blocks|key|3100868|text|您的问题不清楚，但由于您计划接受来自web表单的路径名(?)你可能应该阻止重命名某些东西的尝试；"C:\Program+Files“。这意味着您需要规范化路径名以消除“。和"..“在你做访问检查之前。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3100869|鉴于此，我不会尝试删除非法字符。相反，我会使用"new+File(str).getCanonicalFile()“来生成规范的路径，然后检查它们是否满足沙箱限制，最后使用"File.exists()”、"File.isFile()“等来检查源和目标是不是相同的文件系统对象。我会通过尝试执行操作并捕获异常来处理非法字符。|3100870|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

It is not clear from your question, but since you are planning to accept pathnames from a web form (?) you probably ought block attempts renaming certain things; e.g. "C:\Program Files". This implies that you need to canonicalize the pathnames to eliminate "." and ".." before you make your access checks.

Given that, I wouldn't attempt to remove illegal characters. Instead, I'd use "new File(str).getCanonicalFile()" to produce the canonical paths, next check that they satisfy your sandboxing restrictions, and finally use "File.exists()", "File.isFile()", etc to check that the source and destination are kosher, and are not the same file system object. I'd deal with illegal characters by attempting to do the operations and catching the exceptions.

blocks|key|3100894|text|正如其他地方所建议的，这通常不是您想要做的事情。通常，最好使用安全方法(如File.createTempFile()+)创建临时文件。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3100895|您不应该使用白名单这样做，并且只保留“好”字符。如果文件只由中文字符组成，那么您将剥离其中的所有内容。由于这个原因，我们不能使用白名单，我们必须使用黑名单。|3100896|Linux几乎允许任何令人痛苦的事情。我只会将Linux限制在与Windows相同的列表中，这样就可以省去将来的麻烦。|3100897|在Windows上使用这个C#代码段，我生成了一个在Windows上无效的字符列表。这个列表中的字符比你想象的要多(41个)，所以我不建议创建你自己的列表。|3100898|++++++++foreach+(char+c+in+new+string(Path.GetInvalidFileNameChars()))
++++++++{
++++++++++++Console.Write((int)c);
++++++++++++Console.Write(",");
++++++++}|code-block|syntax|javascript|3100899|这是一个简单的Java类，它“清理”了一个文件名。|3100900|public+class+FileNameCleaner+{
final+static+int[]+illegalChars+=+{34,+60,+62,+124,+0,+1,+2,+3,+4,+5,+6,+7,+8,+9,+10,+11,+12,+13,+14,+15,+16,+17,+18,+19,+20,+21,+22,+23,+24,+25,+26,+27,+28,+29,+30,+31,+58,+42,+63,+92,+47};
static+{
++++Arrays.sort(illegalChars);
}
public+static+String+cleanFileName(String+badFileName)+{
++++StringBuilder+cleanName+=+new+StringBuilder();
++++for+(int+i+=+0;+i+<+badFileName.length();+i%2B%2B)+{
++++++++int+c+=+(int)badFileName.charAt(i);
++++++++if+(Arrays.binarySearch(illegalChars,+c)+<+0)+{
++++++++++++cleanName.append((char)c);
++++++++}
++++}
++++return+cleanName.toString();
}
}|3100901|编辑:正如Stephen建议的那样，您可能还应该验证这些文件访问是否只发生在您允许的目录中。|3100902|下面的答案提供了在Java中建立自定义安全上下文，然后在“沙箱”中执行代码的示例代码。|3100903|How+do+you+create+a+secure+JEXL+(scripting)+sandbox?|offset|length|3100904|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/6551073/how-do-you-create-a-secure-jexl-scripting-sandbox^0|0|0|0|0|0|0|0|0|0|0|1G|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|17|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|18|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|19|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|1A|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|1B|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|1C|8|@]|9|@]|A|$K|L]]|$1|Q|3|R|5|6|7|1D|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1E|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1F|8|@]|9|@$W|1G|X|1H|1|1I]]|A|$]]|$1|Y|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]]|Z|$10|$5|11|12|13|A|$14|15]]]]

As suggested elsewhere, this is not usually what you want to do. It is usually best to create a temporary file using a secure method such as File.createTempFile().

