Develop Reference

Submitting Python file in batch mode in Livy(without Hadoop installed)

i have made a simple python file which want to submit in Livy.Livy is currently running in local mode.Also I have mentioned following property in livy.conf file.
Property name: livy.file.local-dir-whitelist,
value "/usr/local/livy/scripts"
My file is kept in following path "/usr/local/livy/scripts"
import json, pprint, requests, textwrap
host = 'http://localhost:8998'
data = {'kind': 'spark'}
headers = {'Content-Type': 'application/json'}
r = requests.post(host + '/sessions', data=json.dumps(data), headers=headers)
r.json()
I am submitting it using curl as follows:
curl -X POST --data '{"file": "/usr/local/livy/scripts/pi.py"}' -H "Content-Type: application/json" 10.140.178.24:8999/batches
It is giving me following error:
requirement failed: Local path /usr/local/livy/scripts/pi.py cannot be added to user sessions.
My Ubuntu system only have following things:
a)Spark
b)Livy
c)Java
What am I doing wrong here?

For people using incubating mode of livy for first time,kindly check that the template file is renamed with stripping off .template in livy.conf.template.Then make sure that the following configurations are present in it.
livy.spark.master = local
livy.file.local-dir-whitelist = /path/to/script/folder/
Kindly make sure that forward slash is present in end of path
Then write url in following manner for
Python:
curl -v -X POST --data '{"file": "/path/to/script/folder/name-of-python-file.py"}' -H "Content-Type: application/json" localhost:8998/batches
Note:It will not accept relative path,whole absolute path needs to be defined in it.

curl -X POST --data '{"file": "/usr/local/livy/scripts/pi.py"}' -H "Content-Type: application/json" 10.140.178.24:8999/batches
{"id":2,"state":"starting","log":[]}

Cannot translate a working CURL PUT statement into a functioning python script

Here is the working curl statement that I am trying to put into a python script:
curl -L -H 'X-Cisco-Meraki-API-Key: <key>' -X PUT -H'Content-Type: application/json' --data-binary '{"name":"new SSID name", "enabled":true, "authMode":"psk", "encryptionMode":"wpa", "psk":"abcd1234", "ipAssignmentMode":"Bridge mode"}' 'https://dashboard.meraki.com/api/v0/networks/[networkId]/ssids/[number]'
Here is my python code.
import requests
import json
url = "https://dashboard.meraki.com/api/v0/networks/XXXXXXX/ssids/2"
headers = {'X-Cisco-Meraki-API-Key': 'YYYYYYY', 'content-type': 'application/json'}
payload = {"name":"test", "enabled":"true", "authMode":"psk", "encryptionMode":"wpa", "psk":"abcd1234", "ipAssignmentMode":"NAT mode"}
r = requests.put(url, headers=headers, data=payload, allow_redirects=True)
print r.status_code
When I run the curl command, it works, but when I try the python script, I get a 400 error message. My guess is that it has to do with the way that the application is interpreting the payload. Any help would be greatly appreciated.

You need to use:
data=json.dumps(payload)
Instead of:
data=payload
When you add 'content-type': 'application/json' header, you mean that you're sending a json data, that's why you need to use json.dumps() to sent a valide json data instead of a Python dictionary / object.

using testrail api in python script

I have the curl command, but not sure about how to run that in python script.
curl -H "Content-Type: application/json" -u "username:password" -d '{ "name":"something" }' "https://xxxxxxxx"
I'm planning to use subprocess, but the api documents aren't very helpful.
Also does anyone know how to get the sectionId from testrail?

Bill from TestRail here. You can find a link to our Python bindings here:
http://docs.gurock.com/testrail-api2/bindings-python
Regarding getting the section ID, you can use the get_sections method for a project/suite to return all the section details including IDs. You can find more info on that here:
http://docs.gurock.com/testrail-api2/reference-sections#get_sections
If you're looking for the section ID for a specific test case, you can get that from the get_case method.

You probably want to use the requests package for this. The curl command translates to something like this:
import json
import requests
response = requests.post('https://xxxxxxxx',
data=json.dumps({'name': 'something'}),
headers={'Content-Type': 'application/json'},
auth=('username', 'password'))
response_data = response.json()
If you really want to use subprocess, you can do something like this:
import subprocess
curl_args = ['curl', '-H', 'Content-Type: application/json', '-u', 'username:password',
'-d', '{ "name":"something" }', 'https://xxxxxxxx']
curl_output = subprocess.check_output(curl_args)
I consider the latter approach less "Pythonic".

