[regular expression] matches multiple characters

1. Match multiple characters

Code	function
*	Match previous character appears 0 Times or infinite times , You can have it or not
+	Match previous character appears 1 Times or infinite times , At least 1 Time
?	Match previous character appears 1 Time or 0 Time , That is to say, there is either 1 Time , Or not
{m}	Match previous character appears m Time
{m,n}	Match previous character appears from m To n Time

Example 1：*

demand ： Match a string, the first letter is the size character , It's followed by lowercase letters and these lowercase letters can Yes, No

import re

ret = re.match("[A-Z][a-z]*","M")
print(ret.group())

ret = re.match("[A-Z][a-z]*","MnnM")
print(ret.group())

ret = re.match("[A-Z][a-z]*","Aabcdef")
print(ret.group())

Running results ：

M
Mnn
Aabcdef

Example 2：+

demand ： Match a string , The first character is t, The last string is o, At least one character in the middle

import re

match_obj = re.match("t.+o", "two")
if match_obj:
    print(match_obj.group())
else:
    print(" Matching failure ")

Running results ：

two

Example 3：?

demand ： Match such data , however https This s There may be , It could be http This s No,

import re

match_obj = re.match("https?", "http")
if match_obj:
    print(match_obj.group())
else:
    print(" Matching failure ")

Running results ：

https

Example 4：{m}、{m,n}

demand ： Match out ,8 To 20 Bit code , It can be in upper and lower case 、 Numbers 、 Underline

import re


ret = re.match("[a-zA-Z0-9_]{6}","12a3g45678")
print(ret.group())

ret = re.match("[a-zA-Z0-9_]{8,20}","1ad12f23s34455ff66")
print(ret.group())

Running results ：

12a3g4
1ad12f23s34455ff66