You should not do this with a whitelist and only keep 'good' characters. If the file is made up of only Chinese characters then you will strip everything out of it. We can't use a whitelist for this reason, we have to use a blacklist.

Linux pretty much allows anything which can be a real pain. I would just limit Linux to the same list that you limit Windows to so you save yourself headaches in the future.

Using this C# snippet on Windows I produced a list of characters that are not valid on Windows. There are quite a few more characters in this list than you may think (41) so I wouldn't recommend trying to create your own list.

<pre><code> foreach (char c in new string(Path.GetInvalidFileNameChars()))
 {
 Console.Write((int)c);
 Console.Write(",");
 }
</code></pre>

Here is a simple Java class which 'cleans' a file name.

<pre><code>public class FileNameCleaner {
final static int[] illegalChars = {34, 60, 62, 124, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 58, 42, 63, 92, 47};
static {
 Arrays.sort(illegalChars);
}
public static String cleanFileName(String badFileName) {
 StringBuilder cleanName = new StringBuilder();
 for (int i = 0; i &lt; badFileName.length(); i++) {
 int c = (int)badFileName.charAt(i);
 if (Arrays.binarySearch(illegalChars, c) &lt; 0) {
 cleanName.append((char)c);
 }
 }
 return cleanName.toString();
}
}
</code></pre>

EDIT:
As Stephen suggested you probably also should verify that these file accesses only occur within the directory you allow.

The following answer has sample code for establishing a custom security context in Java and then executing code in that 'sandbox'.

<a href="https://stackoverflow.com/questions/6551073/how-do-you-create-a-secure-jexl-scripting-sandbox">How do you create a secure JEXL (scripting) sandbox?</a>

blocks|key|3313761|text|有一个非常好的内置Java解决方案--+Character.isXxx().|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|3313762|试试Character.isJavaIdentifierPart(c)|CODE|3313763|String+name+=+"name.é%2B!@#$%25%5E&*(){}][/=?%2B-_\\%7C;:`~!'\",<>";
StringBuilder+filename+=+new+StringBuilder();

for+(char+c+:+name.toCharArray())+{
++if+(c=='.'+%7C%7C+Character.isJavaIdentifierPart(c))+{
++++filename.append(c);
++}
}|code-block|syntax|javascript|3313764|结果是"name.é$_“。|3313765|entityMap^0|K|H|0|2|X|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@$9|V|A|W|B|H]]|D|@]|E|$]]|$1|I|3|J|5|K|7|X|8|@]|D|@]|E|$L|M]]|$1|N|3|O|5|6|7|Y|8|@]|D|@]|E|$]]|$1|P|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|Q|$]]

There's a pretty good built-in Java solution - Character.isXxx().

Try <code>Character.isJavaIdentifierPart(c)</code>:

<pre><code>String name = "name.é+!@#$%^&amp;*(){}][/=?+-_\\|;:`~!'\",&lt;&gt;";
StringBuilder filename = new StringBuilder();

for (char c : name.toCharArray()) {
 if (c=='.' || Character.isJavaIdentifierPart(c)) {
 filename.append(c);
 }
}
</code></pre>

Result is "name.é$_".

blocks|key|3313829|text|或者干脆这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3313830|String+filename+=+"A20/B22b#öA\\BC#Ä$%25ld_ma.la.xps";
String+sane+=+filename.replaceAll("[%5Ea-zA-Z0-9\\._]%2B",+"_");|code-block|syntax|javascript|3313831|结果：A20_B22b_A_BC_ld_ma.la.xps|offset|length|style|CODE|3313832|解释：|3313833|[a-zA-Z0-9\\._]匹配a-z小写或大写字母、数字、点和下划线|3313834|[%5Ea-zA-Z0-9\\._]是相反的。即与第一个表达式不匹配的所有字符|3313835|[%5Ea-zA-Z0-9\\._]%2B是与第一个表达式不匹配的字符序列|3313836|所以每个字符序列不包含a-z，0-9或。_将被替换。|3313837|entityMap^0|0|0|3|Q|0|0|0|F|0|0|G|0|0|H|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@$I|11|J|12|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|14|8|@$I|15|J|16|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|17|8|@$I|18|J|19|K|L]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1A|8|@$I|1B|J|1C|K|L]]|9|@]|A|$]]|$1|U|3|V|5|6|7|1D|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|X|$]]

or just do this:

