Python findall, regex

Question

I have this text:

  u'times_viewed': 12268,
  u'url': u'/photo/79169307/30-seconds-light',
  u'user': {u'affection': 63962,

How can I just get out this string: "/photo/79169307/30-seconds-light"?

I am trying with regex and findall:

list = re.findall(‘u‘url‘: u‘/photo/"([^"]*)"‘, text)

but it won't go.

That looks like output from a python data structure. If it is, maybe try: object_name['url'] instead of printing it out? If not, then could you post the relevant portion of your failing code? — vch
– vch, Commented Aug 10, 2014 at 13:37

Patrick Collins · Accepted Answer · 2014-08-10 14:09:56Z

2

I assume that by "it won't go," you mean that you get a syntax error, which you should. Here:

list=re.findall(‘u‘url‘: u‘/photo/"([^"]*)"‘,text)

you're using " when you mean '. This is causing a syntax error because " closes the string you're trying to pass re.findall. Try:

list_ = re.findall("u'url': u'/photo/([^']*)'", text)

Additionally, this isn't going to grab the text after photo, so you'll need to add more parens:

list_ = re.findall("u'url': u'(/photo/([^']*))'", text)

and now list_.group(1) should hold your string.

On top of that, it looks like you're dealing with JSON. A better approach might be:

import json
json.loads(text)
list_ = text['url']

edited Aug 10, 2014 at 14:09

answered Aug 10, 2014 at 13:45

Patrick Collins

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2 Comments

xZise Over a year ago

For your information: the regexes need to be closed with a quote charachter: re.findall("u'url': u'/photo/([^']*)'", text) and "u'url': u'(/photo/([^']*))'"

Patrick Collins Over a year ago

@xZise I think I fixed all of the typos. Let me know if there's anything else I missed.

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