在python中获取所有％（name）s占位符

Question

我有一个包含%(name)s占位符的字符串，我想获取所有名称，例如：This is a %(name)s example string %(foo)s I would like %(bar)s to extract all the placeholders from %(place)s

我想要一个包含name，foo，bar，place

的列表

Answer 1

使用正则表达式。

例如：

import re

s = "This is a %(name)s example string %(foo)s I would like %(bar)s to extract all the placeholders from %(place)s"
print(re.findall(r"%\((.*?)\)", s))
# --> ['name', 'foo', 'bar', 'place']

Answer 2

似乎是re.findall的好用法

>>> import re
>>> string = "This is a %(name)s example string %(foo)s I would like %(bar)s to extract all the placeholders from %(place)s"
>>> re.findall(r'%\((.+?)\)', string)
['name', 'foo', 'bar', 'place']

%\((.+?)\)的意思是从%(开始，到下一个)结束，捕获之间的所有内容。