I would like to know if there is any way to activate the webhook for all intents (other than activating it one by one). Thank you!
There is no such functionality as of now, but I had similar problem and this is how I solved it:
Download the zip file of all the intents
Write a program (I wrote in python) to go through all files (ignoring files that ends with usersays
change "webhookUsed": false, to "webhookUsed": true,
Upload the zip file replacing existing intents using Restore from zip option
UPDATE 1:
Below is the code:
import zipfile
import json
import os
import glob
cwd = os.getcwd()
zip_ref = zipfile.ZipFile(cwd + '/filename.zip', 'r')
zip_ref.extractall('zipped')
zip_ref.close()
cwd = cwd + '/zipped/intents'
files = glob.glob(cwd + "/*.json")
for file in files:
print(file)
if "usersay" not in file:
json_data= json.loads(open(file).read())
json_data['webhookUsed'] = True
with open(file, 'w') as outfile:
json.dump(json_data, outfile)
Place the zip file you get from dialogflow in the directory same as where you place above code and run the python program.
After running this code, navigate to directory named zipped and zip all the contents of the file and follow step 4.
UPDATE 2:
Updated the code to make it compatible to multiple languages Dialogflow agent.
Hope it helps.
Aside from activating it one by one, or downloading the zip file, setting it one by tone in the JSON, and uploading the results - no.
#sid8491 thank you so much, this worked for me!
I had to make some changes so that it worked correctly. Your answer has been very helpful. This is my final script:
import zipfile
import json
import os
import glob
cwd = os.getcwd()
zip_ref = zipfile.ZipFile(cwd + '/Bill.zip', 'r')
zip_ref.extractall('zipped')
zip_ref.close()
cwd = cwd + '/zipped/intents'
files = glob.glob(cwd + "/*.json")
for file in files:
print(file)
if "usersay" not in file:
json_data = json.loads(open(file, encoding="utf8").read())
json_data['webhookUsed'] = True
with open(file, 'w') as outfile:
json.dump(json_data, outfile)
else:
print("Usersay file", file)
Related
import os ,fnmatch
import os.path
import os
file_dir= '/home/deeghayu/Desktop/mec_sim/up_folder'
file_type = ['*.py']
for root, dirs,files in os.walk( file_dir ):
for extension in ( tuple(file_type) ):
for filename in fnmatch.filter(files, extension):
filepath = os.path.join(root, filename)
if os.path.isfile( filepath ):
print(filename , 'executing...');
exec(open('/home/deeghayu/Desktop/mec_sim/up_folder/{}'.format(filename)).read())
else:
print('Execution failure!!!')
Hello everyone I am working on this code which execute a python file using a python code. I need to save my output of the code as a text file. Here I have shown my code. Can any one give me a solution how do I save my output into a text file?
Piggybacking off of the original answer since they are close but it isn't a best practice to open and close files that way.
It's better to use a context manager instead of saying f = open() since the context manager will handle closing the resource for you regardless of whether your code succeeds or not.
You use it like,
with open("file.txt","w+") as f:
for i in range(10):
f.write("This is line %d\r\n" % (i+1))
try
Open file
f= open("file.txt","w+")
Insert data into file
for i in range(10):
f.write("This is line %d\r\n" % (i+1))
Close the file
f.close()
I recently started getting into Python and I am having a hard time searching through directories and matching files based on a regex that I have created.
Basically I want it to scan through all the directories in another directory and find all the files that ends with .zip or .rar or .r01 and then run various commands based on what file it is.
import os, re
rootdir = "/mnt/externa/Torrents/completed"
for subdir, dirs, files in os.walk(rootdir):
if re.search('(w?.zip)|(w?.rar)|(w?.r01)', files):
print "match: " . files
import os
import re
rootdir = "/mnt/externa/Torrents/completed"
regex = re.compile('(.*zip$)|(.*rar$)|(.*r01$)')
for root, dirs, files in os.walk(rootdir):
for file in files:
if regex.match(file):
print(file)
CODE BELLOW ANSWERS QUESTION IN FOLLOWING COMMENT
That worked really well, is there a way to do this if match is found on regex group 1 and do this if match is found on regex group 2 etc ? – nillenilsson
import os
import re
regex = re.compile('(.*zip$)|(.*rar$)|(.*r01$)')
rx = '(.*zip$)|(.*rar$)|(.*r01$)'
for root, dirs, files in os.walk("../Documents"):
for file in files:
res = re.match(rx, file)
if res:
if res.group(1):
print("ZIP",file)
if res.group(2):
print("RAR",file)
if res.group(3):
print("R01",file)
It might be possible to do this in a nicer way, but this works.
