正则表达式查找条目的第二次出现

Question

i am no expert in regex. Therefore, my skill set is beaten. Consider the following text:

[SectionTitle0]
...
Name: NameOfTechC
...

[SectionTitle1]
...
Name: NameOfZoneC
...

I am interested in extracting the name of Tech-C and Zone-C using regex. This looks like a config-section-party, though I might use a library to parse configs. But this extract is part of a even bigger file. In consequence, config-parsers does not work here.

Currently, I extract the name with Name:\\s?(.+) . Using re.findall in python returns a list containing both names. Is there a way to use something like

TechC_name: regex1
ZoneC_name: regex2

that returns the list for either the Tech-C name or the Zone-C name?

[Update]
I want to clarify some points. The position of 'Name:' is not fixed, therefore it is possible that same points are listed before, and same after, the entry. I updated my question.

I recognised that sometimes SectionTitle0 (former 'Tech-C') and SectionTitle1 (former 'Zone-C') are identically. That makes it a little bit more complicated. May be there is a way to build a regex that fits the first occurrence of 'Name:' and a regex matching the second (or n-th) occurrence of 'Name:'.

Answer 1

The two regexes you are looking for are:

TechC_name:

re.findall(r"\[Tech-C\]\nName: (.*?)\n", s)

ZoneC_name:

re.findall(r"\[Zone-C\]\nName: (.*?)\n", s)

Answer 2

You can get the output in following format easily:-

[(section1, name1), (section2, name2), ...]

By following regex implementation:-

import re
re.findall(r"\[(\S+)\]\nName: (\w+)", t)

Output will be:-

[('Tech-C', 'NameOfTechC'), ('Zone-C', 'NameOfZoneC')]

Answer 3

Regex:

\[([^\]]*)\][\r\n]+(?:(?!Name:).*[\r\n]+)*?Name:\s*(.*)

Efficiently captures section title and name into groups \\1 and \\2 .

Implementation

import re

data  = """[SectionTitle0]
...
Name: NameOfTechC
...

[SectionTitle1]
...
Name: NameOfZoneC
...
"""

regexStr = r'\[([^\]]*)\][\r\n]+(?:(?!Name:).*[\r\n]+)*?Name:\s*(.*)'
regex    = re.compile(regexStr)
regex.findall(data)