blocks|key|254324|text|一个简单的正则表达式替换会起作用吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|254325|比如：|254326|scala>+"package.file.java".replaceAll("(\\.[%5E\\.]*$)",+".rb")+
scala>+"package.file.rb"|code-block|syntax|javascript|254327|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|$]]

Will a simple regex replacement do the trick?

Like:

<pre><code>scala&gt; "package.file.java".replaceAll("(\\.[^\\.]*$)", ".rb") 
scala&gt; "package.file.rb"
</code></pre>

blocks|key|254337|text|您可以始终在replaceAll上使用java.lang.String方法|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|254338|scala>+"foo.bar.coffee".replaceAll("\\.[%5E.]*$",+".js")
res11:+java.lang.String+=+foo.bar.js|code-block|syntax|javascript|254339|它较短，但可读性较低。|254340|entityMap^0|6|A|J|G|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

You could always use the <code>replaceAll</code> method on <code>java.lang.String</code>

<pre><code>scala&gt; "foo.bar.coffee".replaceAll("\\.[^.]*$", ".js")
res11: java.lang.String = foo.bar.js
</code></pre>

It's shorter but less readable.

blocks|key|254383|text|恐怕你实际上得花更长的时间去做可能是最明智、最有力的事情：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|254384|scala>+"oops".split('.').init+%2B%2B+Seq("js")+mkString+"."++
res0:+String+=+js|code-block|syntax|javascript|254385|有点出乎意料的失去你的文件的名称(至少如果你是一个最终用户)！|254386|让我们试试regex：|254387|scala>+"oops".replaceAll("\\.[%5E.]*$",+".js")
res1:+java.lang.String+=+oops|254388|没有丢失文件名，但也没有扩展名。Ack。|254389|让我们修复它：|254390|def+extensor(orig:+String,+ext:+String)+=+(orig.split('.')+match+{
++case+xs+@+Array(x)+=>+xs
++case+y+=>+y.init
})+:%2B+"js"+mkString+"."

scala>+extensor("oops","js")
res2:+String+=+oops.js

scala>+extensor("oops.txt","js")
res3:+String+=+oops.js

scala>+extensor("oops...um...","js")
res4:+String+=+oops...js|254391|或使用regex：|254392|scala>+"oops".replaceAll("\\.[%5E.]*$",+"")+%2B+".js"+
res5:+java.lang.String+=+oops.js

scala>+"oops.txt".replaceAll("\\.[%5E.]*$",+"")+%2B+".js"
res6:+java.lang.String+=+oops.js

scala>+"oops...um...".replaceAll("\\.[%5E.]*$",+"")+%2B+".js"
res7:+java.lang.String+=+oops...um...js|254393|(请注意角落中文件名以句点结尾的不同行为。)|254394|entityMap^0|0|0|0|0|0|0|0|0|0|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]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|13|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|15|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|18|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1A|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

I'm afraid you actually have to make it longer to do what is probably the most sensible robust thing:

<pre><code>scala&gt; "oops".split('.').init ++ Seq("js") mkString "." 
res0: String = js
</code></pre>

Kinda unexpected to lose the name of your file (at least if you're an end user)!

Let's try regex:

<pre><code>scala&gt; "oops".replaceAll("\\.[^.]*$", ".js")
res1: java.lang.String = oops
</code></pre>

Didn't lose the file name, but there's no extension either. Ack.

Let's fix it:

<pre><code>def extensor(orig: String, ext: String) = (orig.split('.') match {
 case xs @ Array(x) =&gt; xs
 case y =&gt; y.init
}) :+ "js" mkString "."

scala&gt; extensor("oops","js")
res2: String = oops.js

scala&gt; extensor("oops.txt","js")
res3: String = oops.js

scala&gt; extensor("oops...um...","js")
res4: String = oops...js
</code></pre>

Or with regex:

<pre><code>scala&gt; "oops".replaceAll("\\.[^.]*$", "") + ".js" 
res5: java.lang.String = oops.js

scala&gt; "oops.txt".replaceAll("\\.[^.]*$", "") + ".js"
res6: java.lang.String = oops.js

scala&gt; "oops...um...".replaceAll("\\.[^.]*$", "") + ".js"
res7: java.lang.String = oops...um...js
</code></pre>

(Note the different behavior on the corner case where the filename ends with periods.)

blocks|key|275058|text|lastIndexOf怎么了？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|275059|fileName.take(1+%2B+fileName.lastIndexOf("."))+%2B+"js"|code-block|syntax|javascript|275060|当然，如果您想在不包含任何点的情况下保留fileName，则需要做更多的工作|275061|(if+(fileName.contains('.'))+fileName.take(fileName.lastIndexOf("."))+
else+fileName)+%2B+".js"|275062|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|$]]

What's wrong with lastIndexOf?

