Eg-My 字符串是 :
str1="[00:00:0.047] [TestRunner] [Main] [sleep_assoc] > [Iteration:0][00:00:0.063] [TestRunner] [Main] [sleep_assoc] > [!Iteration:0][00:00:0.063]"
我想将[Iteration:0]
和[!Iteration:0]
之间的所有字符存储到一个字符串变量中。
您可以使用.*?
来执行此操作,该匹配从[Iteration:0]
到找到[!Iteration:0]
的所有字符:
In [1]: import re
In [2]: s = '[00:00:0.047] [TestRunner] [Main] [sleep_assoc] > [Iteration:0][00:00:0.063] [TestRunner] [Main] [sleep_as
...: soc] > [!Iteration:0][00:00:0.063]'
In [3]: m = re.search(r'[Iteration:0](.*?)[!Iteration:0]', s)
In [4]: res = m.group(1)
In [5]: res
Out[5]: '[00:00:0.063] [TestRunner] [Main] [sleep_assoc] > '
如果字符串如此简单,则不需要正则表达式。
做个str1.split('[Iteration:0]')[1].split('[!Iteration:0]')[0]
你明白代码还是我解释一下?