How to use Python to execute a cURL command?

I want to execute a curl command in Python.
Usually, I just need to enter the command in the terminal and press the return key. However, I don't know how it works in Python.
The command shows below:
curl -d #request.json --header "Content-Type: application/json" https://www.googleapis.com/qpxExpress/v1/trips/search?key=mykeyhere
There is a request.json file to be sent to get a response.
I searched a lot and got confused. I tried to write a piece of code, although I could not fully understand it and it didn't work.
import pycurl
import StringIO
response = StringIO.StringIO()
c = pycurl.Curl()
c.setopt(c.URL, 'https://www.googleapis.com/qpxExpress/v1/trips/search?key=mykeyhere')
c.setopt(c.WRITEFUNCTION, response.write)
c.setopt(c.HTTPHEADER, ['Content-Type: application/json','Accept-Charset: UTF-8'])
c.setopt(c.POSTFIELDS, '#request.json')
c.perform()
c.close()
print response.getvalue()
response.close()
The error message is Parse Error. How to get a response from the server correctly?

For the sake of simplicity, you should consider using the Requests library.
An example with JSON response content would be something like:
import requests
r = requests.get('https://github.com/timeline.json')
r.json()
If you look for further information, in the Quickstart section, they have lots of working examples.
For your specific curl translation:
import requests
url = 'https://www.googleapis.com/qpxExpress/v1/trips/search?key=mykeyhere'
payload = open("request.json")
headers = {'content-type': 'application/json', 'Accept-Charset': 'UTF-8'}
r = requests.post(url, data=payload, headers=headers)

Use curlconverter.com. It'll convert almost any curl command into Python, Node.js, PHP, R, Go and more.
Example:
curl -X POST -H 'Content-type: application/json' --data '{"text":"Hello, World!"}' https://hooks.slack.com/services/asdfasdfasdf
becomes this in Python
import requests
json_data = {
'text': 'Hello, World!',
}
response = requests.post('https://hooks.slack.com/services/asdfasdfasdf', json=json_data)

curl -d #request.json --header "Content-Type: application/json" https://www.googleapis.com/qpxExpress/v1/trips/search?key=mykeyhere
its Python implementation looks like this:
import requests
headers = {
'Content-Type': 'application/json',
}
params = {
'key': 'mykeyhere',
}
with open('request.json') as f:
data = f.read().replace('\n', '')
response = requests.post('https://www.googleapis.com/qpxExpress/v1/trips/search', params=params, headers=headers, data=data)
Check this link, it will help convert cURL commands to Python, PHP and Node.js

import requests
url = "https://www.googleapis.com/qpxExpress/v1/trips/search?key=mykeyhere"
data = requests.get(url).json
maybe?
if you are trying to send a file
files = {'request_file': open('request.json', 'rb')}
r = requests.post(url, files=files)
print r.text, print r.json
ahh thanks #LukasGraf now i better understand what his original code is doing
import requests,json
url = "https://www.googleapis.com/qpxExpress/v1/trips/search?key=mykeyhere"
my_json_data = json.load(open("request.json"))
req = requests.post(url,data=my_json_data)
print req.text
print
print req.json # maybe?

My answer is WRT python 2.6.2.
import commands
status, output = commands.getstatusoutput("curl -H \"Content-Type:application/json\" -k -u (few other parameters required) -X GET https://example.org -s")
print output
I apologize for not providing the required parameters 'coz it's confidential.

I had this exact question because I had to do something to retrieve content, but all I had available was an old version of Python with inadequate SSL support. If you're on an older MacBook, you know what I'm talking about. In any case, curl runs fine from a shell (I suspect it has modern SSL support linked in) so sometimes you want to do this without using requests or urllib.request.
You can use the subprocess module to execute curl and get at the retrieved content:
import subprocess
# 'response' contains a []byte with the retrieved content.
# use '-s' to keep curl quiet while it does its job, but
# it's useful to omit that while you're still writing code
# so you know if curl is working
response = subprocess.check_output(['curl', '-s', baseURL % page_num])
Python 3's subprocess module also contains .run() with a number of useful options.

