Remove leading dollar sign from data and improve current solution - python-3.x

Question

I have string like so:
"Job 1233:name_uuid (table n_Cars_1234567$20220316) done. Records: 24, with errors: 0."
I'd like to retieve the datte from the table name, so far I use:
"\$[0-9]+"
but this yields $20220316. How do I get only the date, without $?
I'd also like to get the table name: n_Cars_12345678$20220316
So far I have this:
pattern_table_info = "\(([^\)]+)\)"
pattern_table_name = "(?<=table ).*"
table_info = re.search(pattern_table_info, message).group(1)
table = re.search(pattern_table_name, table_info).group(0)
However I'd like to have a more simpler solution, how can I improve this?
EDIT:
Actually the table name should be:
n_Cars_12345678
So everything before the "$" sign and after "table"...how can this part of the string be retrieved?

Answer 1

You can use a regex with two capturing groups:
table\s+([^()]*)\$([0-9]+)
See the regex demo. Details:
table - a word
\s+ - one or more whitespaces
([^()]*) - Group 1: zero or more chars other than ( and )
\$ - a $ char
([0-9]+) - Group 2: one or more digits.
See the Python demo:
import re
text = "Job 1233:name_uuid (table n_Cars_1234567$20220316) done. Records: 24, with errors: 0."
rx = r"table\s+([^()]*)\$([0-9]+)"
m = re.search(rx, text)
if m:
print(m.group(1))
print(m.group(2))
Output:
n_Cars_1234567
20220316

Answer 2

You can write a single pattern with 2 capture groups:
\(table (\w+\$(\d+))\)
The pattern matches:
\(table
( Capture group 1
\w+\$ match 1+ word characters and $
(\d+) Capture group 2, match 1+ digits
) Close group 1
\) Match )
See a Regex demo and a Python demo.
import re
s = "Job 1233:name_uuid (table n_Cars_1234567$20220316) done. Records: 24, with errors: 0."
m = re.search(r"\(table (\w+\$(\d+))\)", s)
if m:
print(m.group(1))
print(m.group(2))
Output
n_Cars_1234567$20220316
20220316