Javascript:在 gulpfile.js 中获取 package.json 数据

2022-01-12 00:00:00 javascript gulp

本身不是一个特定于 gulp 的问题,而是如何从 gulpfile.js 中的 package.json 文件中获取信息;例如,我想获取主页或名称并在任务中使用它.

Not a gulp-specific question per-se, but how would one get info from the package.json file within the gulpfile.js; For instance, I want to get the homepage or the name and use it in a task.

推荐答案

不要将 require('./package.json') 用于 watch 进程,就像使用 require 会将模块解析为第一个请求的结果.

Don't use require('./package.json') for a watch process, as using require will resolve the module as the results of the first request.

因此,如果您正在编辑 package.json,除非您停止监视进程并重新启动它,否则这些编辑将不起作用.

So if you are editing your package.json those edits won't work unless you stop your watch process and restart it.

对于 gulp watch 进程,最好在每次执行任务时重新读取文件并解析它,方法是使用 节点的fs方法

For a gulp watch process it would be best to re-read the file and parse it each time that your task is executed, by using node's fs method

var fs = require('fs')
var json = JSON.parse(fs.readFileSync('./package.json'))

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