<pre><code>String filename = "A20/B22b#öA\\BC#Ä$%ld_ma.la.xps";
String sane = filename.replaceAll("[^a-zA-Z0-9\\._]+", "_");
</code></pre>

Result: <code>A20_B22b_A_BC_ld_ma.la.xps</code>

Explanation:

<code>[a-zA-Z0-9\\._]</code> matches a letter from a-z lower or uppercase, numbers, dots and underscores

<code>[^a-zA-Z0-9\\._]</code> is the inverse. i.e. all characters which do not match the first expression

<code>[^a-zA-Z0-9\\._]+</code> is a sequence of characters which do not match the first expression

So every sequence of characters which does not consist of characters from a-z, 0-9 or . _ will be replaced.

blocks|key|3101034|text|下面是我使用的代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3101035|public+static+String+sanitizeName(+String+name+)+{
++++if(+null+==+name+)+{
++++++++return+"";
++++}

++++if(+SystemUtils.IS_OS_LINUX+)+{
++++++++return+name.replaceAll(+"[\u0000/]%2B",+""+).trim();
++++}

++++return+name.replaceAll(+"[\u0000-\u001f<>:\"/\\\\%7C?*\u007f]%2B",+""+).trim();
}|code-block|syntax|javascript|3101036|SystemUtils来自Apache+commons-lang3|offset|length|style|CODE|3101037|entityMap|0|LINK|mutability|MUTABLE|url|http://commons.apache.org/proper/commons-lang/^0|0|0|0|B|D|K|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]]|9|@$I|Z|J|10|1|11]]|A|$]]|$1|M|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

Here is the code I use:

<pre><code>public static String sanitizeName( String name ) {
 if( null == name ) {
 return "";
 }

 if( SystemUtils.IS_OS_LINUX ) {
 return name.replaceAll( "[\u0000/]+", "" ).trim();
 }

 return name.replaceAll( "[\u0000-\u001f&lt;&gt;:\"/\\\\|?*\u007f]+", "" ).trim();
}
</code></pre>

<code>SystemUtils</code> is from <a href="http://commons.apache.org/proper/commons-lang/" rel="nofollow noreferrer">Apache commons-lang3</a>

blocks|key|3313882|text|这是基于Sarel+Botha接受的答案，只要您不遇到Basic+Multilingual+Plane之外的任何字符，它就可以正常工作。如果你需要完全的Unicode支持(谁不需要呢？)请改用以下代码，它是Unicode安全的：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3313883|public+class+FileNameCleaner+{
++final+static+int[]+illegalChars+=+{34,+60,+62,+124,+0,+1,+2,+3,+4,+5,+6,+7,+8,+9,+10,+11,+12,+13,+14,+15,+16,+17,+18,+19,+20,+21,+22,+23,+24,+25,+26,+27,+28,+29,+30,+31,+58,+42,+63,+92,+47};

