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I am very new to the coding world and have been stuck on this one problem for 3 days now, searching everywhere for an answer, so any help will be greatly appreciated. I am needing to extract a small amount of text from a url-located Pdf file. I'm using sessions.get(chart_PDF) as the driver for locating the URL where chart_PDF is the example url below.
Example url is https://www.airservicesaustralia.com/aip/pending/dap/PADGN01-166_09SEP2021.pdf
I know I am able to write it to my local drive but I don't want to do that, I want to be able to do it remotely, since I only need a couple of numbers from it.
I have tried finding the password from the url page for decrypting, couldn't find. I've tried to use PyPDF2, pdfminer and pikepdf (probably not well).
I only need to retrieve two numbers near the bottom of the PDF that can be used for the rest of my code. Please help, even if it is a simple fix, I'm new to all this and need some help. Thanks.
from io import BytesIO
from pikepdf import Pdf as PDF
from pdfminer import high_level
chart_PDF = https://www.airservicesaustralia.com/aip/pending/dap/PADGN01-166_09SEP2021.pdf
retrieve = s.get(chart_PDF)
content = retrieve.content
response =urllib.request.urlopen(chart_PDF)
p = BytesIO(content)
p.getbuffer()
check = PDFPage.get_pages(p, check_extractable=False)
extract = high_level.extract_text(p)
I'm getting:
PDFTextExtractionNotAllowedWarning: The PDF <_io.BytesIO object at 0x000001B007ABEC20> contains a metadata field indicating that it should not allow text extraction. Ignoring this field and proceeding.warnings.warn(warning_msg, PDFTextExtractionNotAllowedWarning)
Alternately, if I try this:
from pikepdf import Pdf as PDF
from pdfminer.pdfpage import PDFPage
from PyPDF2 import PdfFileReader
new_pdf = PDF.new()
with PDF.open(p) as pdf:
print(len(pdf.pages))
page1 = pdf.pages[0]
if PdfFileReader.getIsEncrypted(pdf):
print(True)
PdfFileReader.decrypt(page1, password='')
pdf.close()
I get:
line 1987, in decrypt
return self._decrypt(password)
AttributeError: _decrypt
UPDATE 3/8/21
Thank you so much K J! You've seriously been a huge help!
from io import BytesIO
from pdfminer.pdfpage import PDFPage
from pdfminer import high_level
retrieve = s.get(chart_PDF)
content = retrieve.content
bytes = BytesIO(content)
bytes.getbuffer()
PDFPage.get_pages(bytes, check_extractable=False)
extract = high_level.extract_text(bytes, password='') #THIS LINE THROWS ERROR
joined = ''.join(extract)
find_txt = re.findall(r'[(]\d*[-]\d[.]\d[)]', joined)
print(find_txt)
bytes.close()
This is now working well and I have been able to pull the numbers that I need (I have basically pulled all numbers from inside brackets off the PDF). I'll sort through that to find which one I need.
Strangely enough, although its giving me what I need, my extract = high_level.extract_text(bytes, password='') line still throws the Warning: (warning_msg, PDFTextExtractionNotAllowedWarning) which is rather annoying. Not sure how this process works but its still letting the info out.
I can't use try except or it skips over it. What is the way around this? how can I stop that error coming up?
FINAL UPDATE
I got around the warning and it works well now.
with warnings.catch_warnings():
warnings.simplefilter("ignore")
extract = high_level.extract_text(bytes)
Cheers fellas for putting up with my ignorance, you've helped so much.

