I want to print a pdf file in python. My code is as below:
def printing_reports():
import os
fp = open("/path-to-file/path.txt",'r')
for line in fp:
os.system('lp -d HPLaserJ %s' % (str(line)))
I am on Fedora 20. path.txt is a file that contain path to the pdf file like '/home/user/a.pdf'
When I run the code it says no such file or directory.
Thanks
Try this code may help:
import os
def printing_reports():
fp = open("/path-to-file/path.txt",'r')
for line in fp:
os.system('lp -d HPLaserJ {0}'.format(line.strip()))
printing_reports()
Make sure the file in every line exists.
Old question, but as I needed a an answer of how to print a pdf file from python, I found this answer more profound:
import cups
conn = cups.Connection()
printers = conn.getPrinters()
printer_name = printers.keys()[0]
ppd_options = {}
cups_job_id = conn.printFile(printer_name,'/path/to/a.pdf',"Title printjob", ppd_connection_options)
It uses the pycups module, which needs CUPS >= 1.7 installed on your system (according to their GitHub page)
The ppd_options dictionary might just be empty. (PPD - Postscript Printer Driver)
Related
update with screenshot of the opened file
i am following along the below tutorials to download a text file from Gutenberg.org, however, when i open the file in finder, the first two characters are not recognizable, others looks all fine.
however, if i open it in terminal with vim, the file looks ok.
Do you have any clues on what may have caused this?
http://automatetheboringstuff.com/chapter11/
i'm on MacOS Mojave 10.14.2, python 3.7.2
#! /usr/bin/env python3
import requests
res = requests.get('http://www.gutenberg.org/cache/epub/1322/pg1322.txt');
res.raise_for_status()
print('book length is: ' + str(len(res.text)//1024) + ' kBytes')
print('book content:\n' + res.text[:300])
bookFile = open('book.txt', 'wb')
for chunk in res.iter_content(100000):
bookFile.write(chunk)
I have a requirement to download and unzip a file from a website. Here is the code I'm using:
#!/usr/bin/python
#geoipFolder = r'/my/folder/path/ ' #Mac/Linux folder path
geoipFolder = r'D:\my\folder\path\ ' #Windows folder path
geoipFolder = geoipFolder[:-1] #workaround for Windows escaping trailing quote
geoipName = 'GeoIPCountryWhois'
geoipURL = 'http://geolite.maxmind.com/download/geoip/database/GeoIPCountryCSV.zip'
import urllib2
response = urllib2.urlopen(geoipURL)
f = open('%s.zip' % (geoipFolder+geoipName),"w")
f.write(repr(response.read()))
f.close()
import zipfile
zip = zipfile.ZipFile(r'%s.zip' % (geoipFolder+geoipName))
zip.extractall(r'%s' % geoipFolder)
This code works on Mac and Linux boxes, but not on Windows. There, the .zip file is written, but the script throws this error:
zipfile.BadZipfile: File is not a zip file
I can't unzip the file using Windows Explorer either. It says that:
The compressed (zipped) folder is empty.
However the file on disk is 6MB large.
Thoughts on what I'm doing wrong on Windows?
Thanks
Your zipfile is corrupt on windows because you're opening the file in write/text mode (line-terminator conversion trashes binary data):
f = open('%s.zip' % (geoipFolder+geoipName),"w")
You have to open in write/binary mode like this:
f = open('%s.zip' % (geoipFolder+geoipName),"wb")
(will still work on Linux of course)
To sum it up, a more pythonic way of doing it, using a with block (and remove repr):
with open('{}{}.zip'.format(geoipFolder,geoipName),"wb") as f:
f.write(response.read())
EDIT: no need to write a file to disk, you can use io.BytesIO, since the ZipFile object accepts a file handle as first parameter.
import io
import zipfile
with open('{}{}.zip'.format(geoipFolder,geoipName),"wb") as f:
outbuf = io.BytesIO(f.read())
zip = zipfile.ZipFile(outbuf) # pass the fake-file handle: no disk write, no temp file
zip.extractall(r'%s' % geoipFolder)
I'm not sure if this can be done or not but I thought I'd ask!
I'm running a windows 10 PC using Python 2.7. I'm wanting to download a file form sharepoint to a folder on my C: drive.
OpenPath = "https://office.test.com/sites/Rollers/New improved files/"
OpenFile = "ABC UK.xlsb"
The downside is the file is uploaded externally & due to human error it can be saved as a .xlsx or ABC_UK. Therefor I only want to use the first 3 characters with a wildcard (ABC*) to open that file. Thankfully the first 3 Characters are unique & there only be one file in the path that should match.
to find the file in your dir:
import os, requests, fnmatch
#loop over dir
for filename in os.listdir(OpenPath):
#find the file
if fnmatch.fnmatch(filename, 'ABC_UK*'):
#download the file
# open file handler
with open('C:\dwnlfile.xls', 'wb') as fh:
#try to get it
result = requests.get(OpenPath+filename)
#check u got it
if not result.ok:
print result.reason # or text
exit(1)
#save it
fh.write(result.content)
print 'got it and saved'
I'm trying to inspect my appengine backup files to work out when a data corruption occured. I used gsutil to locate and download the file:
gsutil ls -l gs://my_backup/ > my_backup.txt
gsutil cp gs://my_backup/LongAlphaString.Mymodel.backup_info file://1.backup_info
I then created a small python program, attempting to read the file and parse it using the appengine libraries.
