java 正则表达式匹配方括号或双引号中的字符串
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Regular expression to match strings enclosed in square brackets or double quotes
提问by Click Upvote
I need 2 simple reg exps that will:
我需要 2 个简单的 reg exp，它们将：
- Match if a string is contained within square brackets (
[]e.g[word]) - Match if string is contained within double quotes (
""e.g"word")
- 如果字符串包含在方括号内，则匹配（
[]例如[word]） - 如果字符串包含在双引号内，则匹配（
""例如"word"）
回答by Peter Boughton
\[\w+\]
"\w+"
Explanation:
解释：
The \[ and \] escape the special bracket characters to match their literals.
\[ 和 \] 转义特殊括号字符以匹配它们的文字。
The \w means "any word character", usually considered same as alphanumeric or underscore.
\w 表示“任何单词字符”，通常被认为与字母数字或下划线相同。
The + means one or more of the preceding item.
+ 表示前面的一项或多项。
The " are literal characters.
" 是文字字符。
NOTE: If you want to ensure the whole string matches (not just part of it), prefix with ^and suffix with $.
注意：如果你想确保整个字符串匹配（不仅仅是它的一部分），前缀^和后缀$.
And next time, you should be able to answer this yourself, by reading regular-expressions.info
下一次，您应该能够通过阅读正则表达式.info自己回答这个问题
Update:
更新：
Ok, so based on your comment, what you appear to be wanting to know is if the first character is [ and the last ] or if the first and last are both " ?
If so, these will match those:
好的，所以根据您的评论，您似乎想知道第一个字符是 [ 和最后一个 ] 还是第一个和最后一个字符都是 " ？
如果是这样，这些将匹配：
^\[.*\]$ (or ^\[.*\]$ in a Java String)
"^.*$"
However, unless you need to do some special checking with the centre characters, simply doing:
但是，除非您需要对中心字符进行一些特殊检查，否则只需执行以下操作：
if ( MyString.startsWith("[") && MyString.endsWith("]") )
and
和
if ( MyString.startsWith("\"") && MyString.endsWith("\"") )
Which I suspect would be faster than a regex.
我怀疑这会比正则表达式更快。
回答by Chas. Owens
Important issues that may make this hard/impossible in a regex:
在正则表达式中可能使这变得困难/不可能的重要问题：
Can
[]be nested (e.g.[foo [bar]])? If so, then a traditional regex cannot help you. Perl's extended regexes can, but it is probably better to write a parser.Can
[,], or"appear escaped (e.g."foo said \"bar\"") in the string? If so, see How can I match double-quoted strings with escaped double-quote characters?Is it possible for there to be more than one instance of these in the string you are matching? If so, you probably want to use the non-greedy quantifier modifier (i.e.
?) to get the smallest string that matches:/(".*?"|\[.*?\])/g
可以
[]嵌套（例如[foo [bar]]）？如果是这样，那么传统的正则表达式帮不了你。Perl 的扩展正则表达式可以，但最好编写一个解析器。可以
[,], 或"出现"foo said \"bar\""在字符串中转义（例如）吗？如果是这样，请参阅如何将双引号字符串与转义的双引号字符匹配？在您匹配的字符串中是否可能有多个这些实例？如果是这样，您可能希望使用非贪婪量词修饰符（即
?）来获取匹配的最小字符串：/(".*?"|\[.*?\])/g
Based on comments, you seem to want to match things like "this is a "long" word"
根据评论，您似乎想要匹配诸如 "this is a "long" word"
#!/usr/bin/perl
use strict;
use warnings;
my $s = 'The non-string "this is a crazy "string"" is bad (has own delimiter)';
print $s =~ /^.*?(".*").*?$/, "\n";
回答by Chas. Owens
Are they two separate expressions?
它们是两个不同的表达式吗？
[[A-Za-z]+]
[[A-Za-z]+]
\"[A-Za-z]+\"
\"[A-Za-z]+\"
If they are in a single expression:
如果它们在单个表达式中：
[[\"]+[a-zA-Z]+[]\"]+
[[\"]+[a-zA-Z]+[]\"]+
Remember that in .net you'll need to escape the double quotes " by ""
请记住，在 .net 中，您需要将双引号 "" 转义为 ""