The whole file has to be downloaded to a device via RAM so the blob as a FILE can be parsed at the very END for one OR more %%EOF and the location of page 0 (it gets converted to 1 or i) it could be ANYWHERE IN THE STREAM,.
THEN you can navigate to other sequential numbered pages in the RANDOM order they are built. Any complaints please contact Adobe.
However it is easiest if it is cached as a physical FILE object. If you dont want that on disk use a ram drive for your browser.
Again those two objects at bottom of page one could be anywhere mixed into the content of "page" 99's objects, or otherwise. each letter in a PDF can in its extreme be more than one object anywhere in the file. but a good authoring editor would try to keep them as lines by lines. (there is no such PDF thing as a word or paragraph.)
We can Print the file as Plain Text to see how it is composited and although (secured) that is allowed.
I tried printing from browser with little success but know that can depend on browser system and OS print drivers. Here I have printed the page as text using Acrobat portable, so we can see the sequential offsets of each text block from Left Hand margin JUST LIKE a PDF VIEWER would need to rebuild them.
UPDATE
You said your target is (1380-4.4) to the RIGHT of ALTERNATE but again A PDF has no concept of Left and Right or BEFORE or AFTER so we find IN THIS FILE the variable target is in 2 separate pieces PRIOR to the KNOWN characters which luckily is a complete single block (alternate). Thus here proximity of plain text could well work if the capture is confined to that nearby locality. However there is no guarantee that ALTERNATE would always be a single block.
It was perhaps not a good Idea To show the way a Printer would be given a stream of sequential data
Here is the way one PDF viewer goes about decrypting the file
As stated on this occasion the word ALTERNATE is defined as text however the next item is the "3" under "B" which is text as a vector path it is not called a "character" although it looks like one but a numbered glyph from a font table. We do see later that some of those numbers are stored as "text" and for your target it is mixed in with similar text in the same object.
Thus you need to call a PDF interpreter to give you a meaningful translation of all bits and pieces of objects so that you can extract the "right" text.
The easiest way for a "simple" one line target in a complex file is to use MuPDF to first tidy up the file
mutool clean -gggg -D infile.pdf outfile.pdf
combined with
PDFTOTXT -layout outfile.pdf outfile.txt
or similar to hopefully export that text on a line by line basis, such that you can consistently find your target instantly before ! or after ALTERNATE.
N.B Mutool convert to HTML would place the target value in a table entry AFTER the key word, and if the lines are consistent in number that would be a simpler way to find or grep.

How do you correctly parse web links to avoid a 403 error when using Wget?

I just started learning python yesterday and have VERY minimal coding skill. I am trying to write a python script that will process a folder of PDFs. Each PDF contains at least 1, and maybe as many as 15 or more, web links to supplemental documents. I think I'm off to a good start, but I'm having consistent "HTTP Error 403: Forbidden" errors when trying to use the wget function. I believe I'm just not parsing the web links correctly. I think the main issue is coming in because the web links are mostly "s3.amazonaws.com" links that are SUPER long.
For reference:
Link copied directly from PDF (works to download): https://s3.amazonaws.com/os_uploads/2169504_DFA%20train%20pass.PNG?AWSAccessKeyId=AKIAIPCTK7BDMEW7SP4Q&Expires=1909634500&Signature=aQlQXVR8UuYLtkzjvcKJ5tiVrZQ=&response-content-disposition=attachment;%20filename*=utf-8''DFA%2520train%2520pass.PNG
Link as it appears after trying to parse it in my code (doesn't work, gives "unknown url type" when trying to download): https%3A//s3.amazonaws.com/os_uploads/2169504_DFA%2520train%2520pass.PNG%3FAWSAccessKeyId%3DAKIAIPCTK7BDMEW7SP4Q%26Expires%3D1909634500%26Signature%3DaQlQXVR8UuYLtkzjvcKJ5tiVrZQ%253D%26response-content-disposition%3Dattachment%253B%2520filename%252A%253Dutf-8%2527%2527DFA%252520train%252520pass.PNG
Additionally if people want to weigh in on how I'm doing this in a stupid way. Each PDF starts with a string of 6 digits, and once I download supplemental documents I want to auto save and name them as XXXXXX_attachY.* Where X is the identifying string of digits and Y just increases for each attachment. I haven't gotten my code to work enough to test that, but I'm fairly certain I don't have it correct either.
Help!
#!/usr/bin/env python3
import os
import glob
import pdfx
import wget
import urllib.parse
## Accessing and Creating Six Digit File Code
pdf_dir = "/users/USERNAME/desktop/worky"
pdf_files = glob.glob("%s/*.pdf" % pdf_dir)
for file in pdf_files:
## Identify File Name and Limit to Digits
filename = os.path.basename(file)
newname = filename[0:6]
## Run PDFX to identify and download links
pdf = pdfx.PDFx(filename)
url_list = pdf.get_references_as_dict()
attachment_counter = (1)
for x in url_list["url"]:
if x[0:4] == "http":
parsed_url = urllib.parse.quote(x, safe='://')
print (parsed_url)
wget.download(parsed_url, '/users/USERNAME/desktop/worky/(newname)_attach(attachment_counter).*')
##os.rename(r'/users/USERNAME/desktop/worky/(filename).*',r'/users/USERNAME/desktop/worky/(newname)_attach(attachment_counter).*')
attachment_counter += 1
for x in url_list["pdf"]:
print (parsed_url + "\n")```