I use os library.
import os
os.system("sh script.sh")
script.sh literally only contains the curl.

PYTHON 3
Only works within UNIX (Linux / Mac) (!)
Executing a cURL with Python 3 and parsing its JSON data.
import shlex
import json
import subprocess
# Make sure that cURL has Silent mode (--silent) activated
# otherwise we receive progress data inside err message later
cURL = r"""curl -X --silent POST http://www.test.testtestest/ -d 'username=test'"""
lCmd = shlex.split(cURL) # Splits cURL into an array
p = subprocess.Popen(lCmd, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
out, err = p.communicate() # Get the output and the err message
json_data = json.loads(out.decode("utf-8"))
print(json_data) # Display now the data
Sometimes you also need to install these dependencies on UNIX if you experience strange errors:
# Dependencies
sudo apt install libcurl4-openssl-dev libssl-dev
sudo apt install curl

use requests lib.. this code is :
curl -LH "Accept: text/x-bibliography; style=apa" https://doi.org/10.5438/0000-0C2G
equal to this:
import requests
headers = {
'Accept': 'text/x-bibliography; style=apa',
}
r = requests.get('https://doi.org/10.5438/0000-0C2G', headers=headers)
print(r.text)

if you os supporting curl you can do something like this:
import os
os.system("curl -d #request.json --header "Content-Type: application/json" https://www.googleapis.com/qpxExpress/v1/trips/search?key=mykeyhere")
I'm using this way... And I think you can use this too!
by the way.. the module "os" is auto-installing when you install python.
soo, you don't need to install packages ;)

This is one approach:
Import os
import requests
Data = os.execute(curl URL)
R= Data.json()

How to POST a local file using urllib2 in Python?

I am a complete Python noob and am trying to run cURL equivalents using urllib2. What I want is a Python script that, when run, will do the exact same thing as the following cURL command in Terminal:
curl -k -F docfile=#myLocalFile.csv http://myWebsite.com/extension/extension/extension
I found the following template on a tutorial page:
import urllib
import urllib2
url = "https://uploadWebsiteHere.com"
data = "{From: 'sender#email.com', To: 'recipient#email.com', Subject: 'Postmark test', HtmlBody: 'Hello dear Postmark user.'}"
headers = { "Accept" : "application/json",
"Conthent-Type": "application/json",
"X-Postmark-Server-Token": "abcdef-1234-46cc-b2ab-38e3a208ab2b"}
req = urllib2.Request(url, data, headers)
response = urllib2.urlopen(req)
the_page = response.read()
but I am completely lost on the 'data' and 'headers' vars. The urllib2 documentation (https://docs.python.org/2/library/urllib2.html) defines the 'data' input as "a string specifying additional data to send to the server" and the 'headers' input as "a dictionary". I am totally out of my depth in trying to follow this documentation and do not see why a dictionary is necessary when I could accomplish this same task in terminal by only specifying the file and URL. Thoughts, please?

The data you are posting doesn't appear to be valid JSON. Assuming the server is expecting valid JSON, you should change that.
Your curl invocation does not pass any optional headers, so you shouldn't need to provide much in the request. If you want to verify the exact headers you could add -vi to the curl invocation and directly match them in the Python code. Alternatively, this works for me:
import urllib2
url = "http://localhost:8888/"
data = '{"From": "sender#email.com", "To": "recipient#email.com", "Subject": "Postmark test", "HtmlBody": "Hello dear Postmark user."}'
headers = {
"Content-Type": "application/json"
}
req = urllib2.Request(url, data, headers)
response = urllib2.urlopen(req)
the_page = response.read()
It probably is in your best interest to switch over to using requests, but for something this simple the standard library urllib2 can be made to work.

What I want is a Python script that, when run, will do the exact same thing as the following cURL command in Terminal:
$ curl -k -F docfile=#myLocalFile.csv https://myWebsite.com/extension...
curl -F sends the file using multipart/form-data content type. You could reproduce it easily using requests library:
import requests # $ pip install requests
with open('myLocalFile.csv','rb') as input_file:
r = requests.post('https://myWebsite.com/extension/...',
files={'docfile': input_file}, verify=False)
verify=False is to emulate curl -k.