Alec, there's no limit.  It makes searching very simple.  Please be
sure to get the most up-to-date version of the API.

On Dec 13, 6:10 am, "Alec Bennett" <[email protected]> wrote:
> Interesting. Do you know if it's limited to 64 results?
>
> FYI, here's how I went about it. It works, but is restricted to the first 64
> results. Note that all I needed was the URLs of the search results. And
> there's lots of duct tape here. To neaten this up research JSON.
>
> def query_google(query, start):
>
>     import urllib2, urllib
>
>     #http://ajax.googleapis.com/ajax/services/search/web?v=1.0&q=Paris%20H...
>
>     # urlencode the query
>     query = urllib.quote(query)
>
>     url = 'http://ajax.googleapis.com/ajax/services/search/web?v=1.0&q='+
> query + '&start=' + str(start) + '&rsz=large'
>
>     try:
>         req = urllib2.Request(url)
>         opener = urllib2.build_opener()
>         data_string = opener.open(req).read()
>
>     except urllib2.URLError:
>         print "------ Error opening " + url + "..... Timed out?"
>         return None
>
>     # Should use json to parse the results, but instead we're converting the
> string to dictionary. Duct tape.
>
>     # replace the "null" with "None" for Python
>     data_string = data_string.replace(": null,", ": None,")
>
>     # convert the string to a dictionary
>     exec("data = " + data_string)
>
>     # simplify the results a bit
>     results = data["responseData"]["results"]
>
>     # build list of urls
>     urls = []
>
>     for i in results:
>         url = i["url"]
>         url = url.split("%")[0] # get rid of some garbage from the url.
> Probably avoidable by using Json.
>         urls.append(url)
>
>     return urls
>
> results = query_google("whatever", start=1)
> print results
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