I am trying to combine two files from the internet, and save the output on my computer. I have the below code,but no made what I try I always get the same result. I get the first URL, and nothing more.
To be exact, I am trying to comebine VideoURL and videoURL1, together into one file called output.mp4...
videoURL= 'http://file-examples.com/wp-content/uploads/2017/04/file_example_MP4_480_1_5MG.mp4'
videoURL1 = 'http://techslides.com/demos/sample-videos/small.mp4'
# print(str(embeddHTMLString).find('sources: ['))
local_filename = videoURL.split('/')[-1]
# NOTE the stream=True parameter
response = urlopen(videoURL)
response1 = urlopen(videoURL1)
with open(local_filename, 'wb') as f:
while True:
chunk = response.read(1024)
if not chunk:
break
f.write(chunk)
with open(local_filename, 'ab+') as d:
while True:
chunk1 = response1.read(1024)
if not chunk1:
break
d.write(chunk1)
You're doing it wrong. The gist of this answer has already been given by #Tempo810, you need to download the files separately and concatenate them into a single file later.
I am assuming you have both video1.mp4 and video2.mp4 downloaded from your urls separately. Now to combine them, you simply cannot use append to concat the files, since video files contains format header and metadata, and combining two media files into one means you need to rewrite new metadata and format header, and remove the old ones.
Instead you can use the the library moviepy to save yourself. Here is a small sample of code how to utilise moviepy's concatenate_videoclips() to concat the files:
from moviepy.editor import VideoFileClip, concatenate_videoclips
# opening the clips
clip1 = VideoFileClip("video1.mp4")
clip3 = VideoFileClip("video2.mp4")
# lets concat them up
final_clip = concatenate_videoclips([clip1,clip2])
final_clip.write_videofile("output.mp4")
Your resulting combined file is output.mp4. Thats it!
Related
I am a little bit confused in how to read all lines in many files where the file names have format from "datalog.txt.98" to "datalog.txt.120".
This is my code:
import json
file = "datalog.txt."
i = 97
for line in file:
i+=1
f = open (line + str (i),'r')
for row in f:
print (row)
Here, you will find an example of one line in one of those files:
I need really to your help
I suggest using a loop for opening multiple files with different formats.
To better understand this project I would recommend researching the following topics
for loops,
String manipulation,
Opening a file and reading its content,
List manipulation,
String parsing.
This is one of my favourite beginner guides.
To set the parameters of the integers at the end of the file name I would look into python for loops.
I think this is what you are trying to do
# create a list to store all your file content
files_content = []
# the prefix is of type string
filename_prefix = "datalog.txt."
# loop from 0 to 13
for i in range(0,14):
# make the filename variable with the prefix and
# the integer i which you need to convert to a string type
filename = filename_prefix + str(i)
# open the file read all the lines to a variable
with open(filename) as f:
content = f.readlines()
# append the file content to the files_content list
files_content.append(content)
To get rid of white space from file parsing add the missing line
content = [x.strip() for x in content]
files_content.append(content)
Here's an example of printing out files_content
for file in files_content:
print(file)
I'm still relatively new to programming and Python. But I am sure this must be possible but my searches are not turning up what I'm looking for.
In my current directory, I have 6 PDF files that I wish to read in via the loop below.
What I would like to do is open each of the PDF's with a new variable name, as you can see it is imaginatively called pdf[1-6]File.pdf.
I can list the files in the console and pull them via the code when I stick breaks in to stop it executing but I can't for the life of me work out how to create the variable name. I thought something like "pdf" + str(i) + "File" would have worked but I'm missing something.
