We read every piece of feedback, and take your input very seriously.
To see all available qualifiers, see our documentation.
Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.
By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.
Already on GitHub? Sign in to your account
Problem: running busted -e "print('hello')" . works but busted -e "print('hello')" --lua "lua" . doesn't.
busted -e "print('hello')" .
busted -e "print('hello')" --lua "lua" .
Solution: quote args at if lua and not ignore-lua
-- busted/modules/cli.lua: 214 if cliArgs['lua'] and not cliArgs['ignore-lua'] then local quoted = {} for _, arg in ipairs(args) do local a = arg:gsub('"', '\\"') table.insert(quoted, a) end local end_args = '' if #quoted >0 then end_args = '"' .. table.concat(quoted,'" "') .. '"' end local _, code = execute( cliArgs["lua"] .. " " .. args[0] .. ' --ignore-lua ' .. end_args ) exit(code) end
The text was updated successfully, but these errors were encountered:
Successfully merging a pull request may close this issue.
Problem:
running
busted -e "print('hello')" .
works butbusted -e "print('hello')" --lua "lua" .
doesn't.Solution: quote args at if lua and not ignore-lua
The text was updated successfully, but these errors were encountered: