Peter Otten wrote:
Dale Amon wrote:
This finds nothing:
import re
import string
card = "abcdef"
DEC029 = re.compile("[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!);\\\]%_>?]")
The regular expression you're actually providing is:
>>> print "[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!);\\\]%_>?]"
[^&0-9A-Z/ $*,.\-:#@'="[<(+\^!);\\]%_>?]
^^^
The backslash is escaped (the "\\") and the set ends at the first "]".
errs = DEC029.findall(card.strip("\n\r"))
print errs
This works correctly:
import re
import string
card = "abcdef"
DEC029 = re.compile("[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!)\\;\]%_>?]")
The regular expression you're actually providing is:
>>> print "[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!)\\;\]%_>?]"
[^&0-9A-Z/ $*,.\-:#@'="[<(+\^!)\;\]%_>?]
^^ ^
The first "]" is escaped (the "\]") and the set ends at the second "]".
errs = DEC029.findall(card.strip("\n\r"))
print errs
They differ only in the positioning of the quoted backslash.
Just in case it is of interest to anyone.
You have to escape twice; once for Python and once for the regular
expression. Or use raw strings, denoted by an r"..." prefix:
re.findall("[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!);\\\]%_>?]", "abc")
[]
re.findall("[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!);\\\\\\]%_>?]", "abc")
['a', 'b', 'c']
re.findall(r"[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!);\\\]%_>?]", "abc")
['a', 'b', 'c']
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