Code is below - not complete but enough so you get what I'm looking at:
#Open the PDF files in the current directory for
#reading in binary mode
def opensource():
listOfFiles = os.listdir('.')
pattern = "*.pdf"
for entry in listOfFiles:
if fnmatch.fnmatch(entry, pattern):
# Works to here perfectly
for i in range(len(entry)):
# print(len(entry))
# Trying to create the variable name with
# an incremental numeral in the file name
"pdf" + i + "File" = open(entry, 'rb')
This bit below is how I'm currently doing it and its a pain in the backside. I'm sure it can be done programmatically
#This is the old way. Monolithic and horrid
#Open the files that have to be merged one by one
pdf1File = open('file1.pdf', 'rb')
pdf2File = open('file2.pdf', 'rb')
pdf3File = open('file3.pdf', 'rb')
pdf4File = open('file4.pdf', 'rb')
pdf5File = open('file5.pdf', 'rb')
pdf6File = open('file6.pdf', 'rb')
All help gratefully received.
Thanks
If you are going to use the file pointer outside this for loop, you can very well use a dictionary to do that..
def opensource():
listOfFiles = os.listdir('.')
pattern = "*.pdf"
file_ptrs = {}
for entry in listOfFiles:
if fnmatch.fnmatch(entry, pattern):
# Works to here perfectly
for i in range(len(entry)):
# print(len(entry))
# Trying to create the variable name with
# an incremental numeral in the file name
file_ptrs["pdf" + str(i) + "File"] = open(entry, 'rb')
Caution: Its always advisable to use the open method alongside of a "with" clause in python.. it takes care of closing the file once the file operation goes out of context.
I extracted an embedded object from an excel spreadsheet that was a pdf but the excel zip file saves embedded objects as binary files.
I am trying to read the binary file and return it to it's original format as a pdf. I took some code from another question with a similar issue but when i try opening the pdf adobe gives error "can't open because file is damaged...not decoded correctly.."
Does anyone know of a way to do this?
with open('oleObject1.bin','rb') as f:
binaryData = f.read()
print(binaryData)
with open(os.path.expanduser('test1.pdf'), 'wb') as fout:
fout.write(base64.decodebytes(binaryData))
Link to the object file on github
Thanks Ryan, I was able to see what you are talking about. Here is solution for future reference.
str1 = b'%PDF-' # Begin PDF
str2 = b'%%EOF' # End PDF
with open('oleObject1.bin', 'rb') as f:
binary_data = f.read()
print(binary_data)
# Convert BYTE to BYTEARRAY
binary_byte_array = bytearray(binary_data)
# Find where PDF begins
result1 = binary_byte_array.find(str1)
print(result1)
# Remove all characters before PDF begins
del binary_byte_array[:result1]
print(binary_byte_array)
# Find where PDF ends
result2 = binary_byte_array.find(str2)
print(result2)
# Subtract the length of the array from the position of where PDF ends (add 5 for %%OEF characters)
# and delete that many characters from end of array
print(len(binary_byte_array))
to_remove = len(binary_byte_array) - (result2 + 5)
print(to_remove)
del binary_byte_array[-to_remove:]
print(binary_byte_array)
with open(os.path.expanduser('test1.pdf'), 'wb') as fout:
fout.write(binary_byte_array)
The bin file contains a valid PDF. There is no decoding required. The bin file though does have bytes before and after the PDF that need to be trimmed.
To get the first byte look for the first occurrence of string %PDF-
To get the final byte look for the last %%EOF.
Note, I do not know what "format" the leading/trailing bytes are, that are added by Excel. The solution above obliviously would not work if either of the ascii strings above could also be in the leading/trailing data.
You should try using a python library that allows you to write pdf files like reportlab or pyPDF
First off i must say i am VERY new to programming (less then a week experience in total). I set out to write a program that generates a series of documents of an .odt template. I want to use a template with a specific keyword lets say "X1234X" and so on. This will then be replaced by values generated from the program. Each document is a little different and the values are entered and calculated via a prompt (dates and other things)
I wrote most of the code so far but i am stuck since 2 days on that problem. I used the ezodf module to generate a new document (with different filenames) from a template but i am stuck on how to edit the content.
