正则表达式查找带引号的字符串中的所有大括号

Question

I have a string:

test_str = 'This is the string and it "contains {0} a" few {1} sets of curly brackets'

I would like to only find {0} and not {1} in this example, that is, the brackets themselves and their contents, if only within a set of double quotes.

I've started to solve this by matching the portion in double quotes:

(?<=").*(?=")

See https://regex101.com/r/qO0pO2/1

but I am having difficulty only matching the {0} portion

How can I extend this regex to match {0} ?

Answer 1

Remove the pipe | it will work great: Live Demo

And here is for multiple char between {}

(?<=)\{[^\}]*\}(?=)

With Live Demo

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Update:

This does the stuff :

".*({[^\}]*\}).*"

Answer 2

You can try word boundary \\B and lookarounds - ie

>>>test_str="This is the string and it contains {0} a few {1} sets of curly brackets"
>>>re.findall(r'(?<=\B){.*?}(?=\B)',test_str)
>>>['{0}', '{1}']

See live DEMO

But if your string does not have word boundary then try lazy quantifier evaluation

>>>test_str="This is the string and it contains {0} a few {1} sets of curly brackets"
>>>re.findall(r'{.*?}',test_str)
>>>['{0}', '{1}']

See live DEMO

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EDIT

If you want only {0} then you have to use escape character( \\ ) before braces, since braces are regex token-try as below.

>>>test_str="This is the string and it contains {0} a few {1} sets of curly brackets"
>>>re.findall(r'\{0\}',test_str)
>>>['{0}']

Answer 3

If the quotes are balanced, you could use a lookahead to check for an uneven amount ahead. If you know, that there is only one quoted substring, check if there occurs only one " until the end $

{[^}]+}(?=[^"]*"[^"]*$)

See demo . But if there could be any amount of quoted parts check for an uneven amount until end.

{[^}]+}(?=[^"]*"(?:[^"]*"[^"]*")*[^"]*$)

• {[^}]+} matches the braced stuff: literal { followed by [^}]+ one or more non } until }
• [^"]*" inside the lookahead matches until the first quote
• (?:[^"]*"[^"]*")* followed by zero or more balanced, preceded by any amount of non quotes
• [^"]*$ followed by any amount of non quotes until end

See demo at regex101

Answer 4

Might be difficult to do in one regex, but it's easy with two:

from re import findall

# First find all quoted strings...
for quoted in findall(r'"[^"]*"', test_str):
    # ...then find all bracketed expressions
    for match in findall(r'\{[^\}]*\}', quoted):
        print(match)

or as a one-liner:

[match for match in findall(r'\{[^\}]*\}', quoted) for quoted in findall(r'"[^"]*"', test_str)]