pyPdf error invalid argument

Question

I'm actually using pyPdf to open, read and write the content of a PDF file.

for that I use these lines of code :

from pyPdf import PdfFileWriter, PdfFileReader

pdf = PdfFileReader(file("/myPdfFile.pdf", "w+b"))
content = pdf.getPage(1).extractText()
print content

But it returns me this error and I don't understand exactly why

File "/usr/local/lib/python2.6/dist-packages/pyPdf/pdf.py", line 374, in __init__
    self.read(stream)
File "/usr/local/lib/python2.6/dist-packages/pyPdf/pdf.py", line 702, in read
    stream.seek(-1, 2)
IOError: [Errno 22] Invalid argument

Anyone can help me?

Facing same issue on Newly Installed Amazon Server-Ubuntu – Tejas Tank Oct 16 '15 at 11:25 — Tejas Tank, Oct 16 '15 at 11:25

score 1 · Answer 1 · answered May 22 '12 at 17:24

1

As it says in the Python docs, the mode 'w+b' opens and truncates the file to 0 bytes, while 'r+b' opens the file without truncation.

answered May 22 '12 at 17:24

MRAB

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Try again after recovering your pdf from backup - it is now 0 bytes long – Nick Craig-Wood May 22 '12 at 17:36
No, I actually get the same error `IOError: [Errno 22] Invalid argument` – kschaeffler May 22 '12 at 17:42
When I try to open my PDF using the open('mypdf.pdf', 'rb') function I retrieve an empty file. I also don't understand why – kschaeffler May 23 '12 at 12:34

pyPdf error invalid argument

1 Answers1