我正在寻找的方式,我如何只能发布“信息”级别(为用户)的控制台,和“调试”级别的帖子到文件。目前,我只找到了一个工作解决方案--使用很少的"getLogger()“函数。例如:
其中,函数的每个参数将配置的类别与配置()函数分开。但是我不喜欢它,并且假设有一个getLogger()函数有更好的选项。最后一次认为我试过的是:
log4js.configure({
appenders: {
everything:{ type: 'stdout' },
file_log: { type: 'file' ,filename: 'all-the-logs.log' },
logLevelFilter: { type:'logLevelFilter',level: 'debug', appender: 'file_log' }
},
categories: {
default: {
appenders: [ 'logLevelFilter','everything'], level: 'info'},
}
});
我在控制台中看到只有“信息”级别,但在文件中,我也只看到“信息”级别。如果我添加到附录级" all“,我会在控制台中看到所有的级别,并且相同的也会出现在文件中。
发布于 2019-06-17 20:47:59
log4js提供了特殊的附加类型的categoryFilter
和logLevelFilter
,参见Log4js -评价者,步骤如下
logLevelFilter
,指向相关的附件;config.categories.default.appenders
;log4js.getLogger()
创建记录器和到处日志记录,日志文件用爱级分隔。书写成绩单
import {Appender, Configuration, LogLevelFilterAppender} from "log4js";
const fs = require('fs');
const path = require('path');
let levels = ['trace', 'debug', 'info', 'warn', 'error', 'fatal'];
let conf: Configuration = {
appenders : {
default: {
type : "console",
layout: {
type : "colored",
pattern: "%m"
},
}
},
categories: {
default: {
appenders: levels,
level : 'all'
}
},
};
let logs = path.join(__dirname, 'logs');
if (!fs.existsSync(logs)) {
fs.mkdirSync(logs);
}
for (let level of levels) {
let appender: Appender = {
type : "file",
filename : path.join(logs, level + '.log'),
maxLogSize: 65536,
backups : 10,
layout : {
type : "pattern",
pattern: "%d{dd/MM hh:mm} %-5p %m"
}
};
conf.appenders[level] = appender;
conf.categories[level] = {
appenders: [level],
level : level.toUpperCase(),
}
}
let prefix = 'only-';
for (let level of levels) {
let appender: LogLevelFilterAppender = {
type : "logLevelFilter",
appender: level,
level : level,
maxLevel: level,
};
let name = prefix + level;
conf.appenders[name] = appender;
conf.categories[name] = {
appenders: [level],
level : level.toUpperCase(),
}
}
conf.categories.default.appenders = levels.map(x => prefix + x);
//console.log(JSON.stringify(conf, null, 2));
export default conf;
样本使用
// const log4js = require('log4js');
import * as log4js from 'log4js';
import log4jsConfig from './log4js.config';
log4js.configure(log4jsConfig );
(function () {
const logger = log4js.getLogger();
logger.trace('Entering cheese testing');
logger.debug('Got cheese.');
logger.info('Cheese is Gouda.');
logger.warn('Cheese is quite smelly.');
logger.error('Cheese is too ripe!');
logger.fatal('Cheese was breeding ground for listeria.');
log4js.shutdown(console.log);
})();
https://stackoverflow.com/questions/54384094
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