2

I have the following code snippet from page source: 

var myPDF = new PDFObject({
url: "http://www.site.com/doc55.pdf",
  id: "pdfObjectContainer",
  width: "100%",
  height: "700px",
  pdfOpenParams: {
    navpanes: 0,
    statusbar: 1,
    toolbar: 1,
    view: "FitH"
  }
}).embed("pdf_placeholder");

the

'PDFObject('

is unique on the page. I want to retreive url content using REGEX. In this case I need to get

http://www.site.com/doc55.pdf

Please help.

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1
  • 1
    Regex should work pretty good for this. Commented Jul 4, 2013 at 20:32

7 Answers 7

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3

Here is an alternative for solving your problem without using regex:

url,in_object = None, False
with open('input') as f:
    for line in f:
        in_object = in_object or 'PDFObject(' in line
        if in_object and 'url:' in line:
            url = line.split('"')[1]
            break
print url
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3 Comments

Why on hearth OPs can't be helped with the tools they want to use? There is always someone to tell them "Hey dude! That's not HOW to think about it..."
you picked this out within all those regex answers?
Good answer, I agree that regex is not the best tool for this. But you should probably break the loop after finding the url (or just put the code into a function and return), otherwise you could have false positives if other lines contain "url:".
0

In order to be able to find "something that happens in the line after something else", you need to match things "including the newline". For this you use the (dotall) modifier - a flag added during the compilation.

Thus the following code works:

import re
r = re.compile(r'(?<=PDFObject).*?url:.*?(http.*?)"', re.DOTALL)
s = '''var myPDF = new PDFObject({
url: "http://www.site.com/doc55.pdf",
  id: "pdfObjectContainer",
  width: "100%",
  height: "700px",
  pdfOpenParams: {
    navpanes: 0,
    statusbar: 1,
    toolbar: 1,
    view: "FitH"
  }
}).embed("pdf_placeholder"); '''

print r.findall(s)

Explanation:

r = re.compile(         compile regular expression
    r'                  treat this string as a regular expression
    (?<=PDFObject)      the match I want happens right after PDFObject
    .*?                 then there may be some other characters...
    url:                followed by the string url:
    .*?                 then match whatever follows until you get to the first instance (`?` : non-greedy match of
    (http:.*?)"         match the string http: up to (but not including) the first "
    ',                  end of regex string, but there's more...
    re.DOTALL)          set the DOTALL flag - this means the dot matches all characters
                        including newlines. This allows the match to continue from one line
                        to the next in the .*? right after the lookbehind
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2 Comments

Thanks a lot Floris. Your code is the shortest and it works just fine:)
Glad it worked for you. Was an opportunity for me to figure out the re.DOTALL thing... I knew it existed, had not used it, this was my chance to learn about it. So we both came out ahead.
0

using a combination of look-behind and look-ahead assertions

import re
re.search(r'(?<=url:).*?(?=",)', s).group().strip('" ')
'http://www.site.com/doc55.pdf'
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Comments

0

This works:

import re

src='''\
var myPDF = new PDFObject({
url: "http://www.site.com/doc55.pdf",
URL: "http://www.site.com/doc52.PDF",
  id: "pdfObjectContainer",
  width: "100%",
  height: "700px",
  pdfOpenParams: {
    navpanes: 0,
    statusbar: 1,
    toolbar: 1,
    view: "FitH"
  }
}).embed("pdf_placeholder"); '''   

print [m.group(1).strip('"') for m in 
        re.finditer(r'^url:\s*(.*)[\W]$',
        re.search(r'PDFObject\(\{(.*)',src,re.M | re.S | re.I).group(1),re.M|re.I)]

prints:

['http://www.site.com/doc55.pdf', 'http://www.site.com/doc52.PDF']
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Comments

0

Regex

new\s+PDFObject\(\{\s*url:\s*"[^"]+"

Regular expression image

Demo

Extract url only

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1 Comment

This doesn't address the "after PDFObject" part. There will be other instances of url: "http:.*" on the page - OP wants a specific one.
0

If 'PDFObject(' is the unique identifier in the page, you only have to match the first next quoted content.

Using the DOTALL flag (re.DOTALL or re.S) and the non-greedy star (*?), you can write:

import re

snippet = '''                                    
var myPDF = new PDFObject({
url: "http://www.site.com/doc55.pdf",
  id: "pdfObjectContainer",
  width: "100%",
  height: "700px",
  pdfOpenParams: {
    navpanes: 0,
    statusbar: 1,
    toolbar: 1,
    view: "FitH"
  }
}).embed("pdf_placeholder");
'''

# First version using unnamed groups
RE_UNNAMED = re.compile(r'PDFObject\(.*?"(.*?)"', re.S)

# Second version using named groups
RE_NAMED = re.compile(r'PDFObject\(.*?"(?P<url>.*?)"', re.S)

RE_UNNAMED.search(snippet, re.S).group(1)
RE_NAMED.search(snippet, re.S).group('url')
# result for both: 'http://www.site.com/doc55.pdf'

If you don't want to compile your regex because it's used once, simply this syntax:

re.search(r'PDFObject\(.*?"(.*?)"', snippet, re.S).group(1)
re.search(r'PDFObject\(.*?"(?P<url>.*?)"', snippet, re.S).group('url')

Four choices, one should match you need and taste!

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0

Although the other answers may appear to work, most do not take into account that the only unique thing on the page is 'PDFObject('. A much better regular expression would be the following:

PDFObject\({\surl: "(http[s]?://(?:[a-zA-Z]|[0-9]|[$-_@.&+]|[!*\(\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+)",

It takes into account that 'PDFObject(' is unique and contains some basic URL verification.

Below is an example of how this regex could be used in python

>>> import re
>>> strs = """var myPDF = new PDFObject({
... url: "http://www.site.com/doc55.pdf",
...   id: "pdfObjectContainer",
...   width: "100%",
...   height: "700px",
...   pdfOpenParams: {
...     navpanes: 0,
...     statusbar: 1,
...     toolbar: 1,
...     view: "FitH"
...   }
... }).embed("pdf_placeholder");"""
>>> re.search(r'PDFObject\({\surl: "(http[s]?://(?:[a-zA-Z]|[0-9]|[$-_@.&+]|[!*\(\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+)",',strs).group(1)
'http://www.site.com/doc55.pdf'

A pure python (no regex) alternative would be:

>>> unique = 'PDFObject({\nurl: "'
>>> start = strs.find(unique) + len(unique)
>>> end = start + strs[start:].find('"')
>>> strs[start:end]
'http://www.site.com/doc55.pdf'

No regex oneliner:

>>> (lambda u:(lambda s:(lambda e:strs[s:e])(s+strs[s:].find('"')))(strs.find(u)+len(u)))('PDFObject({\nurl: "')
'http://www.site.com/doc55.pdf'
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4 Comments

Not sure the link validation is needed, but I appreciate that matching http: in my example was going one character too far, as it would skip any https: links - I have modified my answer, and thanks. Does your regex permit all legal links (even ones with URL encoded queries attached)? It's a bit hard to be sure...
@Floris yes this regex accepts all links, even ones with URL encoded queries, given their protocol is either http or https.
That's cool - I will keep a copy, might come in handy. Of your own making, or did you find it somewhere?
I think I found it somewhere, can't remember where though. I used it in one of my projects a while back, just copied it out of there for this.

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