++static+{
++++Arrays.sort(illegalChars);
++}

++public+static+String+cleanFileName(String+badFileName)+{
++++StringBuilder+cleanName+=+new+StringBuilder();
++++int+len+=+badFileName.codePointCount(0,+badFileName.length());
++++for+(int+i=0;+i<len;+i%2B%2B)+{
++++++int+c+=+badFileName.codePointAt(i);
++++++if+(Arrays.binarySearch(illegalChars,+c)+<+0)+{
++++++++cleanName.appendCodePoint(c);
++++++}
++++}
++++return+cleanName.toString();
++}
}|code-block|syntax|javascript|3313884|此处的关键更改：|3313885|3313886|使用codePointCount+i.c.w。length|unordered-list-item|style|CODE|3313887|use而不是charAt|3313888|use+appendCodePoint需要将chars转换为ints。事实上，您永远不应该处理chars，因为对于BMP以外的任何内容，它们基本上都是损坏的。|3313889|3313890|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/35264/sarel-botha|1|http://en.wikipedia.org/wiki/Plane_(Unicode)#Basic_Multilingual_Plane)|2|http://docs.oracle.com/javase/8/docs/api/java/lang/String.html#codePointCount(int,%2520int)|3|http://docs.oracle.com/javase/8/docs/api/java/lang/StringBuilder.html#appendCodePoint-int-^0|4|B|0|R|O|1|0|0|0|0|N|6|2|E|2|0|6|6|0|M|4|U|3|1B|4|4|F|3|0|0^^$0|@$1|2|3|4|5|6|7|19|8|@]|9|@$A|1A|B|1B|1|1C]|$A|1D|B|1E|1|1F]]|C|$]]|$1|D|3|E|5|F|7|1G|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|1H|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|1I|8|@]|9|@]|C|$]]|$1|L|3|M|5|N|7|1J|8|@$A|1K|B|1L|O|P]]|9|@$A|1M|B|1N|1|1O]]|C|$]]|$1|Q|3|R|5|N|7|1P|8|@$A|1Q|B|1R|O|P]]|9|@]|C|$]]|$1|S|3|T|5|N|7|1S|8|@$A|1T|B|1U|O|P]|$A|1V|B|1W|O|P]|$A|1X|B|1Y|O|P]]|9|@$A|1Z|B|20|1|21]]|C|$]]|$1|U|3|-4|5|6|7|22|8|@]|9|@]|C|$]]|$1|V|3|-4|5|6|7|23|8|@]|9|@]|C|$]]]|W|$X|$5|Y|Z|10|C|$11|12]]|13|$5|Y|Z|10|C|$11|14]]|15|$5|Y|Z|10|C|$11|16]]|17|$5|Y|Z|10|C|$11|18]]]]

This is based on the accepted answer by <a href="https://stackoverflow.com/users/35264/sarel-botha">Sarel Botha</a> which works fine as long as you don't encounter any characters outside of the <a href="http://en.wikipedia.org/wiki/Plane_(Unicode)#Basic_Multilingual_Plane" rel="noreferrer">Basic Multilingual Plane</a>. If you need full Unicode support (and who doesn't?) use this code instead which is Unicode safe:

<pre><code>public class FileNameCleaner {
 final static int[] illegalChars = {34, 60, 62, 124, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 58, 42, 63, 92, 47};

 static {
 Arrays.sort(illegalChars);
 }

 public static String cleanFileName(String badFileName) {
 StringBuilder cleanName = new StringBuilder();
 int len = badFileName.codePointCount(0, badFileName.length());
 for (int i=0; i&lt;len; i++) {
 int c = badFileName.codePointAt(i);
 if (Arrays.binarySearch(illegalChars, c) &lt; 0) {
 cleanName.appendCodePoint(c);
 }
 }
 return cleanName.toString();
 }
}
</code></pre>

Key changes here:

<ul>
<li>Use <a href="http://docs.oracle.com/javase/8/docs/api/java/lang/String.html#codePointCount(int,%20int)" rel="noreferrer">codePointCount</a> i.c.w. <code>length</code> instead of just <code>length</code></li>
<li>use <a href="http://docs.oracle.com/javase/8/docs/api/java/lang/String.html#codePointAt(int)" rel="noreferrer">codePointAt</a> instead of <code>charAt</code></li>
<li>use <a href="http://docs.oracle.com/javase/8/docs/api/java/lang/StringBuilder.html#appendCodePoint-int-" rel="noreferrer">appendCodePoint</a> instead of <code>append</code></li>
<li>No need to cast <code>char</code>s to <code>int</code>s. In fact, you should never deal with <code>char</code>s as they are basically broken for anything outside the BMP.</li>
</ul>

