Error parsing JSON string

Question

I am having difficulty parsing a json string that appears to be valid json.

My code is the following. I'm trying to grab some simple json from a URL.

import urllib2    
import simplejson
req = urllib2.Request("http://www.rentrent.org/RENT/Ads.aspx?xmin=-118.01925659179687&ymin=33.71948521132481&xmax=-117.68142700195314&ymax=33.85644642218431&bd=&ba=&pets=-1&type=2&throwErrorIfOverLimit=false&callback=xxx")  

opener = urllib2.build_opener()    
f = opener.open(req)    
content = f.read();    
print "content = " + content   # Appears to print valid json string
json = simplejson.loads(content)

I get an error on simplejson.loads(content):

raise JSONDecodeError("No JSON object could be decoded", s, idx) 
simplejson.decoder.JSONDecodeError: No JSON object could be decoded: line 1 column 0 (char 0)

This is confusing because content appears to be a perfectly good json string.

I need to get to this data as individual elements. Any ideas how to get rid of this error?

You can get rid of the error by giving it valid json, or ignoring the failure. Need to see json. — jdi
– jdi, Commented Apr 29, 2012 at 2:19
I wanted to correct my last comment by saying we don't need to see the json at this point. I looked at the url and saw the data, though its convenient for everyone if a snippet is posted here. — jdi
– jdi, Commented Apr 29, 2012 at 4:30

Community · Accepted Answer · 2017-05-23 12:21:17Z

3

This is a jsonp api url. Its expecting to return to you a response wrapped in a function call xxx(). Look into how to handle jsonp. What you are currently getting is not actually valid jsonp. Its valid javascript.

Here is another question about the difference:
What are the differences between JSON and JSONP?

You can remove the callback param from your url and get back pure json:
http://www.rentrent.org/RENT/Ads.aspx?xmin=-118.01925659179687&ymin=33.71948521132481&xmax=-117.68142700195314&ymax=33.85644642218431&bd=&ba=&pets=-1&type=2&throwErrorIfOverLimit=false

Update

I overlooked that removing the callback param from the url doesn't actually tell this api to return JSON. It only prevents it from being wrapped into a function. The return value is still not json. It is javascript. To be specific, here is an example:

Javascript From the return value

{Status:'Success',

JSON

{'Status':'Success',

All of the keys need to be quoted. Your options are either:

See if the api supports a json return value instead of jsonp
Do a regex fix on the output to quote all of the keys before decoding

edited May 23, 2017 at 12:21

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answered Apr 29, 2012 at 2:22

jdi

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2 Comments

codingJoe Over a year ago

Thanks! I see how it is no longer wrapped in a function call. However it still won't parse. I get a

raise JSONDecodeError("Expecting property name", s, end) simplejson.decoder.JSONDecodeError: Expecting property name: line 1 column 1 (char 1)

Does that url return a valid json or am I parsing it incorrectly?

btk Over a year ago

Some APIs (like Stripe's for example) will return JSONP if callback argument is provided, JSON otherwise.

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