Re: re.findall() hangs in python

[EMAIL PROTECTED] wrote: I have the following regular expression. It works when 'data' contains the pattern and I see 'match2' get print out. But when 'data' does not contain pattern, it just hangs at 're.findall' pattern = re.compile((.*)img (.*?) src=\(.*?)img(.*?)\(.*?), re.S)

Re: re.findall() hangs in python

On Mar 31, 9:12 pm, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: Hi, I have the following regular expression. It works when 'data' contains the pattern and I see 'match2' get print out. But when 'data' does not contain pattern, it just hangs at 're.findall' pattern = re.compile((.*)img

Re: re.findall() hangs in python

En Sun, 01 Apr 2007 03:58:51 -0300, Peter Otten [EMAIL PROTECTED] escribió: [EMAIL PROTECTED] wrote: I have the following regular expression. It works when 'data' contains the pattern and I see 'match2' get print out. But when 'data' does not contain pattern, it just hangs at

Re: re.findall() hangs in python

On Apr 1, 6:12 am, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: But when 'data' does not contain pattern, it just hangs at 're.findall' pattern = re.compile((.*)img (.*?) src=\(.*?)img(.*?)\(.*?), re.S) That pattern is just really slow to evaluate. What you want is probably something more like

Re: re.findall() hangs in python

On Apr 1, 5:23 am, [EMAIL PROTECTED] wrote: On Apr 1, 6:12 am, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: But when 'data' does not contain pattern, it just hangs at 're.findall' pattern = re.compile((.*)img (.*?) src=\(.*?)img(.*?)\(.*?), re.S) That pattern is just really slow to

re.findall() hangs in python

Hi, I have the following regular expression. It works when 'data' contains the pattern and I see 'match2' get print out. But when 'data' does not contain pattern, it just hangs at 're.findall' pattern = re.compile((.*)img (.*?) src=\(.*?)img(.*?)\(.*?), re.S) print before find all