I have a regex that parses US phone numbers into 3 strings.
import re
s = ' 916-2221111 ' # this also works'(916) 222-1111 '
reg_ph = re.match(r'^s*(?(d{3}))?-? *(d{3})-? *-?(d{4})', s)
if reg_ph:
return reg_ph.groups()
else:
raise ValueError ('not a valid phone number')
it works perfectly on the numbers:
'(916) 222-1111 ' ' 916-2221111 '
Now I need to add an additional regex to generate a Value Error for numbers such as
s = '916 111-2222' # there are white spaces between the area code and a local number and NO ')'
I tried
reg_ph = re.match(r'^s*(?(d{3}))?s*-? *(d{3})-? *-?(d{4})', s)
reg_ph = re.match(r'^s*(?(d{3}))?s*-? *(d{3})-? *-?(d{4})', s)
but non rejects the string in question
I will greatly appreciate any ideas. I am very new to Regex!
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Answer
In Python re you could use a conditional to check for group 1 having the opening parenthesis.
If that is the case match the closing parenthesis, optional spaces and 3 digits. Else match - and 3 digits.
If you use re.match you can omit ^
^s*(()?d+(?(1))s*d{3}|-d{3})-?d{4}
If you want to match the whole string and trailing whitespace chars:
^s*(()?d+(?(1))s*d{3}|-d{3})-?d{4}s*$
In parts, the pattern matches:
^Start of strings*Match optional whitespace chars(()?Optional group 1, match(d+Match 1+ digits(?Conditional(1))s*d{3}If group 1 exist, match the closing), optional whitespace chars and 3 digits|Or-?Match optional –d{3}Match 3 digits
)close conditional-?d{4}Match optional – and 4 digits
See a regex demo
For example, using capture groups in the pattern to get the digits:
import re
strings = [' (916) 111-2222',' 916-2221111 ', '916 111-2222']
pattern =r's*(()?(d+)(?(1))s*(d{3})|-(d{3}))-?(d{4})s*$'
for item in strings:
m=re.match(pattern, item)
if m:
t = tuple(s for s in m.groups() if s is not None and s.isdigit())
print(t)
else:
print("no match for " + item)
Output
('916', '111', '2222')
('916', '222', '1111')
no match for 916 111-2222