[英]Python Regex to find a string in double quotes within a string
I'm looking for a code in python using regex that can perform something like this我正在使用正则表达式在 python 中寻找可以执行类似操作的代码
Input: Regex should return "String 1" or "String 2" or "String3"
输入: Regex should return "String 1" or "String 2" or "String3"
Output: String 1,String2,String3
输出: String 1,String2,String3
I tried r'"*"'
我试过r'"*"'
Here's all you need to do:这是您需要做的所有事情:
def doit(text):
import re
matches = re.findall(r'"(.+?)"',text)
# matches is now ['String 1', 'String 2', 'String3']
return ",".join(matches)
doit('Regex should return "String 1" or "String 2" or "String3" ')
result:结果:
'String 1,String 2,String3'
As pointed out by Li-aung Yip :正如Li-aung Yip所指出的:
To elaborate,
.+?
详细地说,.+?
is the "non-greedy" version of.+
.是.+
的“非贪婪”版本。 It makes the regular expression match the smallest number of characters it can instead of the most characters it can.它使正则表达式匹配它可以匹配的最少字符数,而不是它可以匹配的最多字符数。 The greedy version,.+
, will giveString 1" or "String 2" or "String 3
;贪婪版本.+
将给出String 1" or "String 2" or "String 3
; the non-greedy version.+?
非贪婪版本.+?
givesString 1
,String 2
,String 3
.给出String 1
,String 2
,String 3
。
In addition, if you want to accept empty strings, change .+
to .*
.此外,如果要接受空字符串,请将.+
更改为.*
。 Star *
means zero or more while plus +
means at least one.星号*
表示零或多个,而+
表示至少一个。
The highly up-voted answer doesn't account for the possibility that the double-quoted string might contain one or more double-quote characters (properly escaped, of course).高度赞成的答案没有考虑双引号字符串可能包含一个或多个双引号字符的可能性(当然,正确转义)。 To handle this situation, the regex needs to accumulate characters one-by-one with a positive lookahead assertion stating that the current character is not a double-quote character that is not preceded by a backslash (which requires a negative lookbehind assertion ):为了处理这种情况,正则表达式需要一个一个地累积字符,并带有一个肯定的前瞻断言,说明当前字符不是前面没有反斜杠的双引号字符(这需要一个否定的后瞻断言):
"(?:(?:(?!(?<!\\)").)*)"
import re
import ast
def doit(text):
matches=re.findall(r'"(?:(?:(?!(?<!\\)").)*)"',text)
for match in matches:
print(match, '=>', ast.literal_eval(match))
doit('Regex should return "String 1" or "String 2" or "String3" and "\\"double quoted string\\"" ')
Prints:印刷:
"String 1" => String 1
"String 2" => String 2
"String3" => String3
"\"double quoted string\"" => "double quoted string"
Just try to fetch double quoted strings from the multiline string:只需尝试从多行字符串中获取双引号字符串:
import re
s = """
"my name is daniel" "mobile 8531111453733"[[[[[[--"i like pandas"
"location chennai"! -asfas"aadhaar du2mmy8969769##69869"
@4343453 "pincode 642002""@mango,@apple,@berry"
"""
print(re.findall(r'"(.*?)"', s))
From https://stackoverflow.com/a/69891301/1531728来自https://stackoverflow.com/a/69891301/1531728
My solution is:我的解决方案是:
import re
my_strings = ['SetVariables "a" "b" "c" ', 'd2efw f "first" +&%#$%"second",vwrfhir, d2e u"third" dwedew', '"uno"?>P>MNUIHUH~!@#$%^&*()_+=0trewq"due" "tre"fef fre f', ' "uno""dos" "tres"', '"unu""doua""trei"', ' "um" "dois" "tres" ']
my_substrings = []
for current_test_string in my_strings:
for values in re.findall(r'\"(.+?)\"', current_test_string):
my_substrings.append(values)
#print("values are:",values,"=")
print(" my_substrings are:",my_substrings,"=")
my_substrings = []
Alternate regular expressions to use are:要使用的备用正则表达式是:
The current_test_string.split("\"")
approach works if the strings have patterns in which substrings are embedded within quotation marks. This is because it uses the double quotation mark in this example as a delimiter to tokenize the string, and accepts substrings that are not embedded within double quotation marks as valid substring extractions from the string. current_test_string.split("\"")
方法适用于字符串具有其中子字符串嵌入引号内的模式。这是因为它在此示例中使用双引号作为分隔符来标记字符串,并接受符合以下条件的子字符串不作为从字符串中提取的有效子字符串嵌入在双引号中。
References:参考:
For me the only regex that ever worked right for all the cases of quoted strings with possibly escaped quotes inside of them was:对我来说,唯一适用于所有带引号的字符串的正则表达式,其中可能包含转义引号:
regex=r"""(['"])(?:\\\\|\\\1|[^\1])*?\1"""
This will not fail even if the quoted string ends with an escaped backslash.即使引用的字符串以转义的反斜杠结尾,这也不会失败。
import re
r=r"'(\\'|[^'])*(?!<\\)'|\"(\\\"|[^\"])*(?!<\\)\""
texts=[r'"aerrrt"',
r'"a\"e'+"'"+'rrt"',
r'"a""""arrtt"""""',
r'"aerrrt',
r'"a\"errt'+"'",
r"'aerrrt'",
r"'a\'e"+'"'+"rrt'",
r"'a''''arrtt'''''",
r"'aerrrt",
r"'a\'errt"+'"',
"''",'""',""]
for text in texts:
print (text,"-->",re.fullmatch(r,text))
results:结果:
"aerrrt" --> <_sre.SRE_Match object; span=(0, 8), match='"aerrrt"'>
"a\"e'rrt" --> <_sre.SRE_Match object; span=(0, 10), match='"a\\"e\'rrt"'>
"a""""arrtt""""" --> None
"aerrrt --> None
"a\"errt' --> None
'aerrrt' --> <_sre.SRE_Match object; span=(0, 8), match="'aerrrt'">
'a\'e"rrt' --> <_sre.SRE_Match object; span=(0, 10), match='\'a\\\'e"rrt\''>
'a''''arrtt''''' --> None
'aerrrt --> None
'a\'errt" --> None
'' --> <_sre.SRE_Match object; span=(0, 2), match="''">
"" --> <_sre.SRE_Match object; span=(0, 2), match='""'>
--> None
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