进入 Python 中的方括号内
Get inside of square brackets in Python
我有这个字符串。
"ascascascasc[xx]asdasdasdasd[yy]qweqweqwe"
我想获取括号内的字符串。像这样；
"xx", "yy"
我已经试过了，但没有用：
a = "ascascascasc[xx]asdasdasdasd[yy]qweqweqwe"
listinside = []
for i in range(a.count("[")):
listinside.append(a[a.index("["):a.index("]")])
print (listinside)
输出：
['[xx', '[xx']
你不需要计数，你可以使用正则表达式，re.findall() 可以做到:
>>> s="ascascascasc[xx]asdasdasdasd[yy]qweqweqwe"
>>> import re
>>> re.findall(r'\[(.*?)\]',s)
['xx', 'yy']
\[ matches the character [ literally
*? matches Between zero and unlimited times, as few times as possible, expanding as needed [lazy]
\] matches the character ] literally
我有这个字符串。
"ascascascasc[xx]asdasdasdasd[yy]qweqweqwe"
我想获取括号内的字符串。像这样；
"xx", "yy"
我已经试过了，但没有用：
a = "ascascascasc[xx]asdasdasdasd[yy]qweqweqwe"
listinside = []
for i in range(a.count("[")):
listinside.append(a[a.index("["):a.index("]")])
print (listinside)
输出：
['[xx', '[xx']
你不需要计数，你可以使用正则表达式，re.findall() 可以做到:
>>> s="ascascascasc[xx]asdasdasdasd[yy]qweqweqwe"
>>> import re
>>> re.findall(r'\[(.*?)\]',s)
['xx', 'yy']
\[matches the character [ literally
*?matches Between zero and unlimited times, as few times as possible, expanding as needed [lazy]
\]matches the character ] literally