I prefer to use requests (https://requests.readthedocs.io/en/master/) when trying to grab text or files online. I tried it quickly with wget and I got the same error (might be linked to user-agent HTTP headers used by wget).
wget and HTTP headers issues : download image from url using python urllib but receiving HTTP Error 403: Forbidden
HTTP headers : https://developer.mozilla.org/en-US/docs/Web/HTTP/Headers/User-Agent
The good thing with requests is that it lets you modify HTTP headers the way you want (https://requests.readthedocs.io/en/master/user/quickstart/#custom-headers).
import requests
r = requests.get("https://s3.amazonaws.com/os_uploads/2169504_DFA%20train%20pass.PNG?AWSAccessKeyId=AKIAIPCTK7BDMEW7SP4Q&Expires=1909634500&Signature=aQlQXVR8UuYLtkzjvcKJ5tiVrZQ=&response-content-disposition=attachment;%20filename*=utf-8''DFA%2520train%2520pass.PNG")
with open("myfile.png", "wb") as file:
file.write(r.content)
I'm not sure I understand what you're trying to do, but maybe you want to use formatted strings to build your URLs (https://docs.python.org/3/library/stdtypes.html?highlight=format#str.format) ?
Maybe checking string indexes is fine in your case (if x[0:4] == "http":), but I think you should check python re package to use regular expressions to catch the elements you want in a document (https://docs.python.org/3/library/re.html).
import re
regex = re.compile(r"^http://")
if re.match(regex, mydocument):
<do something>

The reason for this behavior is inside wget library. Inside it encodes the URL with urllib.parse.quote() (https://docs.python.org/3/library/urllib.parse.html#urllib.parse.quote).
Basically it replaces characters with their appropriate %xx escape character. Your URL is already escaped but the library does not know that. When it parses the %20 it sees % as a character that needs to be replaced so the result is %2520 and different URL - therefore 403 error.
You could decode that URL first and then pass it, but then you would have another problem with this library because your URL has parameter filename*= but the library expects filename=.
I would recommend doing something like this:
# get the file
req = requests.get(parsed_url)
# parse your URL to get GET parameters
get_parameters = [x for x in parsed_url.split('?')[1].split('&')]
filename = ''
# find the get parameter with the name
for get_parameter in get_parameters:
if "filename*=" in get_parameter:
# split it to get the name
filename = get_parameter.split('filename*=')[1]
# save the file
with open(<path> + filename, 'wb') as file:
file.write(req.content)
I would also recommend removing the utf-8'' in that filename because I don't think it is actually part of the filename. You could also use regular expressions for getting the filename, but this was easier for me.

