Monday, August 7, 2017

RE: Django Python OSError No such file or directory but file exists

That's a great idea.  How do you do that programmatically?

 

From: django-users@googlegroups.com [mailto:django-users@googlegroups.com] On Behalf Of "????????? ????????? (roboslone)"
Sent: Monday, August 7, 2017 4:32 PM
To: django-users@googlegroups.com
Subject: Re: Django Python OSError No such file or directory but file exists

 

You can wait for subprocess to finish and not rely on time.sleep.

 

On 7 Aug 2017, at 23:57, Ronaldo Bahia <ronaldo@jobconvo.com> wrote:

 

Turns out unoconv takes 2 seconds to perform the file conversion.

So after the file conversion, I had to set time.sleep(3) before upload a file to S3.

 

And after 1 week I got this working using variables.

 

Thanks


Ronaldo Bahia

+55 11 3280 6971

+55 11 963 622 581

 

Materiais gratuitos para o RH:

 

2017-08-07 17:17 GMT-03:00 Ronaldo Bahia <ronaldo@jobconvo.com>:

I don't know why but if I set the string ("cv.pdf"), it process just fine.

If I use a variable instead, it doesn't.

 

Here is the working code:

 

        # convert to PDF

        env = os.environ.copy()

        env['HOME'] = '/tmp'

        subprocess.Popen(["unoconv","-f", "pdf", "-o", "cv.pdf","%s" % (file_in)], env = env)

 

        # Define S3 path

        resume_path = 'resumes/%s/' % str(date.today())

        

        # key is the S3 file name

        key = '%s%s' % (resume_path, file_out)

        

        # delete local file

        subprocess.call("rm -f %s" % user_cv_file, shell=True)

        

        # update the new file format

        user_cv.resume = key

        user_cv.save()

        

        # S3 Connection

        conn = S3Connection(settings.AWS_ACCESS_KEY_ID, settings.AWS_SECRET_ACCESS_KEY)

        bucket_out = Bucket(conn, settings.AWS_STORAGE_BUCKET_NAME)

        k_out = Key(bucket=bucket_out, name=user_cv.resume)

 

        # Upload to S3

        k_out.set_contents_from_filename('cv.pdf')

        k_out.make_public()

 

        # deleta o arquivo localmente

        subprocess.call("rm -f cv.pdf", shell=True)


Ronaldo Bahia

+55 11 963 622 581

 

Materiais gratuitos para o RH:

 

2017-08-07 8:16 GMT-03:00 "Александр Христюхин (roboslone)" <roboslone@gmail.com>:

Well, yeah, you're using local path basically. And you should NEVER use .split('/')[-1] to determine file basename, take a look at os.path module instead.

 

But that's not the point. You should try to use absolute path, I believe it would do the trick.

 

On 4 Aug 2017, at 17:49, Ronaldo Bahia <ronaldo@jobconvo.com> wrote:

 

the method is called in a def post() within a class view:

 

user_cv = CandidateCV.objects.get(user=request.user)

user_cv_file = str(user_cv.resume).split('/')[-1]

s3upload(user_cv_file)

 

The file is converted from doc to pdf in server project folder, where manage.py is.

So probably there is no absolute path.

 

How can I solve it?

 

Thanks

Ronaldo Bahia

+55 11 963 622 581

 

Materiais gratuitos para o RH:

 

2017-08-03 1:36 GMT-03:00 "Александр Христюхин (roboslone)" <roboslone@gmail.com>:

Hi,

 

Are you sure s3file contains absolute path? I can't see where s3upload is being called.

 

Also, you might wanna use os.remove instead of calling subprocess.

You also might wanna check out PEP-8 and Sphinx for your docstrings.

 

On 2 Aug 2017, at 02:28, Ronaldo Bahia <ronaldo@jobconvo.com> wrote:

 

Hi everyone, can you help me?

Thanks in advance

 

 

Code:

 



down votefavorite

I'm converting doc and docx files to pdf in the server using unoconv with LibreOffice. And I need to upload to S3 the converted file.

I can convert with success the files and I can see them in the server.

But when I try to upload the pdf, I get the error. What am I missing?

Thanks in advance

This works just fine:

import subprocess
from boto.s3.connection import S3Connection, Bucket, Key
 
def doc_to_pdf(user):
    '''
    Convert doc or docx to PDF.
 
    parameter user: is a request.user
 
    Usage:
        doc_to_pdf(self.request.user):
    '''
 
    user_cv = CandidateCV.objects.get(user=user)
    user_cv_file = str(user_cv.resume).split('/')[-1] # tem que ser PDF
    user_cv_filetype = user_cv_file.split('.')[-1]
 
    if not user_cv_filetype in settings.PDF_FILE_TYPE:
        # Se não for PDF
        file_in = user_cv.resume.url
        file_name = file_in.split('/')[-1]
        # download
        urllib.request.urlretrieve(file_in, file_name)
        file_out = user_cv_file.split('.')[0] + '.pdf'
 
        # converte para PDF
        env = os.environ.copy()
        env['HOME'] = '/tmp'
        subprocess.Popen(["unoconv","-f", "pdf", "%s" % (file_in)], env = env)
 
        # Define a path para salvar o documento na S3
        resume_path = 'resumes/%s/' % str(date.today())
 
        # key é o nome do arquivo na S3
        key = '%s%s' % (resume_path, file_out)
 
        # deleta o arquivo localmente
        subprocess.call("rm -f %s" % user_cv_file, shell=True)
 
        # Salva o novo formato no banco
        user_cv.resume = key
        user_cv.save()

This is the code in which I get the error in line: k_out.set_contents_from_filename(s3file)

def s3upload(s3file):
 
    # Conecta na AWS S3
    conn = S3Connection(settings.AWS_ACCESS_KEY_ID, settings.AWS_SECRET_ACCESS_KEY)
    bucket_out = Bucket(conn, settings.AWS_STORAGE_BUCKET_NAME)
    k_out = Key(bucket=bucket_out, name=s3file)
 
    # Define a path para salvar o documento na S3
    resume_path = 'resumes/%s/' % str(date.today())
 
    # key é o nome do arquivo na S3
    key = '%s%s' % (resume_path, s3file)
    k_out.key = key
 
    # Salva na AWS S3
    k_out.set_contents_from_filename(s3file)
    k_out.make_public()
 
    # deleta o arquivo localmente
    subprocess.call("rm -f %s" % s3file, shell=True)

 

 

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