<pre><code>fileName.take(1 + fileName.lastIndexOf(".")) + "js"
</code></pre>

Of course if you want to keep the fileName when it doesn't contain any dot, you need to do a little bit more

<pre><code>(if (fileName.contains('.')) fileName.take(fileName.lastIndexOf(".")) 
else fileName) + ".js"
</code></pre>

blocks|key|2395038|text|所以，我会在这里追求速度。碰巧，substring是常数时间，因为它根本不复制字符串。所以,|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2395039|((index:+Int)+=>+(
)+%2B+".js")(fileName+lastIndexOf+'.')|code-block|syntax|javascript|2395040|这使用了一个闭包，这会使它慢一点。更快：|2395041|def+addJS(fileName:+String)+=+{
++++def+addJSAt(index:+Int)+=+(
++++++++if+(index+>=+0)+fileName+substring+(0,+index)+
++++++++else+fileName
++++)+%2B+".js"

++++addJSAt(fileName+lastIndexOf+'.')
}|2395042|编辑：现在确实复制了substring上的字符串。|BOLD|2395043|entityMap^0|G|9|0|0|0|0|0|3|A|9|0^^$0|@$1|2|3|4|5|6|7|T|8|@$9|U|A|V|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|W|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|X|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|Z|8|@$9|10|A|11|B|Q]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|R|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|S|$]]

So, I'll go for speed here. As it happens, <code>substring</code> is constant time because it simply does not copy the string. So,

<pre><code>((index: Int) =&gt; (
) + ".js")(fileName lastIndexOf '.')
</code></pre>

This uses a closure, which will slow it down a bit. Faster:

<pre><code>def addJS(fileName: String) = {
 def addJSAt(index: Int) = (
 if (index &gt;= 0) fileName substring (0, index) 
 else fileName
 ) + ".js"

 addJSAt(fileName lastIndexOf '.')
}
</code></pre>

EDIT: As it happens, Java now does copy the string on <code>substring</code>.

blocks|key|275158|text|使用lastIndexOf非常容易，并且使用包含多个点的文件名。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|275159|def+getFileNameWithoutExtension(fileName:+String):+String+=+{
++fileName.dropRight(fileName.length+-+fileName.lastIndexOf("."))
}

val+fileName+=+"foo.bar.coffee"

getFileNameWithoutExtension(fileName)+%2B+".js"|code-block|syntax|javascript|275160|结果是foo.bar.js|275161|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Very easy with lastIndexOf, and work with file name that contains more than one dot

<pre><code>def getFileNameWithoutExtension(fileName: String): String = {
 fileName.dropRight(fileName.length - fileName.lastIndexOf("."))
}

val fileName = "foo.bar.coffee"

getFileNameWithoutExtension(fileName) + ".js"
</code></pre>

Result is foo.bar.js

blocks|key|2395080|text|如果您知道当前的扩展是什么，那么您可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2395081|def+replaceExtension(fileName:+String,+oldExt:+String,+newExt:+String):+String+=
++fileName.stripSuffix(oldExt)+%2B+newExt

//+Be+sure+to+use+`.`+when+calling:
replaceExtension(fileName,+".javascript",+".js")|code-block|syntax|javascript|2395082|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|$]]

If you know what the current extension is, then you could do this:

<pre><code>def replaceExtension(fileName: String, oldExt: String, newExt: String): String =
 fileName.stripSuffix(oldExt) + newExt

// Be sure to use `.` when calling:
replaceExtension(fileName, ".javascript", ".js")
</code></pre>

Here is a version I wrote using split:

<pre><code>fileName.split('.').init ++ Seq("js") mkString "."
</code></pre>

This transforms e.g. <code>foo.bar.coffee</code> into <code>foo.bar.js</code>.

What I like:

<ul>
<li>it works</li>
<li>it doesn't rely on things like <code>indexOf()</code></li>
<li>it feels functional ;)</li>
</ul>

What I don't like:

<ul>
<li>it's not as short as I would hope</li>
<li>it might confuse some readers</li>
</ul>

How can I write an even simpler / straightforward version?

UPDATE: Great answers below! In short:

<ul>
<li>seems like my original approach above was not bad although it doesn't cover some corner cases, but that's fixable with a longer expression if you need to cover those</li>
<li>another, slightly shorter approach uses regexps, which will be more or less readable depending on your regexp background</li>
<li>a slightly shorter syntax for the original approach (corner cases not covered) reads:

<code>fileName.split('.').init :+ "js" mkString "."</code></li>
</ul>

Scala expression to replace a file extension in a string

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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下面是我用split编写的一个版本：fileName.split('.').init ++ Seq("js") mkString "."这可以将例如foo.bar.coffee转换为foo.bar.js。我喜欢的是：indexOf()it it works它不依赖于像这样的东西我不喜欢的是：，，它不像我希望的那么短，它...

问Scala表达式替换字符串中的文件扩展名
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