Foundation
let fileManager = FileManager.default
let path = fileManager.currentDirectoryPath
func shell(launchPath: String, arguments: [String] = []) -> NSString? {
let task = Process()
task.launchPath = launchPath
task.arguments = arguments
let pipe = Pipe()
task.standardOutput = pipe
task.launch()
let data = pipe.fileHandleForReading.readDataToEndOfFile()
let output = NSString(data: data, encoding: String.Encoding.utf8.rawValue)
return output
}
if let output = shell(launchPath: "/tmp/a.sh", arguments: []) {
print(output)
}
The output is:
... *** Terminating app due to uncaught exception 'NSInvalidArgumentException',
reason: 'launch path not accessible'
change the permission of /tmp/a.sh to 777 (by granting the execution permission) and then re-run the swift script:
aaa

Answer 3

My version that is intended to highlight all sorts of expected results. You can replace ParsingError.encodingFailure and OS.Error.processFailure with your own Error implementations.
/// Getting a `Result` from a shell command
public func process(_ command: String) -> Result<String, Swift.Error> {
let process = Process()
process.launchPath = "/bin/bash"
process.arguments = ["-c", command]
let outputPipe = Pipe()
process.standardOutput = outputPipe
let errorPipe = Pipe()
process.standardError = errorPipe
process.launch()
let outputData = outputPipe.fileHandleForReading.readDataToEndOfFile()
guard let output = String(data: outputData, encoding: .utf8) else {
return .failure(ParsingError.encodingFailure(data: outputData))
}
let errorData = errorPipe.fileHandleForReading.readDataToEndOfFile()
guard let errorOutput = String(data: errorData, encoding: .utf8) else {
return .failure(ParsingError.encodingFailure(data: errorData))
}
process.waitUntilExit()
let status = Int(process.terminationStatus)
if status == 0, errorOutput.isEmpty {
return .success(output)
} else {
return .failure(OS.Error.processFailure(code: status, message: errorOutput))
}
}