#!/usr/bin/python
APPENGINE_PATH='/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/'
ADDITIONAL_LIBS = [
'lib/yaml/lib'
]
import sys
sys.path.append(APPENGINE_PATH)
for l in ADDITIONAL_LIBS:
sys.path.append(APPENGINE_PATH+l)
import logging
from google.appengine.api.files import records
import cStringIO
def parse_backup_info_file(content):
"""Returns entities iterator from a backup_info file content."""
reader = records.RecordsReader(cStringIO.StringIO(content))
version = reader.read()
if version != '1':
raise IOError('Unsupported version')
return (datastore.Entity.FromPb(record) for record in reader)
INPUT_FILE_NAME='1.backup_info'
f=open(INPUT_FILE_NAME, 'rb')
f.seek(0)
content=f.read()
records = parse_backup_info_file(content)
for r in records:
logging.info(r)
f.close()
The code for parse_backup_info_file was copied from
backup_handler.py
When I run the program, I get the following output:
./view_record.py
Traceback (most recent call last):
File "./view_record.py", line 30, in <module>
records = parse_backup_info_file(content)
File "./view_record.py", line 19, in parse_backup_info_file
version = reader.read()
File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/api/files/records.py", line 335, in read
(chunk, record_type) = self.__try_read_record()
File "/Applications/GoogleAppEngineLauncher.app/Contents/Resources/GoogleAppEngine-default.bundle/Contents/Resources/google_appengine/google/appengine/api/files/records.py", line 307, in __try_read_record
(length, len(data)))
EOFError: Not enough data read. Expected: 24898 but got 2112
I've tried with a half a dozen different backup_info files, and they all show the same error (with different numbers.)
I have noticed that they all have the same expected length: I was reviewing different versions of the same model when I made that observation, it's not true when I view the backup files of other Modules.
EOFError: Not enough data read. Expected: 24932 but got 911
EOFError: Not enough data read. Expected: 25409 but got 2220
Is there anything obviously wrong with my approach?
I guess the other option is that the appengine backup utility is not creating valid backup files.
Anything else you can suggest would be very welcome.
Thanks in Advance
There are multiple metadata files created when an AppEngine Datastore backup is run:
LongAlphaString.backup_info is created once. This contains metadata about all of the entity types and backup files that were created in datastore backup.
LongAlphaString.[EntityType].backup_info is created once per entity type. This contains metadata about the the specific backup files created for [EntityType] along with schema information for the [EntityType].
Your code works for interrogating the file contents of LongAlphaString.backup_info, however it seems that you are trying to interrogate the file contents of LongAlphaString.[EntityType].backup_info. Here's a script that will print the contents in a human-readable format for each file type:
import cStringIO
import os
import sys
sys.path.append('/usr/local/google_appengine')
from google.appengine.api import datastore
from google.appengine.api.files import records
from google.appengine.ext.datastore_admin import backup_pb2
ALL_BACKUP_INFO = 'long_string.backup_info'
ENTITY_KINDS = ['long_string.entity_kind.backup_info']
def parse_backup_info_file(content):
"""Returns entities iterator from a backup_info file content."""
reader = records.RecordsReader(cStringIO.StringIO(content))
version = reader.read()
if version != '1':
raise IOError('Unsupported version')
return (datastore.Entity.FromPb(record) for record in reader)
print "*****" + ALL_BACKUP_INFO + "*****"
with open(ALL_BACKUP_INFO, 'r') as myfile:
parsed = parse_backup_info_file(myfile.read())
for record in parsed:
print record
for entity_kind in ENTITY_KINDS:
print os.linesep + "*****" + entity_kind + "*****"
with open(entity_kind, 'r') as myfile:
backup = backup_pb2.Backup()
backup.ParseFromString(myfile.read())
print backup
I'm trying to extract new revisions of Chromium.app from their snapshots, and I can download the file fine, but when it comes to extracting it, ZipFile either extracts the chrome-mac folder within as a file, says that directories don't exist, etc. I am very new to python, so these errors make little sense to me. Here is what I have so far.
