blocks|key|391237|text|使用Path+(Java+7%2B)的替代方法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|391238|Path+p+=+Paths.get("C:\\Hello\\AnotherFolder\\The+File+Name.PDF");
String+file+=+p.getFileName().toString();|code-block|syntax|javascript|391239|请注意，在\\上拆分字符串取决于平台，因为文件分隔符可能会有所不同。Path#getName会为您解决这个问题。|391240|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/7/docs/api/java/nio/file/Path.html^0|2|4|2|4|0|0|0|5|2|Y|C|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

Alternative using <a href="http://docs.oracle.com/javase/7/docs/api/java/nio/file/Path.html" rel="noreferrer"><code>Path</code></a> (Java 7+):

<pre><code>Path p = Paths.get("C:\\Hello\\AnotherFolder\\The File Name.PDF");
String file = p.getFileName().toString();
</code></pre>

Note that splitting the string on <code>\\</code> is platform dependent as the file separator might vary. <code>Path#getName</code> takes care of that issue for you.

blocks|key|391268|text|可以使用path+=+C:\Hello\AnotherFolder\TheFileName.PDF|type|unstyled|depth|inlineStyleRanges|entityRanges|data|391269|String+strPath+=+path.substring(path.lastIndexOf("\\")%2B1,+path.length());|code-block|syntax|javascript|391270|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

you can use path = C:\Hello\AnotherFolder\TheFileName.PDF

<pre><code>String strPath = path.substring(path.lastIndexOf("\\")+1, path.length());
</code></pre>

blocks|key|391298|text|其他的答案对于我的特定场景并不是很有效，我正在读取来自不同于我当前操作系统的路径。为了详细说明，我将从Windows平台保存的电子邮件附件保存在Linux服务器上。从JavaMail接口返回的文件名类似于'C:\temp\hello.xls‘|type|unstyled|depth|inlineStyleRanges|entityRanges|data|391299|我最终得到的解决方案是：|391300|String+filenameWithPath+=+"C:\\temp\\hello.xls";
String[]+tokens+=+filenameWithPath.split("[\\\\%7C/]");
String+filename+=+tokens[tokens.length+-+1];|code-block|syntax|javascript|391301|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The other answers didn't quite work for my specific scenario, where I am reading paths that have originated from an OS different to my current one. To elaborate I am saving email attachments saved from a Windows platform on a Linux server. The filename returned from the JavaMail API is something like 'C:\temp\hello.xls'

The solution I ended up with:

<pre><code>String filenameWithPath = "C:\\temp\\hello.xls";
String[] tokens = filenameWithPath.split("[\\\\|/]");
String filename = tokens[tokens.length - 1];
</code></pre>

blocks|key|391315|text|在Apache+Commons+IO中使用FilenameUtils：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|391316|String+name1+=+FilenameUtils.getName("/ab/cd/xyz.txt");
String+name2+=+FilenameUtils.getName("c:\\ab\\cd\\xyz.txt");|code-block|syntax|javascript|391317|entityMap|0|LINK|mutability|MUTABLE|url|https://commons.apache.org/proper/commons-io/javadocs/api-1.4/org/apache/commons/io/FilenameUtils.html#getName(java.lang.String)^0|L|D|1|H|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

Using <code>FilenameUtils</code> in <a href="https://commons.apache.org/proper/commons-io/javadocs/api-1.4/org/apache/commons/io/FilenameUtils.html#getName(java.lang.String)" rel="noreferrer">Apache Commons IO</a> :

<pre><code>String name1 = FilenameUtils.getName("/ab/cd/xyz.txt");
String name2 = FilenameUtils.getName("c:\\ab\\cd\\xyz.txt");
</code></pre>

blocks|key|173386|text|从1.7开始|type|unstyled|depth|inlineStyleRanges|entityRanges|data|173387|++++Path+p+=+Paths.get("c:\\temp\\1.txt");
++++String+fileName+=+p.getFileName().toString();
++++String+directory+=+p.getParent().toString();|code-block|syntax|javascript|173388|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Since 1.7

<pre><code> Path p = Paths.get("c:\\temp\\1.txt");
 String fileName = p.getFileName().toString();
 String directory = p.getParent().toString();
</code></pre>

blocks|key|173413|text|使用java+regex+*提取文件名。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|173414|public+String+extractFileName(String+fullPathFile){
++++++++try+{
++++++++++++Pattern+regex+=+Pattern.compile("([%5E\\\\/:*?\"<>%7C\r\n]%2B$)");
++++++++++++Matcher+regexMatcher+=+regex.matcher(fullPathFile);
++++++++++++if+(regexMatcher.find()){
++++++++++++++++return+regexMatcher.group(1);
++++++++++++}
++++++++}+catch+(PatternSyntaxException+ex)+{
++++++++++++LOG.info("extractFileName::pattern+problem+<"%2BfullPathFile%2B">",ex);
++++++++}
++++++++return+fullPathFile;
++++}|code-block|syntax|javascript|173415|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

extract file name using java regex *.

