url = 'https://cdn.wsform.com/wp-content/uploads/2021/05/currency.csv'
f = open(url, 'r')
print(f.read())
Invalid argument: 'https://cdn.wsform.com/wp-content/uploads/2021/05/currency.csv'
Is there any way so that i can read file from url, instead of downloading it ??
Yes, the standard urllib.request.urlopen() does exactly that:
from urllib import request
with request.urlopen('https://cdn.wsform.com/wp-content/uploads/2021/05/currency.csv') as f:
print(f.read().decode()) # the decode turns the bytes into a string for printing
Just note that the file is opened as a file of bytes, since the function can't know if it's a text file or not.
Related
I am trying to download a PDF file from a website and save it to disk. My attempts either fail with encoding errors or result in blank PDFs.
In [1]: import requests
In [2]: url = 'http://www.hrecos.org//images/Data/forweb/HRTVBSH.Metadata.pdf'
In [3]: response = requests.get(url)
In [4]: with open('/tmp/metadata.pdf', 'wb') as f:
...: f.write(response.text)
---------------------------------------------------------------------------
UnicodeEncodeError Traceback (most recent call last)
<ipython-input-4-4be915a4f032> in <module>()
1 with open('/tmp/metadata.pdf', 'wb') as f:
----> 2 f.write(response.text)
3
UnicodeEncodeError: 'ascii' codec can't encode characters in position 11-14: ordinal not in range(128)
In [5]: import codecs
In [6]: with codecs.open('/tmp/metadata.pdf', 'wb', encoding='utf8') as f:
...: f.write(response.text)
...:
I know it is a codec problem of some kind but I can't seem to get it to work.
You should use response.content in this case:
with open('/tmp/metadata.pdf', 'wb') as f:
f.write(response.content)
From the document:
You can also access the response body as bytes, for non-text requests:
>>> r.content
b'[{"repository":{"open_issues":0,"url":"https://github.com/...
So that means: response.text return the output as a string object, use it when you're downloading a text file. Such as HTML file, etc.
And response.content return the output as bytes object, use it when you're downloading a binary file. Such as PDF file, audio file, image, etc.
You can also use response.raw instead. However, use it when the file which you're about to download is large. Below is a basic example which you can also find in the document:
import requests
url = 'http://www.hrecos.org//images/Data/forweb/HRTVBSH.Metadata.pdf'
r = requests.get(url, stream=True)
with open('/tmp/metadata.pdf', 'wb') as fd:
for chunk in r.iter_content(chunk_size):
fd.write(chunk)
chunk_size is the chunk size which you want to use. If you set it as 2000, then requests will download that file the first 2000 bytes, write them into the file, and do this again, again and again, unless it finished.
So this can save your RAM. But I'd prefer use response.content instead in this case since your file is small. As you can see use response.raw is complex.
Relates:
How to download large file in python with requests.py?
How to download image using requests
In Python 3, I find pathlib is the easiest way to do this. Request's response.content marries up nicely with pathlib's write_bytes.
from pathlib import Path
import requests
filename = Path('metadata.pdf')
url = 'http://www.hrecos.org//images/Data/forweb/HRTVBSH.Metadata.pdf'
response = requests.get(url)
filename.write_bytes(response.content)
You can use urllib:
import urllib.request
urllib.request.urlretrieve(url, "filename.pdf")
Please note I'm a beginner. If My solution is wrong, please feel free to correct and/or let me know. I may learn something new too.
My solution:
Change the downloadPath accordingly to where you want your file to be saved. Feel free to use the absolute path too for your usage.
Save the below as downloadFile.py.
Usage: python downloadFile.py url-of-the-file-to-download new-file-name.extension
Remember to add an extension!
Example usage: python downloadFile.py http://www.google.co.uk google.html
import requests
import sys
import os
def downloadFile(url, fileName):
with open(fileName, "wb") as file:
response = requests.get(url)
file.write(response.content)
scriptPath = sys.path[0]
downloadPath = os.path.join(scriptPath, '../Downloads/')
url = sys.argv[1]
fileName = sys.argv[2]
print('path of the script: ' + scriptPath)
print('downloading file to: ' + downloadPath)
downloadFile(url, downloadPath + fileName)
print('file downloaded...')
print('exiting program...')
