blocks|key|201859|text|使用path模块的basename方法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|201860|path.basename('/foo/bar/baz/asdf/quux.html')
//+returns
'quux.html'|code-block|syntax|javascript|201861|Here是取自上述示例的文档。|201862|entityMap|0|LINK|mutability|MUTABLE|url|https://nodejs.org/api/path.html#path_path_basename_path_ext^0|2|4|9|8|0|0|0|4|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@$9|11|A|12|1|13]]|E|$]]|$1|M|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

Use the <code>basename</code> method of the <code>path</code> module:

<pre><code>path.basename('/foo/bar/baz/asdf/quux.html')
// returns
'quux.html'
</code></pre>

<a href="https://nodejs.org/api/path.html#path_path_basename_path_ext" rel="noreferrer">Here</a> is the documentation the above example is taken from.

blocks|key|201961|text|要获取文件名的文件名部分，请使用basename方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|201962|var+path+=+require("path");
var+fileName+=+"C:\\Python27\\ArcGIS10.2\\python.exe";
var+file+=+path.basename(fileName);

console.log(file);+//+'python.exe'|code-block|syntax|javascript|201963|如果您想要不带扩展名的文件名，可以将扩展名变量(包含扩展名)传递给basename方法，告诉Node只返回不带扩展名的名称：|201964|var+path+=+require("path");
var+fileName+=+"C:\\Python27\\ArcGIS10.2\\python.exe";
var+extension+=+path.extname(fileName);
var+file+=+path.basename(fileName,extension);

console.log(file);+//+'python'|201965|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

To get the file name portion of the file name, the basename method is used:

<pre><code>var path = require("path");
var fileName = "C:\\Python27\\ArcGIS10.2\\python.exe";
var file = path.basename(fileName);

console.log(file); // 'python.exe'
</code></pre>

If you want the file name without the extension, you can pass the extension variable (containing the extension name) to the basename method telling Node to return only the name without the extension:

<pre><code>var path = require("path");
var fileName = "C:\\Python27\\ArcGIS10.2\\python.exe";
var extension = path.extname(fileName);
var file = path.basename(fileName,extension);

console.log(file); // 'python'
</code></pre>

blocks|key|201987|text|对于那些对从文件名中删除扩展名感兴趣的人，可以使用https://nodejs.org/api/path.html#path_path_basename_path_ext|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|201988|path.basename('/foo/bar/baz/asdf/quux.html',+'.html');|code-block|syntax|javascript|201989|entityMap|0|LINK|mutability|MUTABLE|url|https://nodejs.org/api/path.html#path_path_basename_path_ext^0|P|1O|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

For those interested in removing extension from filename, you can use
<a href="https://nodejs.org/api/path.html#path_path_basename_path_ext" rel="noreferrer">https://nodejs.org/api/path.html#path_path_basename_path_ext</a>

<pre><code>path.basename('/foo/bar/baz/asdf/quux.html', '.html');
</code></pre>

blocks|key|4707855|text|var+path+=+require("path");
var+filepath+=+"C:\\Python27\\ArcGIS10.2\\python.exe";
var+name+=+path.parse(filepath).name;
//+returns
'python'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4707856|上面的代码返回不带扩展名的文件名，如果您需要带扩展名的文件名|unstyled|4707857|var+path+=+require("path");
var+filepath+=+"C:\\Python27\\ArcGIS10.2\\python.exe";
var+name+=+path.basename(filepath);
//+returns
'python.exe'|4707858|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>var path = require(&quot;path&quot;);
var filepath = &quot;C:\\Python27\\ArcGIS10.2\\python.exe&quot;;
var name = path.parse(filepath).name;
// returns
'python'
</code></pre>
Above code returns the name of the file without extension, if you need the name with extention use
<pre><code>var path = require(&quot;path&quot;);
var filepath = &quot;C:\\Python27\\ArcGIS10.2\\python.exe&quot;;
var name = path.basename(filepath);
// returns
'python.exe'
</code></pre>

blocks|key|201915|text|如果您已经知道路径分隔符是/+(即，您正在为特定的平台/环境编写代码)，如问题中的示例所暗示的，您可以保持它的简单性，并使用分隔符拆分字符串：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|201916|'/foo/bar/baz/asdf/quux.html'.split('/').pop()|code-block|syntax|javascript|201917|这将是faster+(和更干净的imo)，而不是替换为正则表达式。|201918|再说一次:只有在为特定环境编写代码时才这样做，否则请使用path模块，因为路径非常复杂。例如，Windows在许多情况下支持/，但支持not，例如用于共享网络文件夹等的\\?\?样式前缀。在Windows上，上述方法迟早会失败。|201919|entityMap|0|LINK|mutability|MUTABLE|url|https://jsperf.com/basename-from-path/1|1|http://msdn.microsoft.com/en-us/library/aa365247.aspx^0|D|1|0|0|3|6|0|0|S|4|1Q|1|2C|5|1V|3|1|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|D|@$9|13|A|14|1|15]]|E|$]]|$1|M|3|N|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@$9|1D|A|1E|1|1F]]|E|$]]|$1|O|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]]]

If you already know that the path separator is <code>/</code> (i.e. you are writing for a specific platform/environment), as implied by the example in your question, you could keep it simple and split the string by separator:

<pre><code>'/foo/bar/baz/asdf/quux.html'.split('/').pop()
</code></pre>

That would be <a href="https://jsperf.com/basename-from-path/1" rel="nofollow noreferrer">faster</a> (and cleaner imo) than replacing by regular expression.

