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why python code for google maps api can not work suddenly

the code can work before but i found it can not work recently. please saw my code as below
def signal_distance(lat1,lng1,lat2,lng2):
import simplejson, urllib
orig_coord =lat1,lng1
dest_coord = lat2,lng2
#API request
url = "https://maps.googleapis.com/maps/api/distancematrix/json?origins={0}&destinations={1}&mode=transit&language=zh-TW&key=AIzaSyBlwZDhGYNTrxXiQblz20v3poJTA7zTVho".format(str(orig_coord),str(dest_coord))
result = simplejson.load(urllib.urlopen(url))
print result
signal_distance(25.082969,121.5549714,24.9988582,121.5788795)

It seems the Distance Matrix API request that generates signal_distance function is not valid since it contains parentheses for origins and destinations parameters.
The specified example returns for me NOT_FOUND Element-level status code:
indicates that the origin and/or destination of this pairing could not
be geocoded
The solution would be to replace request from:
https://maps.googleapis.com/maps/api/distancematrix/json?origins=(<latOrig>,<lngOrig>)&destinations=(<latDest>,<lngDest>)&mode=transit&language=<lang>&key=<Key>
to
https://maps.googleapis.com/maps/api/distancematrix/json?origins=<latOrig>,<lngOrig>&destinations=<latDest>,<lngDest>&mode=transit&language=<lang>&key=<Key>
Modified example
def signal_distance(lat1,lng1,lat2,lng2):
import simplejson, urllib
orig_coord = lat1,lng1
dest_coord = lat2,lng2
orig_coord_str = ' ,'.join(map(str, orig_coord))
dest_coord_str = ' ,'.join(map(str, dest_coord))
#API request
url = "https://maps.googleapis.com/maps/api/distancematrix/json?origins={0}&destinations={1}&mode=transit&language=zh-TW&key=AIzaSyBlwZDhGYNTrxXiQblz20v3poJTA7zTVho".format(orig_coord_str,dest_coord_str)
result = simplejson.load(urllib.urlopen(url))
print result

JSON serialization Error in Python 3.2

I am using JSON library and trying to import a page feed to an CSV file. Tried many a ways to get the result however every time code execute it Gives JSON not serialzable. No Facebook use auth code which I have and used it so connection string will change however if you use a page which has public privacy you will still be able to get the result from below code.
following is the code
import urllib3
import json
import requests
#from pprint import pprint
import csv
from urllib.request import urlopen
page_id = "abcd" # username or id
api_endpoint = "https://graph.facebook.com"
fb_graph_url = api_endpoint+"/"+page_id
try:
#api_request = urllib3.Requests(fb_graph_url)
#http = urllib3.PoolManager()
#api_response = http.request('GET', fb_graph_url)
api_response = requests.get(fb_graph_url)
try:
#print (list.sort(json.loads(api_response.read())))
obj = open('data', 'w')
# write(json_dat)
f = api_response.content
obj.write(json.dumps(f))
obj.close()
except Exception as ee:
print(ee)
except Exception as e:
print( e)
Tried many approach but not successful. hope some one can help

api_response.content is the text content of the API, not a Python object so you won't be able to dump it.
Try either:
f = api_response.content
obj.write(f)
Or
f = api_response.json()
obj.write(json.dumps(f))

requests.get(fb_graph_url).content
is probably a string. Using json.dumps on it won't work. This function expects a list or a dictionary as the argument.
If the request already returns JSON, just write it to the file.

django/python variables inside strings?

I'm new to python and would like some assistance.
I have a variable
q = request.GET['q']
How do I insert the variable q inside this:
url = "http://search.com/search?term="+q+"&location=sf"
Now I'm not sure what the convention is? I'm used to PHP or javascript, but I'm learning python and how do you insert a variable dynamically?

Use the format method of String:
url = "http://search.com/search?term={0}&location=sf".format(q)
But of course you should URL-encode the q:
import urllib
...
qencoded = urllib.quote_plus(q)
url =
"http://search.com/search?term={0}&location=sf".format(qencoded)

One way to do it is using urllib.urlencode(). It accepts a dictionary(or associative array or whatever you call it) taking key-value pairs as parameter and value and you can encode it to form urls
from urllib import urlencode
myurl = "http://somewebsite.com/?"
parameter_value_pairs = {"q":"q_value","r":"r_value"}
req_url = url + urlencode(parameter_value_pair)
This will give you "http://somewebsite.com/?q=q_value&r=r_value"

q = request.GET['q']
url = "http://search.com/search?term=%s&location=sf" % (str(q))
Use this it will be faster...