Given that you are a beginner, I would recommend using glob in place of a quickly written file-walking-regex matcher.
Snippets of functions using glob and a file-walking-regex matcher
The below snippet contains two file-regex searching functions (one using glob and the other using a custom file-walking-regex matcher). The snippet also contains a "stopwatch" function to time the two functions.
import os
import sys
from datetime import timedelta
from timeit import time
import os
import re
import glob
def stopwatch(method):
def timed(*args, **kw):
ts = time.perf_counter()
result = method(*args, **kw)
te = time.perf_counter()
duration = timedelta(seconds=te - ts)
print(f"{method.__name__}: {duration}")
return result
return timed
#stopwatch
def get_filepaths_with_oswalk(root_path: str, file_regex: str):
files_paths = []
pattern = re.compile(file_regex)
for root, directories, files in os.walk(root_path):
for file in files:
if pattern.match(file):
files_paths.append(os.path.join(root, file))
return files_paths
#stopwatch
def get_filepaths_with_glob(root_path: str, file_regex: str):
return glob.glob(os.path.join(root_path, file_regex))
Comparing runtimes of the above functions
On using the above two functions to find 5076 files matching the regex filename_*.csv in a dir called root_path (containing 66,948 files):
>>> glob_files = get_filepaths_with_glob(root_path, 'filename_*.csv')
get_filepaths_with_glob: 0:00:00.176400
>>> oswalk_files = get_filepaths_with_oswalk(root_path,'filename_(.*).csv')
get_filepaths_with_oswalk: 0:03:29.385379
The glob method is much faster and the code for it is shorter.
For your case
For your case, you can probably use something like the following to get your *.zip,*.rar and *.r01 files:
files = []
for ext in ['*.zip', '*.rar', '*.r01']:
files += get_filepaths_with_glob(root_path, ext)
Here's an alternative using glob.
from pathlib import Path
rootdir = "/mnt/externa/Torrents/completed"
for extension in 'zip rar r01'.split():
for path in Path(rootdir).glob('*.' + extension):
print("match: " + path)
I would do it this way:
import re
from pathlib import Path
def glob_re(path, regex="", glob_mask="**/*", inverse=False):
p = Path(path)
if inverse:
res = [str(f) for f in p.glob(glob_mask) if not re.search(regex, str(f))]
else:
res = [str(f) for f in p.glob(glob_mask) if re.search(regex, str(f))]
return res
NOTE: per default it will recursively scan all subdirectories. If you want to scan only the current directory then you should explicitly specify glob_mask="*"
I am running into a problem trying to update an AWS Gamelift script with a python command that zips a directory and uploads it with all its contents as a newer version to AWS Gamelift.
from zipfile import ZipFile
import os
from os.path import basename
import boto3
import sys, getopt
def main(argv):
versInput = sys.argv[1]
#initializes client for updating script in aws gamelift
client = boto3.client('gamelift')
#Where is the directory relative to the script directory. In this case, one folder dir lower and the contents of the RealtimeServer dir
dirName = '../RealtimeServer'
# create a ZipFile object
with ZipFile('RealtimeServer.zip', 'w') as zipObj:
# Iterate over all the files in directory
for folderName, subfolders, filenames in os.walk(dirName):
rootlen = len(dirName) + 1
for filename in filenames:
#create complete filepath of file in directory
filePath = os.path.join(folderName, filename)
# Add file to zip
zipObj.write(filePath, filePath[rootlen:])
response = client.update_script(
ScriptId=SCRIPT_ID_GOES_HERE,
Version=sys.argv[1],
ZipFile=b'--zip-file \"fileb://RealtimeServer.zip\"'
)
if __name__ == "__main__":
main(sys.argv[1])
I plan on using it by giving it a new version number everytime I make changes with:
python updateScript.py "0.1.1"
This is meant to help speed up development. However, I am doing something wrong with the ZipFile parameter of client.update_script()
For context, I can use the AWS CLI directly from the commandline and update a script without a problem by using:
aws gamelift update-script --script-id SCRIPT_STRING_ID_HERE --script-version "0.4.5" --zip-file fileb://RealtimeServer.zip
However, I am not sure what is going on because it fails to unzip the file when I try it:
botocore.errorfactory.InvalidRequestException: An error occurred (InvalidRequestException) when calling the UpdateScript operation: Failed to unzip the zipped file.
