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?
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
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
});
其他回答
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。
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
});
NodeJS公开了一个名为__dirname的全局变量。
__dirname返回JavaScript文件所在文件夹的完整路径。
因此,作为一个例子,对于Windows,如果我们用下面的行创建一个脚本文件:
console.log(__dirname);
然后使用以下命令运行脚本:
node ./innerFolder1/innerFolder2/innerFolder3/index.js
输出将是: C: \用户…<项目目录> \ innerFolder1 \ innerFolder2 \ innerFolder3
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)
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)