python re for exact match number and string

Question

I am trying to match a string if it exactly matches, while ignoring case. Below is the code where my string value is different but still matching.

import re
k = "999"
v = "99"
if (re.search(v, k , re.IGNORECASE)):
   print "xyz" 
k = "AAA"
v = "aa"
if (re.search(v, k , re.IGNORECASE)):
   print "xyz" 

In above code k = 999 , v = 99 but matching and k = AAA, v = aa matching. What I exactly need is if k= 999 and v = 999 then match other all cases should not match. like wise k = AAA and v = aaA should match (Meaning ignore case) if k =AAA and v = aa should not match.

Answer 1

你的意思是..... if k == v:

Answer 2

不知道为什么要为此使用RegEx，但是无论出于何种原因，ypou都可以使用字符串的开头和字符串的末尾进行匹配。

k = re.compile(r"^99$")

Answer 3

The canonical way to do a case insensitive compare is to use lower() or upper() :

Code:

def matches(str1, str2):
    return 'matches' if str1.lower() == str2.lower() else 'does not match'

Test Code:

data = (
    ("999", "99"),
    ("999", "999"),
    ("999X", "999x"),
    ("999Xx", "999x"),
)

def matches(str1, str2):
    return 'matches' if str1.lower() == str2.lower() else 'does not match'

for datum in data:
    print('%s %s %s' % (datum[0], matches(*datum), datum[1]))

Results:

999 does not match 99
999 matches 999
999X matches 999x
999Xx does not match 999x