nodejs变量在express response.write里面

Question

Assume I have a dynamic way to set the value of toggle to true or false without restarting the app. 假设我有一种动态方法可以在不重新启动应用程序的情况下将toggle值设置为true或false 。

Is there a better way to achieve what I'm trying to do here? 有没有更好的方法来实现我在这里要做的事情？ Namely print different results based on the value of toggle ? 即根据toggle值打印不同的结果？

const express = require('express');
const app = express();
const port = 8080;

var toggle = true;

app.get('/', function (req, res) {

  res.write(`<!DOCTYPE html>
        <html>
        <head>
        <title>Demo Page</title>
        </head>
        <body>`);

  if (toggle) {
    res.write(`<p>Hello World. Toggle Value: true</p>`);
  }
  else {
    res.write(`<p>Hello World. Toggle Value: false</p>`);
  }

  res.write(`</body></html>`);
  res.end();
});

app.listen(port, () => console.log(`Example app listening on port ${port}!`));

Answer 1

You can concat strings to write single line return 您可以连接字符串以写入单行返回

res.write(`<p>Hello World. Toggle Value: ` + true + `</p>`);

Or using ES6 features 或者使用ES6功能

res.write(`<p>Hello World. Toggle Value: ${true}</p>`);

I would recommend you using some view engine such as mustache or pug, where you would pass 'toggle' variable and handle rendering inside template. 我建议你使用一些视图引擎，如胡子或哈巴狗，你将传递'toggle'变量并处理模板内的渲染。

nodejs变量在express response.write里面

问题描述

1 个解决方案

解决方案1
0 2019-06-08 17:11:05

nodejs变量在express response.write里面

问题描述

1 个解决方案

解决方案1 0 2019-06-08 17:11:05

解决方案1
0 2019-06-08 17:11:05