<pre><code>public String extractFileName(String fullPathFile){
 try {
 Pattern regex = Pattern.compile("([^\\\\/:*?\"&lt;&gt;|\r\n]+$)");
 Matcher regexMatcher = regex.matcher(fullPathFile);
 if (regexMatcher.find()){
 return regexMatcher.group(1);
 }
 } catch (PatternSyntaxException ex) {
 LOG.info("extractFileName::pattern problem &lt;"+fullPathFile+"&gt;",ex);
 }
 return fullPathFile;
 }
</code></pre>

blocks|key|173432|text|考虑到Java是多平台的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|173433|int+lastPath+=+fileName.lastIndexOf(File.separator);
if+(lastPath!=-1){
++++fileName+=+fileName.substring(lastPath%2B1);
}|code-block|syntax|javascript|173434|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Considere the case that Java is Multiplatform:

<pre><code>int lastPath = fileName.lastIndexOf(File.separator);
if (lastPath!=-1){
 fileName = fileName.substring(lastPath+1);
}
</code></pre>

blocks|key|173473|text|java.nio.file.Path的getFileName()方法，用于返回此path对象所指向的文件或目录的名称。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|173474|路径getFileName()|173475|供参考：|173476|https://www.geeksforgeeks.org/path-getfilename-method-in-java-with-examples/|offset|length|173477|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|0|24|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|T|8|@]|9|@$H|U|I|V|1|W]]|A|$]]|$1|J|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|G]]]]

getFileName() method of java.nio.file.Path used to return the name of the file or directory pointed by this path object.

Path getFileName()

For reference:

<a href="https://www.geeksforgeeks.org/path-getfilename-method-in-java-with-examples/" rel="nofollow noreferrer">https://www.geeksforgeeks.org/path-getfilename-method-in-java-with-examples/</a>

blocks|key|173507|text|您可以使用FileInfo对象来获取文件的所有信息。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|173508|++++FileInfo+f+=+new+FileInfo(@"C:\Hello\AnotherFolder\The+File+Name.PDF");
++++MessageBox.Show(f.Name);
++++MessageBox.Show(f.FullName);
++++MessageBox.Show(f.Extension+);
++++MessageBox.Show(f.DirectoryName);|code-block|syntax|javascript|173509|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can use FileInfo object to get all information of your file.

<pre><code> FileInfo f = new FileInfo(@"C:\Hello\AnotherFolder\The File Name.PDF");
 MessageBox.Show(f.Name);
 MessageBox.Show(f.FullName);
 MessageBox.Show(f.Extension );
 MessageBox.Show(f.DirectoryName);
</code></pre>

blocks|key|173562|text|在c#中，这个答案适用于我：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|173563|using+System.IO;
string+fileName+=+Path.GetFileName("C:\Hello\AnotherFolder\The+File+Name.PDF");|code-block|syntax|javascript|173564|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This answer works for me in c#:
<pre><code>using System.IO;
string fileName = Path.GetFileName(&quot;C:\Hello\AnotherFolder\The File Name.PDF&quot;);
</code></pre>

<code>String</code> variable contains a file name, <code>C:\Hello\AnotherFolder\The File Name.PDF</code>. How do I only get the file name <code>The File Name.PDF</code> as a String?

I planned to split the string, but that is not the optimal solution.

How do I get the file name from a String containing the Absolute file path?

服务器

Windows

Linux

Java

String变量包含文件名C:\Hello\AnotherFolder\The File Name.PDF。如何才能仅获取字符串形式的文件名The File Name.PDF？我计划拆分字符串，但这不是最佳解决方案。

问如何从包含绝对文件路径的字符串中获取文件名？
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问如何从包含绝对文件路径的字符串中获取文件名？
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