blocks|key|3313919|text|如果你想使用更多，比如A-Za-z0-9，那么检查MS+Naming+Conventions，不要忘了过滤掉"...Characters，它的整数表示在从1到31的范围内，..."，就像Aaron+Digulla的例子一样。例如，David+Carboni的代码对于这些字符是不够的。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3313920|包含保留字符列表的摘录：|3313921|3313922|使用当前代码页中的任何字符作为名称，包括Unicode字符和扩展字符集中的字符(128-255)，以下字符除外：|blockquote|3313923|3313924|以下保留字符：|3313925|3313926|3313927|<+(较小的than)|unordered-list-item|style|CODE|3313928|>+(较大的than)|3313929|:+(colon)|3313930|"+)双quote)|3313931|/+(向前slash)|3313932|\+(backslash)|3313933|%7C+(垂直条或pipe)|3313934|?+(问题mark)|3313935|*+(asterisk)|3313936|Integer值零，有时被称为ASCII+character.|3313937|Characters，其整数表示在从1到31的范围内，除了允许这些字符的备用数据流之外。有关文件流的更多信息，请参见file+streams。|3313938|目标文件系统不允许的任何其他字符。|3313939|3313940|entityMap|0|LINK|mutability|MUTABLE|url|https://msdn.microsoft.com/en-us/library/windows/desktop/aa365247%2528v=vs.85%2529.aspx#naming_conventions^0|P|L|0|0|0|0|0|0|0|0|0|0|1|0|0|1|0|0|1|0|0|1|0|0|1|0|0|1|0|0|1|0|0|1|0|0|1|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1O|8|@]|9|@$A|1P|B|1Q|1|1R]]|C|$]]|$1|D|3|E|5|6|7|1S|8|@]|9|@]|C|$]]|$1|F|3|-4|5|6|7|1T|8|@]|9|@]|C|$]]|$1|G|3|H|5|I|7|1U|8|@]|9|@]|C|$]]|$1|J|3|-4|5|6|7|1V|8|@]|9|@]|C|$]]|$1|K|3|L|5|6|7|1W|8|@]|9|@]|C|$]]|$1|M|3|-4|5|6|7|1X|8|@]|9|@]|C|$]]|$1|N|3|-4|5|6|7|1Y|8|@]|9|@]|C|$]]|$1|O|3|P|5|Q|7|1Z|8|@$A|20|B|21|R|S]]|9|@]|C|$]]|$1|T|3|U|5|Q|7|22|8|@$A|23|B|24|R|S]]|9|@]|C|$]]|$1|V|3|W|5|Q|7|25|8|@$A|26|B|27|R|S]]|9|@]|C|$]]|$1|X|3|Y|5|Q|7|28|8|@$A|29|B|2A|R|S]]|9|@]|C|$]]|$1|Z|3|10|5|Q|7|2B|8|@$A|2C|B|2D|R|S]]|9|@]|C|$]]|$1|11|3|12|5|Q|7|2E|8|@$A|2F|B|2G|R|S]]|9|@]|C|$]]|$1|13|3|14|5|Q|7|2H|8|@$A|2I|B|2J|R|S]]|9|@]|C|$]]|$1|15|3|16|5|Q|7|2K|8|@$A|2L|B|2M|R|S]]|9|@]|C|$]]|$1|17|3|18|5|Q|7|2N|8|@$A|2O|B|2P|R|S]]|9|@]|C|$]]|$1|19|3|1A|5|Q|7|2Q|8|@]|9|@]|C|$]]|$1|1B|3|1C|5|Q|7|2R|8|@]|9|@]|C|$]]|$1|1D|3|1E|5|Q|7|2S|8|@]|9|@]|C|$]]|$1|1F|3|-4|5|6|7|2T|8|@]|9|@]|C|$]]|$1|1G|3|-4|5|6|7|2U|8|@]|9|@]|C|$]]]|1H|$1I|$5|1J|1K|1L|C|$1M|1N]]]]

If you want to use more than like [A-Za-z0-9], then check <a href="https://msdn.microsoft.com/en-us/library/windows/desktop/aa365247%28v=vs.85%29.aspx#naming_conventions" rel="nofollow noreferrer">MS Naming Conventions</a>, and dont forget to filter out "...Characters whose integer representations are in the range from 1 through 31,...", like the example of Aaron Digulla does. The code e.g. from David Carboni would not be sufficient for these chars.

