方案中的正则表达式和转义字符
在方案中，
有 "hello hellu-#\"hella.helloo,hallo#\return#\"" 字符串
我想将它们列为 ("hello" "hello" "hella" "helloo " "hallo")
用空格、连字符、双引号、点、逗号、return 分隔
我尝试过
(regexp-split #rx"( +)|(#\-)|(#\")|(#\.)|(,)|(#\return)" string)
，但 #\- , #\. 出错，
有任何提示或解决方案吗？
谢谢
in scheme,
there is "hello hellu-#\"hella.helloo,hallo#\return#\"" string
I want to list them as ("hello" "hellu" "hella" "helloo" "hallo")
separate by space, hyphen, double quote, dot, comma, return
I tried
(regexp-split #rx"( +)|(#\-)|(#\")|(#\.)|(,)|(#\return)" string)
but #\- , #\. make error
any hint or solution?
thanks
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看起来您将字符语法 (
#\foo) 与字符串语法混淆了，并且您在字符串和正则表达式中都这样做了。所以我的猜测是，您要拆分的字符串实际上是：其中
\"代表双引号字符，\n代表换行符。如果这是这种情况，那么（再次，这是猜测你的意图）正则表达式应该是：但这也不起作用，因为
\-和\.是无效的转义符（Racket使用类似 C 的转义符），因此将其更改为：这也不起作用，因为
.在正则表达式中具有通常的“任何字符”含义 - 因此您想用反斜杠对其进行转义，与许多其他字符串语法一样，您会得到一个反斜杠。通过用反斜杠转义它，所以现在我们有了一个最终接近工作版本的版本：首先，可以大大改进正则表达式：拆分不需要括号：
然后，而不是使用一堆单字符和
|s，您可以只使用“字符范围”：请注意，重要的是
-是该范围中的第一个（或最后一个）字符，因此它不会具有一系列字符的通常含义。接下来，您似乎确实希望将此类字符的任何序列用作分隔符，这将避免结果中的一些空字符串：在这种情况下，您也可以将空格放入范围中（小心将其放在
-之后，正如我上面所解释的）。我们现在得到：最后，您可能想要删除最后一个空字符串。从技术上讲，它应该在那里，因为在字符串末尾之前有一系列匹配的字符。 Racket 中解决这个问题的一个简单方法是使用互补的 regexp-match* ，它返回匹配列表，而不是拆分匹配列表：
这显然是错误的，因为它给你的是分隔符而不是它们之间的内容。但由于这个正则表达式是一个字符范围，所以很容易解决——只需否定字符范围，你就会得到你想要的：
It looks like you're confusing the syntax for characters (
#\foo) with the syntax for strings, and you do that in both the string and the regexp. So my guess is that the string that you want to split is actually:where
\"stands for a double quote character, and\nfor a newline. If this is the case, then (again, this is guessing your intention) the regexp should be:But that doesn't work either, since
\-and\.are invalid escapes (Racket uses C-like escapes), so change that to:This doesn't work either, since
.has the usual "any char" meaning in a regexp -- so you want to escape it with a backslash. As with many other string syntaxes, you get a backslash by escaping it with a backslash, so now we have a version that is finally close to a working one:First, the regexp can be improved considerably: the parens are not needed for splitting:
Then, instead of using a bunch of single-characters with
|s, you can just use a "character range":Note that it's important that the
-is the first (or last) character in the range, so it will not have the usual meaning of a range of characters. Next, it seems that you really want any sequence of such characters to be used as a separator, which will avoid some of those empty strings in the result:and in this case you can just as well throw the space into the range too (carefully putting it after the
-, as I explained above). We now get:And finally you'd probably want to get rid of that last empty string. Technically, it should be there, since there is a sequence of matching characters before the end of the string. An easy way in Racket around this is to use the complementary
regexp-match*which returns the list of matches rather than splitting on the list of matches:This is obviously broken, since it gives you the separators rather than what's between them. But since this regexp is a range of characters, it is easy to resolve -- simply negate the character range, and you get what you want: