NODE.JS-如何处理OS和URL样式的“路径"的混合正确吗? [英] NODE.JS - how to handle a mix of OS and URL-style "paths" correctly?

问题描述

在我的node.js应用程序中，我有可以传递的功能

In my node.js app I have functions which can be passed either

OS样式的路径，例如c:\ my \ docs \ mydoc.doc(或/usr/docs/mydoc.doc或任何本地文件)

OS-style paths e.g. c:\my\docs\mydoc.doc (or /usr/docs/mydoc.doc or whatever is local)

文件网址，例如file://c:/my/docs/mydoc.doc(我不确定'\'s in ??的有效性?)

File URLS e.g. file://c:/my/docs/mydoc.doc (which I'm not sure about the validity of '\'s in??)

无论哪种方式，我都需要检查它们是否引用了一个特定的位置，该位置将始终作为本地OS样式路径存在，例如c:\ mydata \ directory \或/usr/mydata/directory

Either way, I need to check to see if they refer to a specific location which will always exist as a local OS-style path e.g. c:\mydata\directory\ or /usr/mydata/directory

显然，对于OS风格的路径，我可以将它们作为字符串进行比较-它们应该始终相同(它们是使用path创建的)，但是FILE://URL不一定要使用path.sep，因此不会字符串匹配"?

Obviously for OS-style paths I can just compare them as strings - they SHOULD always be the same (they're created with path) but FILE:// URLS don't necessarily use path.sep and so won't "string match"?

关于处理此问题的最佳方法的任何建议(我个人倾向于用任何一种或多种斜杠来打破一切，然后检查每件??

Any suggestions as to the best way to handle this (I'm personally tempted to break EVERYTHING by one-or-more slashes of either sort and then check each piece??

推荐答案

只需对字符串进行一些处理，并在校正差异后检查以确保它们相同:

Just do some manipulation of the string and check to make sure they are the same after correcting for the differences:

var path = require('path');
var pathname = "\\usr\\home\\newbeb01\\Desktop\\testinput.txt";
var pathname2 = "file://usr/home/newbeb01/Desktop/testinput.txt"

if(PreparePathNameForComparing(pathname) == PreparePathNameForComparing(pathname2))
{   console.log("Same path");   }
else
{   console.log("Not the same path");   }

function PreparePathNameForComparing(pathname)
{
    var returnString = pathname;
    //try to find the file uri prefix, if there strip it off
    if(pathname.search("file://") != -1 || pathname.search("FILE://") != -1)
    {   returnString = pathname.substring(6, pathname.length);  }

    //now make all slashes the same
    if(path.sep === '\\')    //replace all '/' with '\\'
    {   returnString = returnString.replace(/\//g, '\\');   }
    else    //replace all '\\' with '/'
    {   returnString = returnString.replace(/\\/g, '/');    }

    return returnString;
}

我检查了URI路径名称指示符"file://"是否存在，如果有，我将其从比较字符串中删除.然后我将基于路径分隔符的标准化节点的路径模块给了我.这样，它就可以在Linux或Windows环境中工作.

I checked to see if the URI path name indicator "file://" was there, if so, I deleted it off my compare string. Then I normalized based on the path separator node path module will give me. This way it should work in either Linux or Windows environment.

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