我需要一个正则表达式来选择两个外括号之间的所有文本。
例子: START_TEXT(这里的文本(可能的文本)文本(可能的文本(更多的文本))END_TXT ^ ^
结果: (此处文本(可能的文本)文本(可能的文本(更多的文本)))
我需要一个正则表达式来选择两个外括号之间的所有文本。
例子: START_TEXT(这里的文本(可能的文本)文本(可能的文本(更多的文本))END_TXT ^ ^
结果: (此处文本(可能的文本)文本(可能的文本(更多的文本)))
当前回答
"""
Here is a simple python program showing how to use regular
expressions to write a paren-matching recursive parser.
This parser recognises items enclosed by parens, brackets,
braces and <> symbols, but is adaptable to any set of
open/close patterns. This is where the re package greatly
assists in parsing.
"""
import re
# The pattern below recognises a sequence consisting of:
# 1. Any characters not in the set of open/close strings.
# 2. One of the open/close strings.
# 3. The remainder of the string.
#
# There is no reason the opening pattern can't be the
# same as the closing pattern, so quoted strings can
# be included. However quotes are not ignored inside
# quotes. More logic is needed for that....
pat = re.compile("""
( .*? )
( \( | \) | \[ | \] | \{ | \} | \< | \> |
\' | \" | BEGIN | END | $ )
( .* )
""", re.X)
# The keys to the dictionary below are the opening strings,
# and the values are the corresponding closing strings.
# For example "(" is an opening string and ")" is its
# closing string.
matching = { "(" : ")",
"[" : "]",
"{" : "}",
"<" : ">",
'"' : '"',
"'" : "'",
"BEGIN" : "END" }
# The procedure below matches string s and returns a
# recursive list matching the nesting of the open/close
# patterns in s.
def matchnested(s, term=""):
lst = []
while True:
m = pat.match(s)
if m.group(1) != "":
lst.append(m.group(1))
if m.group(2) == term:
return lst, m.group(3)
if m.group(2) in matching:
item, s = matchnested(m.group(3), matching[m.group(2)])
lst.append(m.group(2))
lst.append(item)
lst.append(matching[m.group(2)])
else:
raise ValueError("After <<%s %s>> expected %s not %s" %
(lst, s, term, m.group(2)))
# Unit test.
if __name__ == "__main__":
for s in ("simple string",
""" "double quote" """,
""" 'single quote' """,
"one'two'three'four'five'six'seven",
"one(two(three(four)five)six)seven",
"one(two(three)four)five(six(seven)eight)nine",
"one(two)three[four]five{six}seven<eight>nine",
"one(two[three{four<five>six}seven]eight)nine",
"oneBEGINtwo(threeBEGINfourENDfive)sixENDseven",
"ERROR testing ((( mismatched ))] parens"):
print "\ninput", s
try:
lst, s = matchnested(s)
print "output", lst
except ValueError as e:
print str(e)
print "done"
其他回答
正则表达式是一个错误的工具,因为你正在处理嵌套结构,即递归。
但是有一个简单的算法可以做到这一点,我在之前的问题的回答中详细描述了它。其要点是编写代码扫描字符串,并对尚未与闭括号匹配的开括号保持计数器。当计数器返回0时,您就知道已经到达了最后的右括号。
因为js regex不支持递归匹配,我不能使平衡括号匹配工作。
这是一个简单的javascript循环版本,将“method(arg)”字符串转换为数组
push(number) map(test(a(a()))) bass(wow, abc)
$$(groups) filter({ type: 'ORGANIZATION', isDisabled: { $ne: true } }) pickBy(_id, type) map(test()) as(groups)
const parser = str => {
let ops = []
let method, arg
let isMethod = true
let open = []
for (const char of str) {
// skip whitespace
if (char === ' ') continue
// append method or arg string
if (char !== '(' && char !== ')') {
if (isMethod) {
(method ? (method += char) : (method = char))
} else {
(arg ? (arg += char) : (arg = char))
}
}
if (char === '(') {
// nested parenthesis should be a part of arg
if (!isMethod) arg += char
isMethod = false
open.push(char)
} else if (char === ')') {
open.pop()
// check end of arg
if (open.length < 1) {
isMethod = true
ops.push({ method, arg })
method = arg = undefined
} else {
arg += char
}
}
}
return ops
}
// const test = parser(`$$(groups) filter({ type: 'ORGANIZATION', isDisabled: { $ne: true } }) pickBy(_id, type) map(test()) as(groups)`)
const test = parser(`push(number) map(test(a(a()))) bass(wow, abc)`)
console.log(test)
结果就像
[ { method: 'push', arg: 'number' },
{ method: 'map', arg: 'test(a(a()))' },
{ method: 'bass', arg: 'wow,abc' } ]
[ { method: '$$', arg: 'groups' },
{ method: 'filter',
arg: '{type:\'ORGANIZATION\',isDisabled:{$ne:true}}' },
{ method: 'pickBy', arg: '_id,type' },
{ method: 'map', arg: 'test()' },
{ method: 'as', arg: 'groups' } ]
在处理嵌套模式和正则表达式是解决这类问题的正确工具时,我也陷入了这种情况。
/(\((?>[^()]+|(?1))*\))/
(?<=\().*(?=\))
如果您想在两个匹配的括号之间选择文本,那么使用正则表达式就不太走运了。这是不可能的。
这个正则表达式只返回字符串中第一个开始括号和最后一个结束括号之间的文本。
(*)除非你的regex引擎有像平衡组或递归这样的特性。支持这些特性的引擎的数量正在缓慢增长,但它们仍然不是普遍可用的。
我写了一个叫做balanced的JavaScript库来帮助完成这个任务。你可以这样做
balanced.matches({
source: source,
open: '(',
close: ')'
});
你甚至可以做替换:
balanced.replacements({
source: source,
open: '(',
close: ')',
replace: function (source, head, tail) {
return head + source + tail;
}
});
下面是一个更复杂的交互式示例JSFiddle。