I googled hard but came up empty hope someone here could help. I tried reading the documentations but i must be honest...its a bit tough to understand. I am not familiar with the "slang"
Thanks
PS: a ezodf method would be great, but any other ways will do too. The program doesnt have to be pretty it just has to work (so i can work less ^_^)
Well i figured it out. nd finished the program. I used a ezodf to create the file, then zipfile to extract and edit the content.xml and then repacked the whole thing via a nice >def thingy< from here. I tried to mess with etree...but i couldnt figure it out...
from ezodf import newdoc
import os
import zipfile
import tempfile
for s in temp2:
input2 = s
input2 = str(s)
input1 = cname[0]
file1 = '.odt'
namef = input2 + input1 + file1
odt = newdoc(doctype='odt', filename=namef, template='template.odt')
odt.save()
a = zipfile.ZipFile('template.odt')
content = a.read('content.xml')
content = str(content.decode(encoding='utf8'))
content = str.replace(content,"XXDATEXX", input2)
content = str.replace(content, 'XXNAMEXX', input1)
def updateZip(zipname, filename, data):
# generate a temp file
tmpfd, tmpname = tempfile.mkstemp(dir=os.path.dirname(zipname))
os.close(tmpfd)
# create a temp copy of the archive without filename
with zipfile.ZipFile(zipname, 'r') as zin:
with zipfile.ZipFile(tmpname, 'w') as zout:
zout.comment = zin.comment # preserve the comment
for item in zin.infolist():
if item.filename != filename:
zout.writestr(item, zin.read(item.filename))
# replace with the temp archive
os.remove(zipname)
os.rename(tmpname, zipname)
# now add filename with its new data
with zipfile.ZipFile(zipname, mode='a', compression=zipfile.ZIP_DEFLATED) as zf:
zf.writestr(filename, data)
updateZip(namef, 'content.xml', content)
How to start creating my own filetype in Python ? I have a design in mind but how to pack my data into a file with a specific format ?
For example I would like my fileformat to be a mix of an archive ( like other format such as zip, apk, jar, etc etc, they are basically all archives ) with some room for packed files, plus a section of the file containing settings and serialized data that will not be accessed by an archive-manager application.
My requirement for this is about doing all this with the default modules for Cpython, without external modules.
I know that this can be long to explain and do, but I can't see how to start this in Python 3.x with Cpython.
Try this:
from zipfile import ZipFile
import json
data = json.dumps(['foo', {'bar': ('baz', None, 1.0, 2)}])
with ZipFile('foo.filetype', 'w') as myzip:
myzip.writestr('digest.json', data)
The file is now a zip archive with a json file (thats easy to read in again in many lannguages) for data you can add files to the archive with myzip write or writestr. You can read data back with:
with ZipFile('foo.filetype', 'r') as myzip:
json_data_read = myzip.read('digest.json')
newdata = json.loads(json_data_read)
Edit: you can append arbitrary data to the file with:
f = open('foo.filetype', 'a')
f.write(data)
f.close()
this works for winrar but python can no longer process the zipfile.
Use this:
import base64
import gzip
import ast
def save(data):
data = "[{}]".format(data).encode()
data = base64.b64encode(data)
return gzip.compress(data)
def load(data):
data = gzip.decompress(data)
data = base64.b64decode(data)
return ast.literal_eval(data.decode())[0]
How to use this with file:
open(filename, "wb").write(save(data)) # save data
data = load(open(filename, "rb").read()) # load data
This might look like this is able to be open with archive program
but it cannot because it is base64 encoded and they have to decode it to access it.
Also you can store any type of variable in it!
example:
open(filename, "wb").write(save({"foo": "bar"})) # dict
open(filename, "wb").write(save("foo bar")) # string
open(filename, "wb").write(save(b"foo bar")) # bytes
# there's more you can store!
This may not be appropriate for your question but I think this may help you.
I have a similar problem faced... but end up with some thing like creating a zip file and then renamed the zip file format to my custom file format... But it can be opened with the winRar.