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防止在未捕获异常后崩溃节点js

node.js

2

Juned Ansari · 技术社区 · 7 年前

我如何防止撞车?

const express = require('express');
const winston = require("winston");
const app = express();

//Logging is responsible to log and display errors
require('./startup/logging')();
//routes will contains all the routes list
require('./startup/routes')(app);

//PORT
const port = process.env.PORT || 3000;
app.listen(port,() => winston.info(`Listening on port ${port}....`));

日志.js

const express = require('express');
const winston = require('winston');
// require('express-async-errors');

module.exports = function() {
  winston.handleExceptions(
    new winston.transports.File({ filename: 'uncaughtExceptions.log' })
  );

  process.on('unhandledRejection', (ex) => {
    throw ex;
  });

  winston.add(winston.transports.File, { filename: 'error.log' });

}

1 回复 | 直到 7 年前

1

4

Estus Flask 7 年前

作为 the documentation

默认情况下,winston将在记录未捕获的异常后退出。如果这不是您想要的行为,请设置exitOnError=false
const logger = winston.createLogger({ exitOnError: false });

//
// or, like this:
//
logger.exitOnError = false;

通常认为在异常之后不退出是一种不好的做法,因为其后果是不可预测的。如果已知只有一些异常是可容忍的,则可以使用谓词具体地处理它们:

const ignoreWarnings = err => !(err instanceof WarningError);

const logger = winston.createLogger({ exitOnError: ignoreWarnings });