在Golang中实现一个最小化的日志记录器。

huangapple go评论87阅读模式
英文:

Implement a minimal logger in Golang

问题

我已经创建了一个简单的 Golang 日志记录器。我试图尽可能保持简单,但是出现了两个问题:

  • 消息字符串显示为用方括号 [] 包裹的数组
  • 使用 log.Lshortfile 标志的文件名始终显示为 logger.go

这是我的代码:

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)
}

我像这样使用它们:logger.Debug("Hello")logger.Info("There")

你如何正确实现它?谢谢。

英文:

I've created a minimal logger in Golang.
I tried to keep it as simple as possible but two problems arise:

  • Message string is shown as array wrapped in brackets []
  • filename using log.Lshortfile flag is always shown as logger.go

在Golang中实现一个最小化的日志记录器。

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.

答案1

得分: 3

Println接收一个参数,参数类型是interface{}的切片,并使用切片表示法进行打印。要修复这个问题,将可变参数作为可变参数传递给被调用函数。

func Debug(a ...interface{}) {
     dlog.Println(a...) // <-- 注意这里的...
}

这里是关于可变参数的文档

调用Output来解决文件名问题。

func Debug(a ...interface{}) {
     dlog.Output(2, fmt.Sprintln(a...))
}
英文:

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...))
}

huangapple
  • 本文由 发表于 2016年10月31日 04:31:03
  • 转载请务必保留本文链接:https://go.coder-hub.com/40333323.html
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定