Python 字符串中的正则表达式

Question

I want to validate a string that satisfies the below three conditions using regular expression

1. The special characters allowed are (. , _, - ).
2. Should contain only lower-case characters.
3. Should not start or end with special character.

To satisfy the above conditions, I have created a format as below

^[^\W_][a-z\.,_-]+

This pattern works fine up to second character. However, this pattern is failing for the 3rd and subsequent characters if those contains any special character or upper cases characters.

Example:

Pattern Works for the string S@yanthan but not for Sa@yanthan. I am expecting that pattern to pass even if the third and subsequent characters contains any special characters or upper case characters. Can you suggest me where this pattern goes wrong please? Below is the snippet of the code.

import re 

a = "Sayanthan"
exp = re.search("^[^\W_][a-z\.,_-]+",a)

if exp:
    print(True)
else:
    print(False)

Answer 1

Based on you initial rules I'd go with:

^[a-z](?:[.,_-]*[a-z])*$

See the online demo .

However, you mentioned in the comments:

"Also the third condition is "should not start with Special character" instead of "should not start or end with Special character""

In that case you could use:

^[a-z][-.,_a-z]*$

See the online demo

Answer 2

The pattern that you tried ^[^\W_][az.,_-]+ starts with [^\W_] which will match any word char except an underscore, so it could also be an uppercase char.

Then [az.,_-]+ will match 1+ times any of the listed, which means the string can also end with a comma for example.

Looking at the conditions listed, you could use:

^[a-z](?:[a-z.,_-]*[a-z])?\Z

• ^ Start of string
• [az] Match a lower case char az
• (?: Non capture group
  • [az.,_-]*[az] Match 0+ occurrences of the listed ending with az
• )? Close group and make it optional
• \Z End of string

Regex demo