[SOLVED] Regular expression question in python...

zinon75 · 11-09-2015, 07:01 AM

Code:

#!/usr/bin/python

import re

abc = """
random text
random text
random text
flashvars="flvsource=http://zzz.tvstation.tv/uploads/abcd345mngjrUeoKSDfd.flv&preview_image=http://a.jpg&backgcolor=FFFFFF&autoplay=true&url_logo=http://logo-player.png&logo=top_right&floating_navbar=false&color_nav_bar_top=0x478dc2&color_nav_bar_bottom=0xE7EBEC&ads_background_color=0x00CCFF&ads_border_color=0xCCCCCC&scrubber_position_color=0x6AA1CE&scrubber_load_color=0x888888&scrubber_background_color=0xBBBBBB&volume_bar_color=0xBBBBBB&aspect_ratio=stretch"></embed>' onClick="javascript:document.code.embed.focus();document.code.embed.select(); return false;" />
random text
random text
dsfgdsgdgdgdgdgdfgdgdfretrtsdgdf
"""

flvsource = re.findall('http://zzz.tvstation.tv/uploads/abcd345mngjrUeoKSDfd.flv', abc)
flvsource2 = re.findall('http://zzz.tvstation.tv/uploads/*.flv', abc)

print flvsource
print flvsource2

i need to extract this kind of string:
"http://zzz.tvstation.tv/uploads/abcd345mngjrUeoKSDfd.flv"

with something similar to this:
"http://zzz.tvstation.tv/uploads/*.flv"

how can i do this ?

grail · 11-09-2015, 08:00 AM

"http://zzz.tvstation.tv/uploads/*.flv" :- looking at this I would imagine it is a glob and not a regex, as it currently says:

find the string "http://zzz.tvstation.tv/uploads" followed by zero or more slashes (/*) followed by any character (.) and then the string "flv"

As you can see this would never match your previous string. If you want to practice your regex skills, may i suggest the following site:

http://www.rubular.com/

This should help you to build the necessary matching pattern

zinon75 · 11-09-2015, 01:19 PM

grail,

i got this working:
"http:\/\/zzz.tvstation.tv\/uploads\/.....................flv"

19 "." dots of 20 for the characters and numbers and the last dot for the ".flv".

I can't figure out how to do the "*.flv" without the dots.

grail · 11-09-2015, 02:06 PM

Did you go to the site I recommended? If so, check out the quick reference at the bottom and specifically the last column on the right.

I will show you a gotcha line that will bugger up your regex and maybe something you have to be careful of:

Code:

flashvars="flvsource=http://zzz.tvstation.tv/uploads/abcd345mngjrUeoKSDfd.flv&preview_image=http://a.jpg&backgcolor=FFFFFF&autoplay=true&url_logo=http://logo-player.png_flv&logo=top_right&floating_navbar=false&color_nav_bar_top=0x478dc2&color_nav_bar_bottom=0xE7EBEC&ads_background_color=0x00CCFF&ads_border_color=0xCCCCCC&scrubber_position_color=0x6AA1CE&scrubber_load_color=0x888888&scrubber_background_color=0xBBBBBB&volume_bar_color=0xBBBBBB&aspect_ratio=stretch"></embed>' onClick="javascript:document.code.embed.focus();document.code.embed.select(); return false;" />

HMW · 11-10-2015, 02:51 AM

Quote:

Originally Posted by zinon75

I can't figure out how to do the "*.flv" without the dots.

It's ok buddy, you are actually really close. Try to invert the sequence. Instead of *. try .*

Since you've been working hard at this, I'll show you one possible approach. With this file (flv.txt):

Code:

http://zzz.tvstation.tv/uploads/abcd345mngjrUeoKSDfd.flv&preview_image=http://a.jpg&backgcolor=FFFFFF&autoplay=true&url_logo=http://logo-player.png&logo=top_right&floating_navbar=false&color_nav_bar_top=0x478dc2&color_nav_bar_bottom=0xE7EBEC&ads_background_color=0x00CCFF&ads_border_color=0xCCCCCC&scrubber_position_color=0x6AA1CE&scrubber_load_color=0x888888&scrubber_background_color=0xBBBBBB&volume_bar_color=0xBBBBBB&aspect_ratio=stretch"></embed>' onClick="javascript:document.code.embed.focus();document.code.embed.select(); return false;

I get the result you want with this regex (in bold):

Code:

grep -o 'http.*flv' flv.txt 
http://zzz.tvstation.tv/uploads/abcd345mngjrUeoKSDfd.flv

Check out grail's link, regular expressions can be a pain in the b.tt, but the are very useful!

Best regards,
HMW

zinon75 · 11-10-2015, 02:53 AM

grail,

"http:\/\/zzz.tvstation.tv\/uploads\/.{1,20}.flv"

ok, this is working right.

"check out the quick reference at the bottom and specifically the last column on the right."
that helped me out

zinon75 · 11-10-2015, 03:04 AM

HMW,

that works too.

I use the "?" and "*" in bash all the time,
but i'm not familiar with the regular expressions
in python or in ruby.

HMW · 11-10-2015, 04:10 AM

Quote:

Originally Posted by zinon75

HMW,

that works too.

I use the "?" and "*" in bash all the time,
but i'm not familiar with the regular expressions
in python or in ruby.

They are pretty much the same.
.* means (roughly) "repeat any character any number of times", whereas
.{1,20} means repeat any character between one and 20 times.

So, as you have noticed, they both work for you in this particular case.

Best regards,
HMW

grail · 11-10-2015, 10:23 AM

I would also point out, that had you used HMW's suggested solution on the extra link I provided, this is where you might get tripped up, however, part of writing regular expressions also relies on how well you know the data, ie. will you never have a file name longer than 20 characters??

HMW · 11-10-2015, 10:54 AM

Quote:

Originally Posted by grail

I would also point out, that had you used HMW's suggested solution on the extra link I provided, this is where you might get tripped up

^True. In that case, you'd have to use this regex:

Code:

grep -o 'http.*\.flv' flv.txt 
http://zzz.tvstation.tv/uploads/abcd345mngjrUeoKSDfd.flv

Best regards,
HMW

grail · 11-10-2015, 11:31 AM

hmmm ... yes in that case it would work and i am also not familiar with what alterations there might be, ie. there could be another '.flv' file name further along the string.
My point was to show the issue of greediness can sometimes be a limiting factor