Generally, this should work in Python3:
import urllib.request
..
urllib.request.get(url)
Remember that urllib and urllib2 don't work properly after Python2.
If in some mysterious cases requests don't work (happened with me), you can also try using
wget.download(url)
Related:
Here's a decent explanation/solution to find and download all pdf files on a webpage:
https://medium.com/#dementorwriter/notesdownloader-use-web-scraping-to-download-all-pdfs-with-python-511ea9f55e48
regarding Kevin answer to write in a folder tmp, it should be like this:
with open('./tmp/metadata.pdf', 'wb') as f:
f.write(response.content)
he forgot . before the address and of-course your folder tmp should have been created already
I am trying to download a PDF file from a website and save it to disk. My attempts either fail with encoding errors or result in blank PDFs.
In [1]: import requests
In [2]: url = 'http://www.hrecos.org//images/Data/forweb/HRTVBSH.Metadata.pdf'
In [3]: response = requests.get(url)
In [4]: with open('/tmp/metadata.pdf', 'wb') as f:
...: f.write(response.text)
---------------------------------------------------------------------------
UnicodeEncodeError Traceback (most recent call last)
<ipython-input-4-4be915a4f032> in <module>()
1 with open('/tmp/metadata.pdf', 'wb') as f:
----> 2 f.write(response.text)
3
UnicodeEncodeError: 'ascii' codec can't encode characters in position 11-14: ordinal not in range(128)
In [5]: import codecs
In [6]: with codecs.open('/tmp/metadata.pdf', 'wb', encoding='utf8') as f:
...: f.write(response.text)
...:
I know it is a codec problem of some kind but I can't seem to get it to work.
You should use response.content in this case:
with open('/tmp/metadata.pdf', 'wb') as f:
f.write(response.content)
From the document:
You can also access the response body as bytes, for non-text requests:
>>> r.content
b'[{"repository":{"open_issues":0,"url":"https://github.com/...
So that means: response.text return the output as a string object, use it when you're downloading a text file. Such as HTML file, etc.
And response.content return the output as bytes object, use it when you're downloading a binary file. Such as PDF file, audio file, image, etc.
You can also use response.raw instead. However, use it when the file which you're about to download is large. Below is a basic example which you can also find in the document:
import requests
url = 'http://www.hrecos.org//images/Data/forweb/HRTVBSH.Metadata.pdf'
r = requests.get(url, stream=True)
with open('/tmp/metadata.pdf', 'wb') as fd:
for chunk in r.iter_content(chunk_size):
fd.write(chunk)
chunk_size is the chunk size which you want to use. If you set it as 2000, then requests will download that file the first 2000 bytes, write them into the file, and do this again, again and again, unless it finished.
So this can save your RAM. But I'd prefer use response.content instead in this case since your file is small. As you can see use response.raw is complex.
Relates:
How to download large file in python with requests.py?
How to download image using requests
In Python 3, I find pathlib is the easiest way to do this. Request's response.content marries up nicely with pathlib's write_bytes.
from pathlib import Path
import requests
filename = Path('metadata.pdf')
url = 'http://www.hrecos.org//images/Data/forweb/HRTVBSH.Metadata.pdf'
response = requests.get(url)
filename.write_bytes(response.content)
You can use urllib:
import urllib.request
urllib.request.urlretrieve(url, "filename.pdf")
Please note I'm a beginner. If My solution is wrong, please feel free to correct and/or let me know. I may learn something new too.
My solution:
Change the downloadPath accordingly to where you want your file to be saved. Feel free to use the absolute path too for your usage.
Save the below as downloadFile.py.
Usage: python downloadFile.py url-of-the-file-to-download new-file-name.extension
Remember to add an extension!