Again: Only do this if you're writing for a specific environment, otherwise use the <code>path</code> module, as paths are surprisingly complex. Windows, for instance, supports <code>/</code> in many cases but <a href="http://msdn.microsoft.com/en-us/library/aa365247.aspx" rel="nofollow noreferrer">not</a> for e.g. the <code>\\?\?</code> style prefixes used for shared network folders and the like. On Windows the above method is doomed to fail, sooner or later.

blocks|key|4707906|text|在NodeJS中，__filename.split(/%7C//).pop()仅返回任何OS平台上绝对文件路径中的文件名。为什么需要关心记住/导入API，而这种正则表达式方法也让我们回忆起我们的正则表达式技能。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4707907|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

In NodeJS, __filename.split(/\|//).pop() returns just the file name from the absolute file path on any OS platform.
Why need to care about remembering/importing an API while this regex approach also letting us recollect our regex skills.

blocks|key|202187|text|所以Nodejs附带了一个名为'__fileName'的缺省全局变量，该变量保存正在执行的当前文件。我的建议是将__fileName从任何文件传递给一个服务，这样fileName的检索就是动态的|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|202188|下面，我使用fileName字符串，然后根据path.sep拆分它。注意:+path.sep避免了posix文件分隔符和windows文件分隔符的问题(+'/‘和’\‘的问题)。它干净多了。得到子串，只得到最后一个分离的名字，然后用实际长度减去3就不言而喻了。|202189|你可以像这样写一个服务(注意这是用typescript写的，但是你也可以用js写它)|202190|export+class+AppLoggingConstants+{

++++constructor(){

++++}
++++++//+Here+make+sure+the+fileName+param+is+actually+'__fileName'
++++getDefaultMedata(fileName:+string,+methodName:+string)+{
++++++++const+appName+=+APP_NAME;
++++++++const+actualFileName+=+fileName.substring(fileName.lastIndexOf(path.sep)%2B1,+fileName.length+-+3);
++++++++//const+actualFileName+=+fileName;
+++++return+appName%2B+'+--+'%2BactualFileName;
++++}


}

export+const+AppLoggingConstantsInstance+=+new+AppLoggingConstants();|code-block|syntax|javascript|202191|entityMap^0|F|C|0|M|8|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@$9|U|A|V|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|W|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|X|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

So Nodejs comes with the default global variable called <code>'__fileName'</code> that holds the current file being executed
My advice is to pass the __fileName to a service from any file , so that the retrieval of the fileName is made dynamic

Below, I make use of the fileName string and then split it based on the <code>path.sep</code>. Note path.sep avoids issues with posix file seperators and windows file seperators (issues with '/' and '\'). It is much cleaner. Getting the substring and getting only the last seperated name and subtracting it with the actulal length by 3 speaks for itself.

You can write a service like this (Note this is in typescript , but you can very well write it in js )

<pre><code>export class AppLoggingConstants {

 constructor(){

 }
 // Here make sure the fileName param is actually '__fileName'
 getDefaultMedata(fileName: string, methodName: string) {
 const appName = APP_NAME;
 const actualFileName = fileName.substring(fileName.lastIndexOf(path.sep)+1, fileName.length - 3);
 //const actualFileName = fileName;
 return appName+ ' -- '+actualFileName;
 }


}

export const AppLoggingConstantsInstance = new AppLoggingConstants();
</code></pre>

If there any API could retrieve file name from an absolute file path?

e.g. <code>"foo.txt"</code> from <code>"/var/www/foo.txt"</code>

I know it works with string operation, like <code>fullpath.replace(/.+\//, '')</code>
but I want to know is there a more 'formal' way, like <code>file.getName()</code> in java, could do it.

NodeJS get file name from absolute path?

nodejs get file name from absolute path?

Windows

是否有任何API可以从绝对文件路径中检索文件名？例如，来自"/var/www/foo.txt"的"foo.txt"我知道它适用于字符串操作，比如fullpath.replace(/.+\//, '')，但我想知道有没有一种更“正式”的方式，比如java中的file.getName()。NodeJS是否从绝对路径获取文件名？

问nodejs从绝对路径获取文件名？
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