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
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。
其他回答
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。
var settings =
JSON.parse(
require('fs').readFileSync(
require('path').resolve(
__dirname,
'settings.json'),
'utf8'));
如果你想在shell脚本中使用类似$0的东西,试试这个:
var path = require('path');
var command = getCurrentScriptPath();
console.log(`Usage: ${command} <foo> <bar>`);
function getCurrentScriptPath () {
// Relative path from current working directory to the location of this script
var pathToScript = path.relative(process.cwd(), __filename);
// Check if current working dir is the same as the script
if (process.cwd() === __dirname) {
// E.g. "./foobar.js"
return '.' + path.sep + pathToScript;
} else {
// E.g. "foo/bar/baz.js"
return pathToScript;
}
}
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
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
});