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

如果你使用pkg来打包你的应用,你会发现这个表达式很有用:

appDirectory = require('path').dirname(process.pkg ? process.execPath : (require.main ? require.main.filename : process.argv[0]));

process.pkg tells if the app has been packaged by pkg. process.execPath holds the full path of the executable, which is /usr/bin/node or similar for direct invocations of scripts (node test.js), or the packaged app. require.main.filename holds the full path of the main script, but it's empty when Node runs in interactive mode. __dirname holds the full path of the current script, so I'm not using it (although it may be what OP asks; then better use appDirectory = process.pkg ? require('path').dirname(process.execPath) : (__dirname || require('path').dirname(process.argv[0])); noting that in interactive mode __dirname is empty. For interactive mode, use either process.argv[0] to get the path to the Node executable or process.cwd() to get the current directory.

你可以使用process.env.PWD来获取当前应用程序的文件夹路径。

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)

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
 });

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;
    }
}