由于眼睛的问题,我不得不将控制台背景色改为白色,但字体是灰色的,它使消息无法阅读。我怎样才能改变呢?


当前回答

我重载了控制台方法。

var colors={
Reset: "\x1b[0m",
Red: "\x1b[31m",
Green: "\x1b[32m",
Yellow: "\x1b[33m"
};

var infoLog = console.info;
var logLog = console.log;
var errorLog = console.error;
var warnLog = console.warn;

console.info= function(args)
{
    var copyArgs = Array.prototype.slice.call(arguments);
    copyArgs.unshift(colors.Green);
    copyArgs.push(colors.Reset);
    infoLog.apply(null,copyArgs);
};

console.warn= function(args)
{
    var copyArgs = Array.prototype.slice.call(arguments);
    copyArgs.unshift(colors.Yellow);
    copyArgs.push(colors.Reset);
    warnLog.apply(null,copyArgs);
};
console.error= function(args)
{
    var copyArgs = Array.prototype.slice.call(arguments);
    copyArgs.unshift(colors.Red);
    copyArgs.push(colors.Reset);
    errorLog.apply(null,copyArgs);
};

// examples
console.info("Numeros",1,2,3);
console.warn("pares",2,4,6);
console.error("reiniciandooo");

输出为。

其他回答

日志/索引.js

const colors = {
    Reset : "\x1b[0m",
    Bright : "\x1b[1m",
    Dim : "\x1b[2m",
    Underscore : "\x1b[4m",
    Blink : "\x1b[5m",
    Reverse : "\x1b[7m",
    Hidden : "\x1b[8m",

    FgBlack : "\x1b[30m",
    FgRed : "\x1b[31m",
    FgGreen : "\x1b[32m",
    FgYellow : "\x1b[33m",
    FgBlue : "\x1b[34m",
    FgMagenta : "\x1b[35m",
    FgCyan : "\x1b[36m",
    FgWhite : "\x1b[37m",

    BgBlack : "\x1b[40m",
    BgRed : "\x1b[41m",
    BgGreen : "\x1b[42m",
    BgYellow : "\x1b[43m",
    BgBlue : "\x1b[44m",
    BgMagenta : "\x1b[45m",
    BgCyan : "\x1b[46m",
    BgWhite : "\x1b[47m",
};

module.exports = () => {
    Object.keys(colors).forEach(key => {
        console['log' + key] = (strg) => {
            if(typeof strg === 'object') strg = JSON.stringify(strg, null, 4);
            return console.log(colors[key]+strg+'\x1b[0m');
        }
    });
}

app.js

require('./logger')();

然后这样使用它:

console.logBgGreen(" grüner Hintergrund ")

Sindre Sorhus设计的这个图书馆是目前最好的:

粉笔

高性能 不扩展String.prototype 富有表现力的API 嵌套样式的能力 干净和专注 自动检测颜色支持 积极维护 5500+模块使用

我发现上面的答案(https://stackoverflow.com/a/41407246/4808079)非常有用,但不完整。如果你只想给某样东西上色一次,我想这没问题,但我认为以可运行的函数形式共享它更适用于现实生活中的用例。

const Color = {
  Reset: "\x1b[0m",
  Bright: "\x1b[1m",
  Dim: "\x1b[2m",
  Underscore: "\x1b[4m",
  Blink: "\x1b[5m",
  Reverse: "\x1b[7m",
  Hidden: "\x1b[8m",
  
  FgBlack: "\x1b[30m",
  FgRed: "\x1b[31m",
  FgGreen: "\x1b[32m",
  FgYellow: "\x1b[33m",
  FgBlue: "\x1b[34m",
  FgMagenta: "\x1b[35m",
  FgCyan: "\x1b[36m",
  FgWhite: "\x1b[37m",
  FgGray: "\x1b[90m",
  
