Java 正则表达式错误：\( 不是有效字符
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Java regexp error: \( is not a valid character
提问by Marthin
I was using java regexp today and found that you are not allowed to use the following regexp sequence
我今天在用java regexp，发现你不允许使用下面的regexp序列
String pattern = "[a-zA-Z\s\.-\)\(]*";
if I do use it it will fail and tell me that \( is not a valid character.
如果我确实使用它，它会失败并告诉我 \( 不是有效字符。
But if I change the regexp to
但是如果我将正则表达式更改为
String pattern = "[[a-zA-Z\s\.-]|[\(\)]]*";
Then it will work. Is this a bug in the regxp engine or am I not understanding how to work with the engine?
然后它会起作用。这是 regxp 引擎中的错误还是我不了解如何使用引擎？
EDIT:I've had an error in my string: there shouldnt be 2 starting [[, it should be only one. This is now corrected
编辑：我的字符串中有一个错误：不应该有 2 个 [[，它应该只有一个。现在已更正
回答by codaddict
Your regex has two problems.
您的正则表达式有两个问题。
You've not closed the character class.
The
-is acting as a range operator with.on LHS and(on RHS. But(comes before.in unicode, so this results in an invalid range.
你还没有关闭字符类。
将
-充当范围内运营商.在LHS和(在RHS。但在 unicode(之前.出现，所以这会导致无效范围。
To fix problem 1, close the char class or if you meant to not include [in the allowed characters delete one of the [.
要解决问题 1，请关闭 char 类，或者如果您打算不包含[在允许的字符中，请删除[.
To fix problem 2, either escape the -as \\-or move the -to the beginning or to the end of the char class.
要解决问题 2，请转义-as\\-或将 移至-char 类的开头或结尾。
So you can use:
所以你可以使用：
String pattern = "[a-zA-Z\s\.\-\)\(]*";
or
或者
String pattern = "[a-zA-Z\s\.\)\(-]*";
or
或者
String pattern = "[-a-zA-Z\s\.\)\(]*";
回答by Tim
You should only use the dash -at the end of the character class, since it is normally used to show a range (as in a-z). Rearrange it:
您应该只-在字符类的末尾使用破折号，因为它通常用于显示范围（如 中所示a-z）。重新排列：
String pattern = "[[a-zA-Z\s\.\)\(-]*";
Also, I don't think you have to escape (.)characters inside brackets.
另外，我认为您不必对(.)括号内的字符进行转义。
Update: As others pointed out, you must also escape the [in a java regex character class.
更新：正如其他人指出的那样，您还必须转义[java 正则表达式字符类中的 。
回答by Joachim Sauer
The problem here is that \.-\)("\\.-\\)"in a Java string literal) tries to define a range from .to ). Since the Unicode codepoint of .(U+002E) is higher than that of )(U+0029) this is an error.
这里的问题是\.-\)（"\\.-\\)"在 Java 字符串文字中）试图定义从.到的范围)。由于.(U+002E)的 Unicode 代码点高于)(U+0029)的 Unicode 代码点，这是一个错误。
Try using this pattern and you'll see: [z-a].
尝试使用此模式，您将看到：[z-a].
The correct solution is to either put the dash -at the end of the character group (at which point it will lose its special meaning) or to escape it.
正确的解决方案是将破折号-放在字符组的末尾（此时它将失去其特殊含义）或将其转义。
You also need to close the unclosed open square bracket or escape it, if it was not intended for grouping.
如果不用于分组，您还需要关闭未闭合的开放方括号或将其转义。
Also, escaping the fullstop .is not necessary inside a character group.
此外，.在字符组内不需要转义句号。
回答by Denis Tulskiy
You have to escape the dash and close the unmatched square bracket. So you are going to get two errors with this regex:
您必须避开破折号并关闭不匹配的方括号。所以这个正则表达式会出现两个错误：
java.util.regex.PatternSyntaxException: Illegal character range near index 14
because the dash is used to specify a range, and \) is obviously a not valid range character. If you escape the dash, making it [[a-zA-Z\s\.\-\)\(]*you'll get
因为破折号用于指定范围，而 \) 显然是无效的范围字符。如果你逃脱了破折号，成功了，[[a-zA-Z\s\.\-\)\(]*你会得到
java.util.regex.PatternSyntaxException: Unclosed character class near index 19
which means that you have an extra opening square bracket that is used to specify character class. I don't know what you meant by putting an extra bracket here, but either escaping or removing it will make it a valid regex.
这意味着您有一个额外的左方括号用于指定字符类。我不知道你在这里放置一个额外的括号是什么意思，但是转义或删除它会使它成为有效的正则表达式。
