blocks|key|1530626|text|看看basename命令：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1530627|NAME="$(basename+/foo/fizzbuzz.bar+.bar)"|code-block|syntax|javascript|1530628|指示它删除后缀.bar，结果为NAME=fizzbuzz|offset|length|style|CODE|1530629|entityMap^0|0|0|7|4|F|D|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

look at the basename command:
<pre><code>NAME=&quot;$(basename /foo/fizzbuzz.bar .bar)&quot;
</code></pre>
instructs it to remove the suffix <code>.bar</code>, results in <code>NAME=fizzbuzz</code>

blocks|key|1745585|text|perl+-pe+'s/\..*$//;s{%5E.*/}{}'|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1745586|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

blocks|key|1745630|text|basename和dirname函数就是您想要的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1745631|mystring=/foo/fizzbuzz.bar
echo+basename:+$(basename+"${mystring}")
echo+basename+%2B+remove+.bar:+$(basename+"${mystring}"+.bar)
echo+dirname:+$(dirname+"${mystring}")|code-block|syntax|javascript|1745632|具有输出：|1745633|basename:+fizzbuzz.bar
basename+%2B+remove+.bar:+fizzbuzz
dirname:+/foo|1745634|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|$]]

The basename and dirname functions are what you're after:

<pre><code>mystring=/foo/fizzbuzz.bar
echo basename: $(basename "${mystring}")
echo basename + remove .bar: $(basename "${mystring}" .bar)
echo dirname: $(dirname "${mystring}")
</code></pre>

Has output:

<pre><code>basename: fizzbuzz.bar
basename + remove .bar: fizzbuzz
dirname: /foo
</code></pre>

blocks|key|1745662|text|下面是如何在Bash中使用#和%25运算符完成此操作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1745663|$+x="/foo/fizzbuzz.bar"
$+y=${x%25.bar}
$+echo+${y##*/}
fizzbuzz|code-block|syntax|javascript|1745664|也可以使用${x%25.*}来删除点之后的所有内容，或者使用${x%25%25.*}来删除第一个点之后的所有内容。|offset|length|style|CODE|1745665|示例：|1745666|$+x="/foo/fizzbuzz.bar.quux"
$+y=${x%25.*}
$+echo+$y
/foo/fizzbuzz.bar
$+y=${x%25%25.*}
$+echo+$y
/foo/fizzbuzz|1745667|文档可以在Bash+manual中找到。查找${parameter%25word}和${parameter%25%25word}尾部部分匹配部分。|1745668|entityMap|0|LINK|mutability|MUTABLE|url|https://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html#Shell-Parameter-Expansion^0|0|0|5|7|S|8|0|0|0|M|H|14|I|5|B|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@$I|13|J|14|K|L]|$I|15|J|16|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|17|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|18|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|19|8|@$I|1A|J|1B|K|L]|$I|1C|J|1D|K|L]]|9|@$I|1E|J|1F|1|1G]]|A|$]]|$1|S|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

Here's how to do it with the # and % operators in Bash.

<pre><code>$ x="/foo/fizzbuzz.bar"
$ y=${x%.bar}
$ echo ${y##*/}
fizzbuzz
</code></pre>

<code>${x%.bar}</code> could also be <code>${x%.*}</code> to remove everything after a dot or <code>${x%%.*}</code> to remove everything after the first dot.

Example:

<pre><code>$ x="/foo/fizzbuzz.bar.quux"
$ y=${x%.*}
$ echo $y
/foo/fizzbuzz.bar
$ y=${x%%.*}
$ echo $y
/foo/fizzbuzz
</code></pre>

Documentation can be found in the <a href="https://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html#Shell-Parameter-Expansion" rel="noreferrer">Bash manual</a>. Look for <code>${parameter%word}</code> and <code>${parameter%%word}</code> trailing portion matching section.

blocks|key|1530759|text|使用basename假设您知道文件扩展名是什么，不是吗？|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1530760|我认为，各种正则表达式建议不能处理包含多个“”的文件名。|1530761|下面的代码似乎可以处理两个点。哦，还有包含"/“本身的文件名(只是为了好玩)|1530762|用Pascal的话来说，“对不起，这个脚本太长了，我没有时间把它变短。”|1530763|++#!/usr/bin/perl
++$fullname+=+$ARGV[0];
++($path,$name)+=+$fullname+=~+/%5E(.*[%5E\\]\/)*(.*)$/;
++($basename,$extension)+=+$name+=~+/%5E(.*)(\.[%5E.]*)$/;
++print+$basename+.+"\n";|code-block|syntax|javascript|1530764|entityMap^0|2|8|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|W|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|X|8|@]|D|@]|E|$]]|$1|L|3|M|5|N|7|Y|8|@]|D|@]|E|$O|P]]|$1|Q|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|R|$]]

Using <code>basename</code> assumes that you know what the file extension is, doesn't it?

And I believe that the various regular expression suggestions don't cope with a filename containing more than one "."