UPDATE:
After reading more documentation about the ZipFile parameter:
https://docs.aws.amazon.com/gamelift/latest/apireference/API_UpdateScript.html
https://boto3.amazonaws.com/v1/documentation/api/latest/reference/services/gamelift.html#GameLift.Client.update_script
I tried sending a base64 encoded version of the zip file. However, that didn't work. I put the following code before the client_update part of the script and used b64EncodedZip as the ZipFile parameter.
with open("RealtimeServer.zip", "rb") as f:
bytes = f.read()
b64EncodedZip = base64.b64encode(bytes)
I was able to get it to work by having some help from a maintainer of boto3 over at https://github.com/boto/boto3/issues/2646
(Thanks #swetashre)
Here is the code and it will only work up to 5mb and requires use of an s3 bucket if you want to upload a zip file any larger than that.
from zipfile import ZipFile
import os
from os.path import basename
import boto3
import sys, getopt
def main(argv):
versInput = sys.argv[1]
#initializes client for updating script in aws gamelift
client = boto3.client('gamelift')
#Where is the directory relative to the script directory. In this case, one folder dir lower and the contents of the RealtimeServer dir
dirName = '../RealtimeServer'
# create a ZipFile object
with ZipFile('RealtimeServer.zip', 'w') as zipObj:
# Iterate over all the files in directory
for folderName, subfolders, filenames in os.walk(dirName):
rootlen = len(dirName) + 1
for filename in filenames:
#create complete filepath of file in directory
filePath = os.path.join(folderName, filename)
# Add file to zip
zipObj.write(filePath, filePath[rootlen:])
with open('RealtimeServer.zip','rb') as f:
contents = f.read()
response = client.update_script(
ScriptId="SCRIPT_ID_GOES_HERE",
Version=sys.argv[1],
ZipFile=contents
)
if __name__ == "__main__":
main(sys.argv[1])
I got the script working but I did it by avoiding the use of boto3. I don't like it but it works.
os.system("aws gamelift update-script --script-id \"SCRIPT_ID_GOES_HERE\" --script-version " + sys.argv[1] + " --zip-file fileb://RealtimeServer.zip")
If anyone knows how to get boto3 to work for updating an AWS Gamelift script then please let me know.
I am able to write file in directory using python with following code.
How can I pass the dynamic file name in below code. Right now i am passing hard coded name -'abc.xml'. I have a for loop and file name will be different for each loop iteration. How can i open the directory and then write each file with different file name in it?
import os
if not os.path.exists(directory):
os.makedirs(directory)
with open(directory +"\\"+'abc.xml', 'wb') as file:
file.write(a.content)
Try formatting file name:
import os
if not os.path.exists(directory):
os.makedirs(directory)
for i in range(10):
with open(directory + "\\" + f'abc_{i}.xml', 'wb') as file:
file.write(a.content)
the simple way to get the filename as an argument is by using argv
sys.argv[0] - is the script name - python
sys.argv[1] - is the first argument to the scripts
so you can do the following:
with open(directory +"\\"+sys.argv[1] , 'wb') as file:
file.write(a.content)
import os
directory = os.listdir("E:\append_some_line_to_the_start_of_a_file")
for file in directory:
print(file)
afer this I want to write in all the files in the same time.
You should use ExitStack for this:
from contextlib import ExitStack
import os
dir = 'E:\append_some_line_to_the_start_of_a_file'
with ExitStack() as stack:
files = [stack.enter_context(open(os.path.join(dir,fname), 'w')) for fname in os.listdir('./dir')]
for file in files:
file.write('test')