Read JSON response in Python

Question

I am trying to read json response from this link. But its not working! I get the following error:

ValueError: No JSON object could be decoded.

Here is the code I've tried:

import urllib2, json
a = urllib2.urlopen('https://www.googleapis.com/pagespeedonline/v3beta1/mobileReady?key=AIzaSyDkEX-f1JNLQLC164SZaobALqFv4PHV-kA&screenshot=true&snapshots=true&locale=en_US&url=https://www.economicalinsurance.com/en/&strategy=mobile&filter_third_party_resources=false&callback=_callbacks_._DElanZU7Xh1K')
data = json.loads(a)

I made these changes:

import requests, json
r=requests.get('https://www.googleapis.com/pagespeedonline/v3beta1/mobileReady?key=AIzaSyDkEX-f1JNLQLC164SZaobALqFv4PHV-kA&screenshot=true&snapshots=true&locale=en_US&url=https://www.economicalinsurance.com/en/&strategy=mobile&filter_third_party_resources=false')
json_data = json.loads(r.text)
print json_data['ruleGroups']['USABILITY']['score']

A Quick question - Construct Image link .

I able to get here : -

from selenium import webdriver

txt = json_data['screenshot']['data']
txt = str(txt).replace('-','/').replace('_','/')
#then in order to construct the image link i tried : -
image_link = 'data:image/jpeg;base64,'+txt
driver = webdriver.Firefox()
driver.get(image_link)

The problem is i am not getting the image, also the len(object_original) as compared len(image_link) differs . Could anybody please advise the right elements missing in my constructed image link ?. Thank you

Here is API link - https://www.google.co.uk/webmasters/tools/mobile-friendly/ Sorry added it late .

I assume the server returns JSONP because of `callback=_callbacks_._DElanZU7Xh1K`. Remove it and it will likely return JSON. — Felix Kling, Oct 22 '15 at 13:23
Thanks felix-kling , removing the 'callback' parameter from url solved my problem , I will modify the requests parameter import requests, json r = requests.get('https://www.googleapis.com/pagespeedonline/v3beta1/mobileReady?key=AIzaSyDkEX-f1JNLQLC164SZaobALqFv4PHV-kA&screenshot=true&snapshots=true&locale=en_US&url=https://www.economicalinsurance.com/en/&strategy=mobile&filter_third_party_resources=false') json_data = json.loads(r.text) print json_data['ruleGroups']['USABILITY']['score'] Could add your answer below, i will accept it now . — Shekhar Samanta, Oct 22 '15 at 13:44

score 2 · Accepted Answer · edited May 23 '17 at 11:45

2

Two corrections need to be made to your code:

The url was corrected (as mentioned by Felix Kling here). You have to remove the callback parameter from the GET request you were sending.
Also, if you check the type of the response that you were fetching earlier you'll notice that it wasn't a string. It was <type 'instance'>. And since json.loads() accepts a string as a parameter variable you would've got another error. Therefore, use a.read() to fetch the response data in string.

Hence, this should be your code:

import urllib2, json
a = urllib2.urlopen('https://www.googleapis.com/pagespeedonline/v3beta1/mobileReady?key=AIzaSyDkEX-f1JNLQLC164SZaobALqFv4PHV-kA&screenshot=true&snapshots=true&locale=en_US&url=https://www.economicalinsurance.com/en/&strategy=mobile&filter_third_party_resources=false')
data = json.loads(a.read())

Answer to your second query (regarding the image) is:

from base64 import decodestring

arr = json_data['screenshot']['data']
arr = arr.replace("_", "/")
arr = arr.replace("-","+")

fh = open("imageToSave.jpeg", "wb")
fh.write(str(arr).decode('base64'))
fh.close()

Here, is the image you were trying to fetch - Link

edited May 23 '17 at 11:45

Community

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answered Oct 22 '15 at 13:30

JRodDynamite

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hey Jason, That was amazing . i will change the "imageToSave.jpeg" string to list elements to iterate over a list of domains. Thanks – Shekhar Samanta Oct 23 '15 at 06:31
Awesome. Next time if you have another query please post it as a separate question especially if the second query has no correlation to the first query. Anyways, Enjoy! :) – JRodDynamite Oct 23 '15 at 06:39
You are a life saver man , Could you please help me on this - http://stackoverflow.com/questions/33304357/write-json-elements-to-csv-file-python – Shekhar Samanta Oct 23 '15 at 14:15
Will have a look into it. – JRodDynamite Oct 23 '15 at 14:58
Okay. Do notify me like this if you have any other issue. – JRodDynamite Oct 23 '15 at 15:04
@ShekharSamanta - Sorry, I'm not familiar with tkinter. But will have a look though. – JRodDynamite Oct 30 '15 at 14:23

score 0 · Answer 2 · answered Oct 22 '15 at 13:37

Felix Kling is right about the address, but I also created a variable that holds the URL. You can try this out to and it should work:

import urllib2, json
url = "https://www.googleapis.com/pagespeedonline/v3beta1/mobileReady?key=AIzaSyDkEX-f1JNLQLC164SZaobALqFv4PHV-kA&screenshot=true&snapshots=true&locale=en_US&url=https://www.economicalinsurance.com/en/&strategy=mobile&filter_third_party_resources=false"
response = urllib2.urlopen(url)
data = json.loads(response.read())
print data

Read JSON response in Python

2 Answers2

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