  BgBlack: "\x1b[40m",
  BgRed: "\x1b[41m",
  BgGreen: "\x1b[42m",
  BgYellow: "\x1b[43m",
  BgBlue: "\x1b[44m",
  BgMagenta: "\x1b[45m",
  BgCyan: "\x1b[46m",
  BgWhite: "\x1b[47m"
  BgGray: "\x1b[100m",
}

function colorString(color, string) {
  return `${color}${string}${Color.Reset}`;
}

function colorLog(color, ...args) {
  console.log(...args.map(
   (it) => typeof it === "string" ? colorString(color, string) : it
  ));
}

像这样使用它:

colorLog(Color.FgYellow, "Some Yellow text to console log", { someObj: true });

console.log([
  colorString(Color.FgRed, "red"),
  colorString(Color.FgGreen, "green"),
  colorString(Color.FgBlue, "blue"),
].join(", "));

你也可以使用颜色。

用法:

var cw = require('colorworks').create();
console.info(cw.compile('[[red|Red message with a [[yellow|yellow]] word.]]'));

为了简化工作,您还可以使用它创建一个函数。

function say(msg) {
  console.info(cw.compile(msg));
}

现在你可以做:

say(`[[yellow|Time spent: [[green|${time}]]ms.]]`);

我在我的snippet目录中创建了一个名为styles.js的文件,我认为它可以帮助任何想要导入单个文件的人。

这是对color.js的styles.js文件的一个小修改,帮助了我很多。

以下是该文件的内容:

// Original: https://github.com/Marak/colors.js/blob/master/lib/styles.js

const styleCodes = {
    // Reset all styles.
    reset: [0, 0],
    
    // Text styles.
    bold: [1, 22],
    dim: [2, 22],
    italic: [3, 23],
    underline: [4, 24],
    inverse: [7, 27],
    hidden: [8, 28],
    strikethrough: [9, 29],
    
    // Foregound classic colours.
    fgBlack: [30, 39],
    fgRed: [31, 39],
    fgGreen: [32, 39],
    fgYellow: [33, 39],
    fgBlue: [34, 39],
    fgMagenta: [35, 39],
    fgCyan: [36, 39],
    fgWhite: [37, 39],
    fgGray: [90, 39],
    
    // Foreground bright colours.
    fgBrightRed: [91, 39],
    fgBrightGreen: [92, 39],
    fgBrightYellow: [93, 39],
    fgBrightBlue: [94, 39],
    fgBrightMagenta: [95, 39],
    fgBrightCyan: [96, 39],
    fgBrightWhite: [97, 39],

    // Background basic colours.
    bgBlack: [40, 49],
    bgRed: [41, 49],
    bgGreen: [42, 49],
    bgYellow: [43, 49],
    bgBlue: [44, 49],
    bgMagenta: [45, 49],
    bgCyan: [46, 49],
    bgWhite: [47, 49],
    bgGray: [100, 49],
    bgGrey: [100, 49],
    
    // Background bright colours.
    bgBrightRed: [101, 49],
    bgBrightGreen: [102, 49],
    bgBrightYellow: [103, 49],
    bgBrightBlue: [104, 49],
    bgBrightMagenta: [105, 49],
    bgBrightCyan: [106, 49],
    bgBrightWhite: [107, 49],
};

// This object will contain the string representation for all style codes.
const styles = {};

// Loop over all the style codes and assign them to the `styles` object.
// 
// The a `styleCode` in the `styleCodes` object consists of two numbers:
// Index 0: The opening style code (In HTML this can be the opening <b> tag).
// Index 1: The closing style code (In HTML this can be the closing </b> tag).
for (let styleCode of Object.keys(styleCodes)) {
    styles[styleCode] = {
        open: `\x1B[${styleCodes[styleCode][0]}m`,
        close: `\x1B[${styleCodes[styleCode][1]}m`,
    };
}

module.exports = styles;

其实用起来很简单。

const styles = require("/path/to/styles.js");

// Let's say we've got an error:
const errorOpen = styles.bold.open + styles.bgRed.open + styles.fgWhite.open;
const errorClose = styles.reset.close; // Close everything
console.log(errorOpen, "ERROR", errorClose, ": Missing semicolon at line 9.");