Python: How to get the Content-Type of an URL?

I need to get the content-type of an internet(intranet) resource not a local file. How can I get the MIME type from a resource behind an URL:
I tried this:
res = urllib.urlopen("http://www.iana.org/assignments/language-subtag-registry")
http_message = res.info()
message = http_message.getplist()
I get:
['charset=UTF-8']
How can I get the Content-Type, can be done using urllib and how or if not what is the other way?

A Python3 solution to this:
import urllib.request
with urllib.request.urlopen('http://www.google.com') as response:
info = response.info()
print(info.get_content_type()) # -> text/html
print(info.get_content_maintype()) # -> text
print(info.get_content_subtype()) # -> html

res = urllib.urlopen("http://www.iana.org/assignments/language-subtag-registry" )
http_message = res.info()
full = http_message.type # 'text/plain'
main = http_message.maintype # 'text'

Update: since info() function is deprecated in Python 3.9, you can read about the preferred type called headers here
import urllib
r = urllib.request.urlopen(url)
header = r.headers # type is email.message.EmailMessage
contentType = header.get_content_type() # or header.get('content-type')
contentLength = header.get('content-length')
filename = header.get_filename()
also, a good way to quickly get the mimetype without actually loading the url
import mimetypes
contentType, encoding = mimetypes.guess_type(url)
The second method does not guarantee an answer but is a quick and dirty trick since it's just looking at the URL string rather than actually opening the URL.

Extracting data from a URL result with special formatting

I have a URL:
http://somewhere.com/relatedqueries?limit=2&query=seedterm
where modifying the inputs, limit and query, will generate wanted data. Limit is the max number of term possible and query is the seed term.
The URL provides text result formatted in this way:
oo.visualization.Query.setResponse({version:'0.5',reqId:'0',status:'ok',sig:'1303596067112929220',table:{cols:[{id:'score',label:'Score',type:'number',pattern:'#,##0.###'},{id:'query',label:'Query',type:'string',pattern:''}],rows:[{c:[{v:0.9894380670262618,f:'0.99'},{v:'newterm1'}]},{c:[{v:0.9894380670262618,f:'0.99'},{v:'newterm2'}]}],p:{'totalResultsCount':'7727'}}});
I'd like to write a python script that takes two arguments (limit number and the query seed), go fetch the data online, parse the result and return a list with the new terms ['newterm1','newterm2'] in this case.
I'd love some help, especially with the URL fetching since I have never done this before.