import urllib2
response = urllib2.urlopen('http://build.chromium.org/buildbot/snapshots/chromium-rel-mac/LATEST')
latestRev = response.read()
print latestRev
# we have the revision, now we need to download the zip and extract it
latestZip = urllib2.urlopen('http://build.chromium.org/buildbot/snapshots/chromium-rel-mac/%i/chrome-mac.zip' % (int(latestRev)), '~/Desktop/ChromiumUpdate/%i-update' % (int(latestRev)))
#declare some vars that hold paths n shit
workingDir = '/Users/slehan/Desktop/ChromiumUpdate/'
chromiumZipPath = '%s%i-update.zip' % (workingDir, (int(latestRev)))
chromiumAppPath = 'chrome-mac/' #the path of the chromium executable within the zip file
chromiumAppExtracted = '%s/Chromium.app' % (workingDir) # path of the extracted executable
output = open(chromiumZipPath, 'w') #delete any current file there
output.write(latestZip.read())
output.close()
# we have the .zip now we need to extract the Chromium.app file, it's in ziproot/chrome-mac/Chromium.app
import zipfile, os
zippedFile = open(chromiumZipPath)
zippedChromium = zipfile.ZipFile(zippedFile, 'r')
zippedChromium.extract(chromiumAppPath, workingDir)
#print zippedChromium.namelist()
zippedChromium.close()
#zippedChromium.close()
Any ideas?
It seems you have encountered a bug in Python. This other question details the problem and workarounds. You can elect to use one of those workarounds, or update to Python 2.6.5 or 2.7b2.
One of the workarounds suggests copying the patched zipfile.py module from the fixed Python.
Best of luck!
This seems to be working for me:
import os
import urllib2
import zipfile
from StringIO import StringIO
response = urllib2.urlopen('http://build.chromium.org/buildbot/snapshots/chromium-rel-mac/LATEST')
latestRev = response.read()
print 'getting revision', latestRev
# we have the revision, now we need to download the zip and extract it
locRef='http://build.chromium.org/buildbot/snapshots/chromium-rel-mac/%i/chrome-mac.zip' % (int(latestRev))
latestZip = StringIO(urllib2.urlopen(locRef).read())
# we have the .zip now we need to extract the Chromium.app file, it's in chrome-mac/Chromium.app/
zippedChromium = zipfile.ZipFile(latestZip)
# find all zip members in chrome-mac/Chromium.app
members = [m for m in zippedChromium.namelist() if m.startswith('chrome-mac/Chromium.app/')]
#zippedChromium.extract(chromiumAppPath, workingDir)
target = 'chromium-%s' % latestRev
if os.path.isdir(target):
print 'destination already exists, exiting'
raise SystemExit(1)
os.makedirs(target)
zippedChromium.extractall(target, members)
#zippedChromium.close()
Here's another cut - this is the same technique, but it walks the result to demonstrate that it works.
import os
import urllib2
import zipfile
from StringIO import StringIO
response = urllib2.urlopen('http://build.chromium.org/buildbot/snapshots/chromium-rel-mac/LATEST')
latestRev = response.read()
print 'getting revision', latestRev
# we have the revision, now we need to download the zip and extract it
locRef='http://build.chromium.org/buildbot/snapshots/chromium-rel-mac/%i/chrome-mac.zip' % (int(latestRev))
latestZip = StringIO(urllib2.urlopen(locRef).read())
# we have the .zip now we need to extract the Chromium.app file, it's in chrome-mac/Chromium.app/
zippedChromium = zipfile.ZipFile(latestZip)
# find all zip members in chrome-mac/Chromium.app
members = [m for m in zippedChromium.namelist() if m.startswith('chrome-mac/Chromium.app/')]
#zippedChromium.extract(chromiumAppPath, workingDir)
target = 'chromium-%s' % latestRev
if os.path.isdir(target):
print 'destination already exists, exiting'
raise SystemExit(1)
os.makedirs(target)
zippedChromium.extractall(target, members)
lengths = [
(len(dirnames), len(filenames))
for dirpath, dirnames, filenames in os.walk(target)
]
dirlengths, filelengths = zip(*lengths)
ndirs = sum(dirlengths)
nfiles = sum(filelengths)
print 'extracted %(nfiles)d files in %(ndirs)d dirs' % vars()
#zippedChromium.close()
The output I get when I run it is
> .\getapp.py
getting revision 48479
extracted 537 files in 184 dirs
There is another problem extracting an .app from a zip in Python (which doesn't happen with a typically zip utility). No one else seems to have mentioned this...
The .app can ceases to function post extraction this way, as a result of losing the execution permission bit on the nested binary. You can fix this though, by simply granting that again.
Here's a loose snippet of code that I'm using. Revise this as needed for your purposes (or write a more generic function to handle this situation in a more universal manner):
import os, zipfile
...
ZIP_PATH = APP_PATH + ".zip"
APP_BIN_DIR = os.path.join( APP_PATH, "Contents/MacOS" )
zipfile.ZipFile( ZIP_PATH, 'r' ).extractall( WORK_DIR )
BIN_PATH = os.path.join( APP_BIN_DIR, os.listdir( APP_BIN_DIR )[0] )
os.chmod( BIN_PATH, 0o777 )
My program already knew where to expect the APP_PATH to be found (i.e. within the WORK_DIR). I had to zip it up though, and shoe horn that detail in after the fact. I name my zip like XXXXX.app.zip. I resolve the BIN_PATH here pretty simply without the need to know the name of binary inside the .app, because I know there is only going to be one file in there for my use case. I grant full (777) permissions to it, because I simply delete the .app at the end of my script anyway.