python判断电话号码是否合理_python判断号码是否可用(号码过滤)

def delCustomer(customer): # 返回两个 第一个用来标记是否可用&＃xff0c;第二个标记号码

f &＃61; open(AbnormalPhone,"r")

src &＃61; f.readlines()

f.close()

abnormalSet &＃61; set()

for line in src:

phone &＃61; str(line.split("\n")[0].strip())

abnormalSet.add(phone)

matchObj &＃61; re.match(&＃39;0?1[3-9]\d{9}$&＃39;, customer)

if (matchObj!&＃61;None): # 如果是手机号

if len(customer) &＃61;&＃61; 11:

return 1,customer

elif len(customer) &＃61;&＃61; 12:

return 1,customer[1:]

else: #如果不是手机号

matchObj &＃61; re.match(&＃39;0(10|2\d)&＃39;, customer) # 区号是3位 customer[0:3], customer[3:] 区号&＃xff0c;号码

if (matchObj !&＃61; None):

if (len(customer[3:]) &＃61;&＃61; 7 or len(customer[3:]) &＃61;&＃61; 8) and customer not in abnormalSet:

return 1,customer

else:

return 0,customer

matchObj &＃61; re.match(&＃39;0([3-9]\d{2})&＃39;, customer) # 区号是4位 customer[0:4], customer[4:] 区号&＃xff0c;号码

if (matchObj !&＃61; None):

if (len(customer[4:]) &＃61;&＃61; 7 or len(customer[4:]) &＃61;&＃61; 8) and customer not in abnormalSet:

return 1,customer

else:

return 0, customer

if (len(customer) &＃61;&＃61; 7 or len(customer) &＃61;&＃61; 8) and customer not in abnormalSet:

return 1,customer

else:

return 0,customer