It sounds like you can break this problem up into several subproblems.
Subproblems
There are a handful of problems that need to be solved before composing the completed script:
Forming the request URL: Creating a configured request URL from a template
Retrieving data: Actually making the request
Unwrapping JSONP: The returned data appears to be JSON wrapped in a JavaScript function call
Traversing the object graph: Navigating through the result to find the desired bits of information
Forming the request URL
This is just simple string formatting.
url_template = 'http://somewhere.com/relatedqueries?limit={limit}&query={seedterm}'
url = url_template.format(limit=2, seedterm='seedterm')
Python 2 Note
You will need to use the string formatting operator (%) here.
url_template = 'http://somewhere.com/relatedqueries?limit=%(limit)d&query=%(seedterm)s'
url = url_template % dict(limit=2, seedterm='seedterm')
Retrieving data
You can use the built-in urllib.request module for this.
import urllib.request
data = urllib.request.urlopen(url) # url from previous section
This returns a file-like object called data. You can also use a with-statement here:
with urllib.request.urlopen(url) as data:
# do processing here
Python 2 Note
Import urllib2 instead of urllib.request.
Unwrapping JSONP
The result you pasted looks like JSONP. Given that the wrapping function that is called (oo.visualization.Query.setResponse) doesn't change, we can simply strip this method call out.
result = data.read()
prefix = 'oo.visualization.Query.setResponse('
suffix = ');'
if result.startswith(prefix) and result.endswith(suffix):
result = result[len(prefix):-len(suffix)]
Parsing JSON
The resulting result string is just JSON data. Parse it with the built-in json module.
import json
result_object = json.loads(result)
Traversing the object graph
Now, you have a result_object that represents the JSON response. The object itself be a dict with keys like version, reqId, and so on. Based on your question, here is what you would need to do to create your list.
# Get the rows in the table, then get the second column's value for
# each row
terms = [row['c'][2]['v'] for row in result_object['table']['rows']]
Putting it all together
#!/usr/bin/env python3
"""A script for retrieving and parsing results from requests to
somewhere.com.
This script works as either a standalone script or as a library. To use
it as a standalone script, run it as `python3 scriptname.py`. To use it
as a library, use the `retrieve_terms` function."""
import urllib.request
import json
import sys
E_OPERATION_ERROR = 1
E_INVALID_PARAMS = 2
def parse_result(result):
"""Parse a JSONP result string and return a list of terms"""
prefix = 'oo.visualization.Query.setResponse('
suffix = ');'
# Strip JSONP function wrapper
if result.startswith(prefix) and result.endswith(suffix):
result = result[len(prefix):-len(suffix)]
# Deserialize JSON to Python objects
result_object = json.loads(result)
# Get the rows in the table, then get the second column's value
# for each row
return [row['c'][2]['v'] for row in result_object['table']['rows']]
def retrieve_terms(limit, seedterm):
"""Retrieves and parses data and returns a list of terms"""
url_template = 'http://somewhere.com/relatedqueries?limit={limit}&query={seedterm}'
url = url_template.format(limit=limit, seedterm=seedterm)
try:
with urllib.request.urlopen(url) as data:
data = perform_request(limit, seedterm)
result = data.read()
except:
print('Could not request data from server', file=sys.stderr)
exit(E_OPERATION_ERROR)
terms = parse_result(result)
print(terms)
def main(limit, seedterm):
"""Retrieves and parses data and prints each term to standard output"""
terms = retrieve_terms(limit, seedterm)
for term in terms:
print(term)
if __name__ == '__main__'
try:
limit = int(sys.argv[1])
seedterm = sys.argv[2]
except:
error_message = '''{} limit seedterm
limit must be an integer'''.format(sys.argv[0])
print(error_message, file=sys.stderr)
exit(2)
exit(main(limit, seedterm))
Python 2.7 version
#!/usr/bin/env python2.7
"""A script for retrieving and parsing results from requests to
somewhere.com.
This script works as either a standalone script or as a library. To use
it as a standalone script, run it as `python2.7 scriptname.py`. To use it
as a library, use the `retrieve_terms` function."""
import urllib2
import json
import sys
E_OPERATION_ERROR = 1
E_INVALID_PARAMS = 2
def parse_result(result):
"""Parse a JSONP result string and return a list of terms"""
prefix = 'oo.visualization.Query.setResponse('
suffix = ');'
# Strip JSONP function wrapper
if result.startswith(prefix) and result.endswith(suffix):
result = result[len(prefix):-len(suffix)]
# Deserialize JSON to Python objects
result_object = json.loads(result)
# Get the rows in the table, then get the second column's value
# for each row
return [row['c'][2]['v'] for row in result_object['table']['rows']]
def retrieve_terms(limit, seedterm):
"""Retrieves and parses data and returns a list of terms"""
url_template = 'http://somewhere.com/relatedqueries?limit=%(limit)d&query=%(seedterm)s'
url = url_template % dict(limit=2, seedterm='seedterm')
try:
with urllib2.urlopen(url) as data:
data = perform_request(limit, seedterm)
result = data.read()
except:
sys.stderr.write('%s\n' % 'Could not request data from server')
exit(E_OPERATION_ERROR)
terms = parse_result(result)
print terms
def main(limit, seedterm):
"""Retrieves and parses data and prints each term to standard output"""
terms = retrieve_terms(limit, seedterm)
for term in terms:
print term
if __name__ == '__main__'
try:
limit = int(sys.argv[1])
seedterm = sys.argv[2]
except:
error_message = '''{} limit seedterm
limit must be an integer'''.format(sys.argv[0])
sys.stderr.write('%s\n' % error_message)
exit(2)
exit(main(limit, seedterm))

i didn't understand well your problem because from your code there it seem to me that you use Visualization API (it's the first time that i hear about it by the way).
But well if you are just searching for a way to fetch data from a web page you could use urllib2 this is just for getting data, and if you want to parse the retrieved data you will have to use a more appropriate library like BeautifulSoop
if you are dealing with another web service (RSS, Atom, RPC) rather than web pages you can find a bunch of python library that you can use and that deal with each service perfectly.
import urllib2
from BeautifulSoup import BeautifulSoup
result = urllib2.urlopen('http://somewhere.com/relatedqueries?limit=%s&query=%s' % (2, 'seedterm'))
htmletxt = resul.read()
result.close()
soup = BeautifulSoup(htmltext, convertEntities="html" )
# you can parse your data now check BeautifulSoup API.