Python 3.x 获取字符串中倒数第二个出现的字符

Python 3.x 获取字符串中倒数第二个出现的字符,python-3.x,Python 3.x,如何获取字符串中字符的倒数第二次出现? 我想用它从www.example.com/example/abc/aaa获取aaa,我喜欢正则表达式: import re p = re.compile('a') all_positions = [oMatch.span() for oMatch in p.finditer(sText)] second_last_position_span = all_positions[-2] second_last_position_start_index =

如何获取字符串中字符的倒数第二次出现?
我想用它从
www.example.com/example/abc/aaa
获取
aaa
,我喜欢正则表达式:

import re
p = re.compile('a')
all_positions = [oMatch.span() for oMatch in p.finditer(sText)]
second_last_position_span = all_positions[-2]   
second_last_position_start_index = second_last_position_span[0]

我注意到你没有接受任何答案。。。查看以下内容以了解其工作原理: