Reputation: 443
I've created a minimal logger in Golang. I tried to keep it as simple as possible but two problems arise:
log.Lshortfile
flag is always shown as logger.go
Here's my code:
package logger
import (
"log"
"os"
)
var (
dlog = log.New(os.Stdout, "\x1B[36mDEBUG: \x1B[0m", log.Ldate|log.Ltime|log.Lshortfile)
wlog = log.New(os.Stdout, "\x1B[35mWARN: \x1B[0m", log.Ldate|log.Ltime|log.Lshortfile)
ilog = log.New(os.Stdout, "\x1B[32mINFO: \x1B[0m", log.Ldate|log.Ltime|log.Lshortfile)
elog = log.New(os.Stderr, "\x1B[31mERROR: \x1B[0m", log.Ldate|log.Ltime|log.Lshortfile)
)
// Debug Log
func Debug(a ...interface{}) {
dlog.Println(a)
}
// Warn log
func Warn(a ...interface{}) {
wlog.Println(a)
}
// Info Log
func Info(a ...interface{}) {
ilog.Println(a)
}
// Error Log
func Error(a ...interface{}) {
elog.Println(a)
}
I use them like this: logger.Debug("Hello") or logger.Info("There")
How do I implement this correctly? Thank you.
Upvotes: 0
Views: 674
Reputation: 121119
Println is passed a single parameter, a slice of interface{}, and prints it using slice notation. To fix this, pass the variadic parameters to your functions as variadic arguments to the called functions.
func Debug(a ...interface{}) {
dlog.Println(a...) // <-- note ... here
}
Here's the doc on variadic args.
Call Output to workaround the file name issue.
func Debug(a ...interface{}) {
dlog.Output(2, fmt.Sprintln(a...))
}
Upvotes: 3