The following seems to cope with double dots. Oh, and filenames that contain a "/" themselves (just for kicks)

To paraphrase Pascal, "Sorry this script is so long. I didn't have time to make it shorter"

<pre><code>
 #!/usr/bin/perl
 $fullname = $ARGV[0];
 ($path,$name) = $fullname =~ /^(.*[^\\]\/)*(.*)$/;
 ($basename,$extension) = $name =~ /^(.*)(\.[^.]*)$/;
 print $basename . "\n";
</code> </pre>

blocks|key|1530831|text|basename完成此操作，删除路径。如果给定后缀并且后缀与文件的后缀匹配，它还会删除后缀，但您需要知道要为命令指定的后缀。否则，您可以使用mv并以其他方式确定新名称应该是什么。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1530832|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|$]]

The basename does that, removes the path. It will also remove the suffix if given and if it matches the suffix of the file but you would need to know the suffix to give to the command. Otherwise you can use mv and figure out what the new name should be some other way.

blocks|key|1745822|text|注意推荐的perl解决方案:它删除了第一个点之后的所有内容。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1745823|$+echo+some.file.with.dots+%7C+perl+-pe+'s/\..*$//;s{%5E.*/}{}'
some|code-block|syntax|javascript|1745824|如果你想用perl来做这件事，这是可行的：|1745825|$+echo+some.file.with.dots+%7C+perl+-pe+'s/(.*)\..*$/$1/;s{%5E.*/}{}'
some.file.with|1745826|但是，如果您使用的是Bash，使用y=${x%25.*}+(或者basename+"$x"+.ext，如果您知道这个扩展)的解决方案要简单得多。|offset|length|style|CODE|1745827|entityMap^0|0|0|0|0|H|9|U|I|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]|$M|Z|N|10|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|R|$]]

Beware of the suggested perl solution: it removes anything after the first dot.

<pre><code>$ echo some.file.with.dots | perl -pe 's/\..*$//;s{^.*/}{}'
some
</code></pre>

If you want to do it with perl, this works:

<pre><code>$ echo some.file.with.dots | perl -pe 's/(.*)\..*$/$1/;s{^.*/}{}'
some.file.with
</code></pre>

But if you are using Bash, the solutions with <code>y=${x%.*}</code> (or <code>basename "$x" .ext</code> if you know the extension) are much simpler.

blocks|key|1530916|text|通过使用basename，我使用了以下方法来实现此目的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1530917|for+file+in+*;+do
++++ext=${file##*.}
++++fname=`basename+$file+$ext`

++++#+Do+things+with+$fname
done;|code-block|syntax|javascript|1530918|这不需要文件扩展名的先验知识，即使您的文件名中有点(在其扩展名前面)，它也可以工作；它确实需要程序basename，但这是GNU核心的一部分，所以它应该随任何发行版一起提供。|offset|length|style|CODE|1530919|entityMap^0|0|0|1D|8|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Using basename I used the following to achieve this:

<pre><code>for file in *; do
 ext=${file##*.}
 fname=`basename $file $ext`

 # Do things with $fname
done;
</code></pre>

This requires no a priori knowledge of the file extension and works even when you have a filename that has dots in it's filename (in front of it's extension); it does require the program <code>basename</code> though, but this is part of the GNU coreutils so it should ship with any distro.

blocks|key|1530950|text|将排名靠前的答案与排名靠前的答案组合，得到不含完整路径的文件名：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1530951|$+x="/foo/fizzbuzz.bar.quux"
$+y=(`basename+${x%25%25.*}`)
$+echo+$y
fizzbuzz|code-block|syntax|javascript|1530952|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|$]]

Combining the top-rated answer with the second-top-rated answer to get the filename without the full path: 

<pre><code>$ x="/foo/fizzbuzz.bar.quux"
$ y=(`basename ${x%%.*}`)
$ echo $y
fizzbuzz
</code></pre>

blocks|key|1530959|text|您可以使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1530960|mv+*<PATTERN>.jar+"$(basename+*<PATTERN>.jar+<PATTERN>.jar).jar"|code-block|syntax|javascript|1530961|例如：-我想从我的文件名中删除-SNAPSHOT。对于下面使用的命令|offset|length|style|CODE|1530962|+mv+*-SNAPSHOT.jar+"$(basename+*-SNAPSHOT.jar+-SNAPSHOT.jar).jar"|1530963|entityMap^0|0|0|F|9|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

You can use
<pre><code>mv *&lt;PATTERN&gt;.jar &quot;$(basename *&lt;PATTERN&gt;.jar &lt;PATTERN&gt;.jar).jar&quot;
</code></pre>
For e.g:- I wanted to remove <code>-SNAPSHOT</code> from my file name. For that used below command
<pre><code> mv *-SNAPSHOT.jar &quot;$(basename *-SNAPSHOT.jar -SNAPSHOT.jar).jar&quot;
</code></pre>

Given a string file path such as <code>/foo/fizzbuzz.bar</code>, how would I use bash to extract just the <code>fizzbuzz</code> portion of said string?

How do I remove the file suffix and path portion from a path string in Bash?

给定一个字符串文件路径，比如/foo/fizzbuzz.bar，我该如何使用bash来提取字符串的fizzbuzz部分呢？

问如何在Bash中删除路径字符串中的文件后缀和路径部分？
EN

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问如何在Bash中删除路径字符串中的文件后缀和路径部分？
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