正则表达式：
import re
s1 = 'python ui web 张三'
print(s1.find('web'))
print(re.search('web',s1)) #正则提供的精准查找
print(re.findall('web',s1)) #正则提供的模糊查找
t = '\d+'
print(re.findall(t,s1)) #通过正则查找字符串里的数字
#:t通配符：万能符，可以匹配\n以外的所有符号
s1 = 'tom 200 bom 308 do\nm 400 5e'
res1 = re.findall('[0-9]+',s1)
res2 = re.findall('[0-9]{1,}',s1)
res3 = re.findall('\d+',s1)
res4 = re.findall('\\d+',s1)
res5 = re.findall(r'\d+',s1)
res6 = re.findall(r'd|b+',s1)
res7 = re.findall(r'[tbd]om',s1)
res8 = re.findall(r'\n',s1)
res9 = re.findall(r'[0-9a-zA-Z]',s1)
print(res1)
print(res2)
print(res3)
print(res4)
print(res5)
print(res6)
print(res7)
print(res8)
print(res9)
s1 = 'aeeee age agree ape are a\ne ape a9e aFe'
reg = 'a.e'
res = re.findall(reg,s1)
res1 = re.findall('a.e',s1)
print(res1)
#2[]:表示字符串只能匹配一个
s2 = 'aeeee age agree ape are a\ne ape a9e aFe'
res1 = re.findall('a[p]e',s1)
res2 = re.findall('a[pg]e',s1)
res3 = re.findall('a[0-9]e',s1)
res4 = re.findall('a[a-z]e',s1)
res5 = re.findall('a[A-Z]e',s1)
res6 = re.findall('a[a-zA-Z]e',s1)
res7 = re.findall('a[^0-9]e',s1)
res8 = re.findall('a[^a-z]e',s1)
print(res1)
print(res2)
print(res3)
print(res4)
print(res5)
print(res6)
print(res7)
print(res8)
#{}制定左侧可以重复的范围
s3 = 'aeeee age agree ape are a\ne ape a9e aFe'
res1 = re.findall('a...e',s3)
res2 = re.findall('a{3}e',s3)
#贪婪匹配规则，默认按照最大数量匹配
res6 = re.findall('a.{1,3}e',s3)
#取消贪婪匹配规则，在{}加？
res7 = re.findall('a.{1,3}?',s3)
#如果第二个值不写，默认无穷大
res8 = re.findall('a.{1,}',s3)
#更换匹配模式，匹配所有
res9 = re.findall('a.{1,}e',s3,re.S)
print(res6)
print(res7)
print(res8)
print(res9)
#?:制定左侧符号出现0次或1次，等同于{0，1}，默认贪婪
s4 = 'aeeee age agree ape are a\ne ape a9e aFe'
res10 = re.findall('a.{0,1}?e',s4)
res11 = re.findall('a.??e',s4)
print(res10)
print(res11)
#*:制定左侧符号出现0次或多次，等同于{0，}，默认贪婪
s4 = 'aeeee age agree ape are a\ne ape a9e aFe'
res10 = re.findall('a.{0,}?e',s4)
res11 = re.findall('a.*?e',s4)
print(res10)
print(res11)
#+:制定左侧符号出现1次或多次，等同于{1，}，默认贪婪
s4 = 'aeeee age agree ape are a\ne ape a9e aFe'
res13 = re.findall('a.+?e',s4)
res12 = re.findall('[0-9]+?',s4)
print(res12)
print(res13)
s7 = ['122223333', '13432416666','11011112222']
def test():
for i in num:
rees = re.findall(r'^1[3-9]\d{9}$',n)
print(res)
test()
^与$
$以某个规则结尾
^以某个规则开头
s5 = '王哥王弟王嫂'
res12 = re.findall('王',s5)
res13 = re.findall('^王.',s5)
res14 = re.findall('王.$',s5)