如何在Node.js中获取脚本的路径?

我知道有流程。Cwd,但它只引用调用脚本的目录,而不是脚本本身。例如,假设我在/home/kyle/目录下,然后运行以下命令:

node /home/kyle/some/dir/file.js

如果我调用process.cwd(),我会得到/home/kyle/,而不是/home/kyle/some/dir/。有办法得到那个目录吗?


当前回答

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

其他回答

你可以使用process.env.PWD来获取当前应用程序的文件夹路径。

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。

当涉及到主脚本时,它非常简单:

process.argv[1]

来自Node.js文档:

process.argv 包含命令行参数的数组。第一个元素是“node”,第二个元素是JavaScript文件的路径。下一个元素将是任何额外的命令行参数。

如果你需要知道模块文件的路径,那么使用__filename。

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