问题
I want to retrieve points of interests of a place/country through google places api say 'points of interests in london'. I want the results to be same as I google search it like [here][1. I've used something like this
'https://maps.googleapis.com/maps/api/place/radarsearch/json?location='+str(lat)+','+str(long)+'&radius=500&type=tourist&rankby=prominence&key=API_KEY')
where lat,long are respective latitude,longitude of that place/country.
But the resulting json file does not retrieve points of interests of that place rather it just outputs nearby places.
i also tried this
'https://maps.googleapis.com/maps/api/place/nearbysearch/json?location='+str(lat)+','+str(long)+'&radius=500&type=tourist&&rankby=prominence&key=API_KEY'
which ended up in same results
is there any option to do it without using radius parameter
please do help me
回答1:
Try this google places API url. You will get the point of interest/Attraction/Tourists places in (For Eg:) New York City. You have to use the CITY NAME with the keyword Point Of Interest
https://maps.googleapis.com/maps/api/place/textsearch/json?query=new+york+city+point+of+interest&language=en&key=API_KEY
These API results are same as the results of the below google search results.
https://www.google.com/search?sclient=psy-ab&site=&source=hp&btnG=Search&q=New+York+point+of+interest
回答2:
Try this google place API url. https://maps.googleapis.com/maps/api/place/textsearch/json?query=point+of+interest+in+cityName&key=API_KEY
回答3:
It looks like you may be looking for an API that surfaces Google Search results, and there is one: Google Custom Search https://developers.google.com/custom-search/docs/overview
You could use Nearby or Radar search, if you'd be looking for more specific, supported types: https://developers.google.com/places/supported_types#table1
回答4:
Please tied to use google API
- Step 1 : Create Google API key Get API Key
- Step 2: Add Places API
- Step 3: Implement
$url= "https://maps.googleapis.com/maps/api/place/textsearch/json?query=dubai+point+of+interest&language=en&radius=2000&key=API_KEY";
$json=file_get_contents($url);
$obj = json_decode($json);
print_r($obj);
来源:https://stackoverflow.com/questions/37874018/retrieve-points-of-interests-attractions-tourist-places-of-a-place-country-usi