I am trying to read/get data from a json file. This json file is stored in the project > Requests > request1.json. In a script i am trying to read data from the json file and failing badly. This is the code i'm trying to use to open file in read mode.
Trying to replace(in windows)
f = open('D:\\Test\\projectname\\RequestJson\\request1.json', 'r') with
f = open(os.path.expanduser('~user') + "Requests/request1.json", 'r')
Any help would be greatly appreciated.
Using current directory path (assuming that is in the project) and appending the remaining static file path:
import os
current_dir = os.path.abspath(os.getcwd())
path = current_dir + "/RequestJson/request1.json"
with open(path, 'r') as f:
f.write(data)
Related
I am trying to find files with .desktop extension in a specific directory in Python3. I tried the code snippet below but it didn't work as I wanted. I want it to be a single string value.
import os, fnmatch
desktopfile = configparser.ConfigParser ()
def find(pattern, path):
result = []
for root, dirs, files in os.walk(path):
for name in files:
if fnmatch.fnmatch(name, pattern):
result.append(os.path.join(root, name))
return result
script_tmp_dir = "/tmp/appiload/appinstall" # Geçici dizin (dosyalar burada ayıklanıyor)
desktopfilea=f"{script_tmp_dir}/squashfs-root/{str(find ('*.desktop', f'{script_tmp_dir}/squashfs-root/')}"
print(desktopfilea)
desktopfile.items()
Result:
/tmp/appiload/appinstall/squashfs-root/['/tmp/appiload/appinstall/squashfs-root/helloworld.desktop']
Use glob.glob instead of writing a function to do this job.
import os, glob
desktopfile = configparser.ConfigParser ()
script_tmp_dir = "/tmp/appiload/appinstall" # Geçici dizin (dosyalar burada ayıklanıyor)
desktopfilea = glob.glob(f'{script_tmp_dir}/squashfs-root/*.desktop')
# desktopfilea = " ".join(desktopfilea) # Join them in one string, using space as seperator
print(str(desktopfilea))
desktopfile.items()
I don't exactly understand what do you mean but I made a simple program that will print all the files with the .desktop extension and save them to 2 files: applications.json in an array and applications.txt just written one after another.
I also have 2 versions of the program, one that will only print and save only the file names with extensions and other one that will print and save the whole path.
File names only:
import os
import json
strApplications = ""
applications = []
for file in os.listdir(os.path.dirname(os.path.realpath(__file__))):
if file.endswith(".desktop"):
applications.append(file)
with open("applications.json", "w") as f:
json.dump(applications, f)
strApplications = strApplications + file + "\n"
with open("applications.txt", "w") as f:
f.write(strApplications)
print(strApplications)
Full file path:
import os
import json
cwd = os.getcwd()
files = [cwd + "\\" + f for f in os.listdir(cwd) if f.endswith(".desktop")]
with open("applications.json", "w") as f:
json.dump(files, f)
with open("applications.txt", "w") as f:
f.write("\n".join(files))
print("\n".join(files))
Current code:
#!/usr/bin/python
import boto3
s3=boto3.client('s3')
list=s3.list_objects(Bucket='my_bucket_name')['Contents']
for key in list:
s3.download_file('my_bucket_name', key['Key'], key['Key'])
In the specific path I have N files. This way I download them and then I have also N local files. I just want one single file.
I did:
data=""
files = []
for file in glob.glob("*.json"):
files.append(file)
for file in files:
with open(file) as fp:
data += fp.read()
data += "\n"
with open ('output.json', 'w') as fp:
fp.write(data)
Is there a way to do it faster or even using boto to stream downloaded bytes to a file?
Initially I want to create zip file dynamically and return it in http response. I use python 3.7 lib zipfile.
I tried both io buffer and tmp dir, neither one of them creates valid zip archive. Archive is only opened if its saved on disc
import zipfile
import io
#==============================================
# V1
file_like_object = io.BytesIO()
myZipFile = zipfile.ZipFile(file_like_object, "w", compression=zipfile.ZIP_DEFLATED)
myZipFile.writestr(u'test.py', b'test')
tmparchive="zip1.zip"
out = open(tmparchive,'wb') ## Open temporary file as bytes
out.write(file_like_object.getvalue())
out.close()
r = open(tmparchive, 'rb')
print (r.read())
r.close()
#==============================================
# V2
tmparchive2 = 'zip2.zip'
myZipFile2 = zipfile.ZipFile(tmparchive2, "w", compression=zipfile.ZIP_DEFLATED)
myZipFile2.writestr(u'test.py', b'test')
r2 = open(tmparchive2, 'rb')
print (r2.read())
r2.close()
#====================================================
It's preferable to use a context manager like so:
import zipfile, io
file_like_object = io.BytesIO()
with zipfile.ZipFile(file_like_object, "w", compression=zipfile.ZIP_DEFLATED) as myZipFile:
myZipFile.writestr(u'test.txt', b'test')
# file_like_object.getvalue() are the bytes you send in your http response.
I wrote it to file. It's definitely a valid zip file.
If you want to open the archive, you need to save it to disk. Applications like Explorer and 7-Zip have no way to read the BytesIO object that exists in the python process. They can only open archives saved to disk.
Calling print(r.read()) isn't going to open the archive. It's just going to print the bytes that make up the tiny zip file you just created.
I am trying to save the variable in python which will create a text file. Am i going wrong? I want to know where the file would be created/
Here is the code:
import pickle as pk
f = open("featuresvmt.txt", "w")
pk.dump(feature_svmt, f)
pk.dump(out_val, f)
f.close()
The file should be opened in binary writing mode: f = open("featuresvmt.txt", "wb").
The file featuresvmt.txt will be created in the current working directory. You can find the current working directory using os.getcwd(). Or, simply supply an absolute path: f = open("/path/to/featuresvmt.txt", "wb").
import pickle as pk
feature_svmt, out_val = 'foo', 12.34
with open("featuresvmt.txt", "wb") as f:
pk.dump(feature_svmt, f)
pk.dump(out_val, f)
with open("featuresvmt.txt", "rb") as f:
print(pk.load(f))
# foo
print(pk.load(f))
# 12.34
Your code can run and create a file where you run iPython. (Of course you must initialized both of used variable)
If you run iPython from terminal or command prompt, which directory is your actual path will contain the new file.
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.