I know this is pretty old, and the original question I was responding to is marked as duplicate and directed here, but I ran into an issue trying to get jasmine-reporters to work and didn't like the idea that I had to downgrade in order for it to work. I found out that jasmine-reporters wasn't resolving the savePath correctly and was actually putting the reports folder output in jasmine-reporters directory instead of the root directory of where I ran gulp. In order to make this work correctly I ended up using process.env.INIT_CWD to get the initial Current Working Directory which should be the directory where you ran gulp. Hope this helps someone.

var reporters = require('jasmine-reporters');
var junitReporter = new reporters.JUnitXmlReporter({
  savePath: process.env.INIT_CWD + '/report/e2e/',
  consolidateAll: true,
  captureStdout: true
 });

我又看了一遍文件后找到的。我要找的是__filename和__dirname模块级变量。

__filename是当前模块的文件名。这是当前模块文件的解析绝对路径。(例:/ home /凯尔/一些/ dir / file.js) __dirname是当前模块的目录名。(例:/ home /凯尔/一些/ dir)

每个Node.js程序在其环境中都有一些全局变量，这些变量表示关于进程的一些信息，其中一个是__dirname。

基本上你可以这样做:

fs.readFile(path.resolve(__dirname, 'settings.json'), 'UTF-8', callback);

使用resolve()而不是连接'/'或'\'，否则您将遇到跨平台问题。

注意:__dirname是模块或包含脚本的本地路径。如果你正在编写一个插件，需要知道主脚本的路径，它是:

require.main.filename

或者，获取文件夹名称:

require('path').dirname(require.main.filename)

var settings = 
    JSON.parse(
        require('fs').readFileSync(
            require('path').resolve(
                __dirname, 
                'settings.json'),
            'utf8'));

Index.js中包含要导出的模块的任何文件夹

const entries = {};
for (const aFile of require('fs').readdirSync(__dirname, { withFileTypes: true }).filter(ent => ent.isFile() && ent.name !== 'index.js')) {
  const [ name, suffix ] = aFile.name.split('.');
  entries[name] = require(`./${aFile.name}`);
}

module.exports = entries;

这将找到当前目录的根目录下的所有文件，要求并导出与文件名干相同的导出名称的每个文件。