In Python, urllib.urlretrieve downloads a file which says "Go away"

I'm trying to download the (APK) files from links such as https://www.apkmirror.com/wp-content/themes/APKMirror/download.php?id=215041. When you enter the link in your browser, it brings up a dialog to open or save the file (see below).
I would like to save the file using a Python script. I've tried the following:
import urllib
download_link = 'https://www.apkmirror.com/wp-content/themes/APKMirror/download.php?id=215041'
download_file = '/tmp/apkmirror_test/youtube.apk'
if __name__ == "__main__":
urllib.urlretrieve(url=download_link, filename=download_file)
but the resulting youtube.apk contains only the words "Go away".
Since I am able to download the file by pasting the link in my browser's address bar, there must be some difference between that and urllib.urlretrieve that makes this not work. Can someone explain this difference and how to eliminate it?

You should not programmatically access that download page as it is disallowed in the robots.txt:
https://www.apkmirror.com/robots.txt
That being said, your request header is different. Python by default sets User-Agent to something like "Python...". That is the most likely cause of detection.

urllib2 download HTML file

Using urllib2 in Python 2.7.4, I can readily download an Excel file:
output_file = 'excel.xls'
url = 'http://www.nbmg.unr.edu/geothermal/GEOTHERM-30Jun11.xls'
file(output_file, 'wb').write(urllib2.urlopen(url).read())
This results in the expected file that I can use as I wish.
However, trying to download just an HTML file gives me an empty file:
output_file = 'webpage.html'
url = 'http://www.nbmg.unr.edu/geothermal/mapfiles/nvgeowel.html'
file(output_file, 'wb').write(urllib2.urlopen(url).read())
I had the same results using urllib. There must be something simple I'm missing or don't understand. How do I download an HTML file from a URL? Why doesn't my code work?

If you want to download files or simply save a webpage you can use urlretrieve(from urllib library)instead of use read and write.
import urllib
urllib.urlretrieve("http://www.nbmg.unr.edu/geothermal/mapfiles/nvgeowel.html","doc.html")
#urllib.urlretrieve("url","save as..")
If you need to set a timeout you have to put it at the start of your file:
import socket
socket.setdefaulttimeout(25)
#seconds

It also Python 2.7.4 in my OS X 10.9, and the codes work well on it.
So I think there maybe other problems prevent its working. Can you open "http://www.nbmg.unr.edu/geothermal/GEOTHERM-30Jun11.xls" in your browser?

This may not directly answer the question, but if you're working with HTTP and have sufficient privileges to install python packages, I'd really recommend doing this with 'requests'. There's a related answered here - https://stackoverflow.com/a/13137873/45698

Creating pdfs in Python with Pisa / xhtml2pdf

I know there are a lot of questions based on pdf creation in Python but I haven't seen anything based on creating pdfs with Pisa or xhtml2pdf.
Here is my code.
pisa.pisaDocument(cStringIO.StringIO(a).encode('utf-8'),file('mypdf.pdf','wb'))
and then
pisa.startViewer('mypdf.pdf')
I assembled this over a couple different tutorials and examples but every single thing that I've tried always results in the pdf being corrupted and I get this message when trying to open the pdf.
"Adobe Reader could not open 'awesomer.pdf' because it is either not a supported file type or because the file has been damaged (for example, it was sent as an email attachment and wasn't correctly decoded)."
This message occurs even when I don't use the .encode('utf-8') on the string.
What am I doing wrong? Does the encoding on my Mac have to do with this?

I'd suggest closing the file manually, had a simmilar problem. Try this:
f = file('mypdf.pdf', 'wb')
pisa.pisaDocument(cStringIO.StringIO(a).encode('utf-8'),f)
f.close()

I recommend doing the following:
pdf = pisa.pisaDocument(cStringIO.StringIO(a).encode('utf-8'),file('mypdf.pdf','wb'))
if pdf.err:
print "*** %d ERRORS OCCURED" % pdf.err
And then see what the error output is.
I'm not sure what string you are encoding but this might also help:
pdf = pisa.pisaDocument(cStringIO.StringIO(html.encode(a)).encode('utf-8'),file('mypdf.pdf','wb'))
It depends on if a needs to be html encoded