CODEWITHSUNDEEP
pythonregex
Aaron Conway
Aaron Conway

Reputation: 127

Python and regex to find a number of length 5

So currently I've got the string

'JavaScript:doEdit('41228', '', '', 2);'

I'd like to do regex in python on it to filter out ONLY the 41228. I've tried two methods and came up with issues on both. The first was trying to find things that aren't digits of length 5 by using

re.sub('^\d{5}', string )

Then I tried re.match and re.compile which gives me the error TypeError: unsupported operand type(s) for &: 'str' and 'int'.

The only thing close I've done is use re.sub{'\D', string} but then it sticks in the extra 2 I don't want.

I guess it wouldn't be an issue to just find the 19th through 24th characters in the string. since the string should never change composition. When I generate a new id.

SOLVED: working code is

screen_id = 'JavaScript:doEdit('41228', '', '', 2);'
reduced_screenc_id = re.search(r'\d{5}', screenc_id)
print (reduced_screenc_id.group())

Upvotes: 0

Views: 2952

Answers (2)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626728

If your string is always in this format, you may use a mere split operation with ':

s = "JavaScript:doEdit('41228', '', '', 2);"
print(s.split("'")[1])
# => 41228

If you plan to study regex, you may use re.search with either \d{5} or doEdit'(\d{5})' regex.

import re
s = "JavaScript:doEdit('41228', '', '', 2);"
res = re.search(r"\d{5}", s)
if res:
    print(res.group())    # Here, we need the whole match
res = re.search(r"doEdit\('(\d{5})'", s)
if res:
    print(res.group(1)) # Grabbing Group 1 value only

Since you need only the first match, there is no point using re.findall.

See this Python demo.

Upvotes: 0

L3viathan
L3viathan

Reputation: 27273

Use re.findall to find all (non-overlapping) instances of the pattern in your string:

>>> import re
>>> string = "JavaScript:doEdit('41228', '', '', 2);"
>>> pattern = '\d{5}'  # 5 digits
>>> number = re.findall(pattern, string)[0]
>>> number
'41228'

You might want to cast the "number" to an actual number using number = int(number) then.

Upvotes: 3

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