Excerpt containing the list of reserved characters:

<blockquote>
 Use any character in the current code page for a name, including Unicode characters and characters in the extended character set (128–255), except for the following:
 
 The following reserved characters:
 
 <ul>
 <li><code>&lt;</code> (less than)</li>
 <li><code>&gt;</code> (greater than)</li>
 <li><code>:</code> (colon)</li>
 <li><code>"</code> (double quote)</li>
 <li><code>/</code> (forward slash)</li>
 <li><code>\</code> (backslash)</li>
 <li><code>|</code> (vertical bar or pipe)</li>
 <li><code>?</code> (question mark)</li>
 <li><code>*</code> (asterisk)</li>
 <li>Integer value zero, sometimes referred to as the ASCII NUL character.</li>
 <li>Characters whose integer representations are in the range from 1 through 31, except for alternate data streams where these characters are allowed. For more information about file streams, see File Streams.</li>
 <li>Any other character that the target file system does not allow.</li>
 </ul>
</blockquote>

blocks|key|335620|text|Paths.get(...)对非法字符的位置抛出一个详细的异常。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|335621|public+static+String+removeInvalidChars(final+String+fileName)
{
++try
++{
++++Paths.get(fileName);
++++return+fileName;
++}
++catch+(final+InvalidPathException+e)
++{
++++if+(e.getInput()+!=+null+&&+e.getInput().length()+>+0+&&+e.getIndex()+>=+0)
++++{
++++++final+StringBuilder+stringBuilder+=+new+StringBuilder(e.getInput());
++++++stringBuilder.deleteCharAt(e.getIndex());
++++++return+removeInvalidChars(stringBuilder.toString());
++++}
++++throw+e;
++}
}|code-block|syntax|javascript|335622|entityMap^0|0|E|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

<code>Paths.get(...)</code> throws a detailed exception with the position of the illegal character.

<pre><code>public static String removeInvalidChars(final String fileName)
{
 try
 {
 Paths.get(fileName);
 return fileName;
 }
 catch (final InvalidPathException e)
 {
 if (e.getInput() != null &amp;&amp; e.getInput().length() &gt; 0 &amp;&amp; e.getIndex() &gt;= 0)
 {
 final StringBuilder stringBuilder = new StringBuilder(e.getInput());
 stringBuilder.deleteCharAt(e.getIndex());
 return removeInvalidChars(stringBuilder.toString());
 }
 throw e;
 }
}
</code></pre>

I'm making a cross-platform application that renames files based on data retrieved online. I'd like to sanitize the Strings I took from a web API for the current platform.

I know that different platforms have different file-name requirements, so I was wondering if there's a cross-platform way to do this?

Edit: On Windows platforms you cannot have a question mark '?' in a file name, whereas in Linux, you can. The file names may contain such characters and I would like for the platforms that support those characters to keep them, but otherwise, strip them out.

Also, I would prefer a standard Java solution that doesn't require third-party libraries.

Is there a cross-platform Java method to remove filename special chars?

文件系统

数据流

Windows

Linux

Java

我正在制作一个跨平台的应用程序，它根据在线检索的数据重命名文件。我想要清理我从当前平台的web API中获取的字符串。我知道不同的平台有不同的文件名要求，所以我想知道是否有跨平台的方法来做到这一点？Windows：在平台上不能有问号'?‘在文件名中，而在Linux中，您可以。文件名可能包含这样的字符，我希望支持这些字符...

问有没有一种跨平台的Java方法来删除文件名的特殊字符？
EN

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问有没有一种跨平台的Java方法来删除文件名的特殊字符？
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