Example usage: python downloadFile.py http://www.google.co.uk google.html
import requests
import sys
import os
def downloadFile(url, fileName):
with open(fileName, "wb") as file:
response = requests.get(url)
file.write(response.content)
scriptPath = sys.path[0]
downloadPath = os.path.join(scriptPath, '../Downloads/')
url = sys.argv[1]
fileName = sys.argv[2]
print('path of the script: ' + scriptPath)
print('downloading file to: ' + downloadPath)
downloadFile(url, downloadPath + fileName)
print('file downloaded...')
print('exiting program...')
Generally, this should work in Python3:
import urllib.request
..
urllib.request.get(url)
Remember that urllib and urllib2 don't work properly after Python2.
If in some mysterious cases requests don't work (happened with me), you can also try using
wget.download(url)
Related:
Here's a decent explanation/solution to find and download all pdf files on a webpage:
https://medium.com/#dementorwriter/notesdownloader-use-web-scraping-to-download-all-pdfs-with-python-511ea9f55e48
regarding Kevin answer to write in a folder tmp, it should be like this:
with open('./tmp/metadata.pdf', 'wb') as f:
f.write(response.content)
he forgot . before the address and of-course your folder tmp should have been created already
I'm having difficulty with the following code (which is simplified from a larger application I'm working on in Python).
from io import StringIO
import gzip
jsonString = 'JSON encoded string here created by a previous process in the application'
out = StringIO()
with gzip.GzipFile(fileobj=out, mode="w") as f:
f.write(str.encode(jsonString))
# Write the file once finished rather than streaming it - uncomment the next line to see file locally.
with open("out_" + currenttimestamp + ".json.gz", "a", encoding="utf-8") as f:
f.write(out.getvalue())
When this runs I get the following error:
File "d:\Development\AWS\TwitterCompetitionsStreaming.py", line 61, in on_status
with gzip.GzipFile(fileobj=out, mode="w") as f:
File "C:\Python38\lib\gzip.py", line 204, in __init__
self._write_gzip_header(compresslevel)
File "C:\Python38\lib\gzip.py", line 232, in _write_gzip_header
self.fileobj.write(b'\037\213') # magic header
TypeError: string argument expected, got 'bytes'
PS ignore the rubbish indenting here...I know it doesn't look right.
What I'm wanting to do is to create a json file and gzip it in place in memory before saving the gzipped file to the filesystem (windows). I know I've gone about this the wrong way and could do with a pointer. Many thanks in advance.
You have to use bytes everywhere when working with gzip instead of strings and text. First, use BytesIO instead of StringIO. Second, mode should be 'wb' for bytes instead of 'w' (last is for text) (samely 'ab' instead of 'a' when appending), here 'b' character means "bytes". Full corrected code below:
Try it online!
from io import BytesIO
import gzip
jsonString = 'JSON encoded string here created by a previous process in the application'
out = BytesIO()
with gzip.GzipFile(fileobj = out, mode = 'wb') as f:
f.write(str.encode(jsonString))
currenttimestamp = '2021-01-29'
# Write the file once finished rather than streaming it - uncomment the next line to see file locally.
with open("out_" + currenttimestamp + ".json.gz", "wb") as f:
f.write(out.getvalue())
I'm testing some things and i keep getting the error "write() argument must be str, not HTTPResponse" Here's my code:
import requests
image="http://www.casperdenhaan.nl/wp-content/uploads/2020/03/Logo.jpg"
savefile=open("image.png","w+")
savefile.write(requests.get(image).raw)
savefile.close()
I can get the raw data, but I can't write it to a new file. Is there a way i could get around this problem?
When you call .raw on the response object it returns an HTTPResponse object. You need to call .content to get a bytes object.
type(requests.get(image).raw)
urllib3.response.HTTPResponse
type(requests.get(image).content)
bytes
You need to open the file in write binary mode:
open("image.png","wb")
I suggest using a "with" block, this way you don't need to explicitly close the file. Here is a working version of the code:
import requests
url = "http://www.casperdenhaan.nl/wp-content/uploads/2020/03/Logo.jpg"
with open('image.png', 'wb') as f:
f.write(requests.get(url).content)
try this way of doing it
import requests
img_url = "http://www.casperdenhaan.nl/wp-content/uploads/2020/03/Logo.jpg"
img = requests.get(img_url)
with open('image.png', 'wb') as save_file:
save_file.write(img.raw)
This way you don't have to deal with closing the file. Also, the 'wb' opens the file in a writable binary mode.
I am trying to download file from GitHub(raw file) and then run this file as .sql file.
import snowflake.connector
from codecs import open
import logging
import requests
from os import getcwd
import os
import sys
#logging
logging.basicConfig(
filename='C:/Users/abc/Documents/Test.log',
level=logging.INFO
)
url = "https://github.com/raw/abc/master/file_name?token=Anvn3lJXDks5ciVaPwA%3D%3D"
directory = getcwd()
filename = os.path.join(getcwd(),'VIEWS.SQL')
r = requests.get(url)
filename.decode("utf-8")
f = open(filename,'w')
f.write(str(r.content))
with open(filename,'r') as theFile, open(filename,'w') as outFile:
data = theFile.read().split('\n')
data = theFile.read().replace('\n','')
data = theFile.read().replace("b'","")
data = theFile.read()
outFile.write(data)
However I get this error
syntax error line 1 at position 0 unexpected 'b'
My converted sql file has b at the beginning and bunch of newline \n characters in the file. Also the entire output file is in single quotes 'text'. Can anyone help me get rid of these? Looks like replace isn't working.
OS: Windows
Python Version: 3.7.0
You introduced a b'.. prefix by converting the response.content bytes value to a string with str():
>>> import requests
>>> r = requests.get("https://github.com/raw/abc/master/file_name?token=Anvn3lJXDks5ciVaPwA%3D%3D")
>>> r.content
b'Not Found'
>>> str(r.content)
"b'Not Found'"
Of course, the specific dummy URL you gave in your question produces a 404 Not Found response, hence the Not Found content of the response body:
>>> r.status_code
404
so the contents in this demonstration are not actually all that useful. However, even for your real URL you probably want to test for a 200 status code before moving to write the data to a file!
What is going wrong in the above is that str(bytesvalue) converts a bytes object to its representation. You'd normally want to decode a bytes value with a text codec, using the bytes.decode() method. But because you are writing the data to a file here, you should instead just open the file in binary mode and write the bytes object without decoding:
r = requests.get(url)
if r.status_code == 200:
with open(filename, 'wb') as f:
f.write(r.content)
The 'wb' mode opens the file for writing in binary mode. Writing binary content to a binary file is the most efficient; decoding it first then writing to a text file requires that it is encoded again. Better to avoid doing double work.
As a side note: there is no need to join a local filename with getcwd(); relative paths always end up in the current working directory, and otherwise it's better to use os.path.abspath(filename).
You could also trust that GitHub sets the correct character set in the Content-Type headers and have response decode the value to str for you in the form of the response.text attribute:
r = requests.get(url)
if r.status_code == 200:
with open(filename, 'w') as f:
f.write(r.text)
but again, that's really doing extra work for nothing, first decoding the binary content from the request, then encoding again when writing to a text file.
Finally, for larger file responses it is better to stream the data and copy it directly to a file. The shutil.copyfileobj() function can take a raw response fileobject directly, provided you enable transparent transport decompression:
import shutil
r = requests.get(url, stream=True)
if r.status_code == 200:
with open(filename, 'wb') as f:
# enable transparent transport decompression handling
r.raw.decode_content = True
shutil.copyfileobj(r.raw, f)
Depending on your version of Python/OS it could be as simple as changing the file to read/write in binary (and if they're still there then altering where you have the replaces):
with open(filename,'rb') as theFile, open(filename,'wb') as outFile:
outfile.write(str(r.content))
data = theFile.read().split('\n')
data = data.replace('\n','')
data = data.replace("b'","")
outFile.write(data)
It would help to have a copy of the file and the line the error is occurring on.