解析从Google Maps API返回的JSON

Question

我试图对地址进行地址解析，在这种情况下，Google地址是HQ。

当我试图让jQuery的$.ajax()请求得到JSON，它说：未捕获的SyntaxError：意外的标记：我好像

无法揣摩出意外：是。我想要做的是将地址作为一个字符串解码lat和long，并使用Places API找到这些坐标附近的机制。

$.ajax({ 
    url: 'https://maps.googleapis.com/maps/api/geocode/json?address=1600+Amphitheatre+Parkway,+Mountain+View,+CA&key=myKey', 
    dataType: 'jsonp', 
    jsonp: 'callback', 
    method: 'GET', 
    success: function(results){ 
     console.log(results); 
     queryLat = results['results']['geometry']['location']['lat']; 
     queryLong = results['results']['geometry']['location']['lng']; 
     console.log('latitude: '+queryLat); 
     console.log('longitude: '+queryLong); 


     function callback(mapsResults, status){ 
      if(status == google.maps.places.PlacesServiceStatus.OK){ 
       for(var i = 0; i < mapResults.length; i++){ 
        var place = mapResultsi]; 
        createMarker(mapResults[i]); 
       } 
      } 
     } 
     var map; 
     var service; 

     map = new google.maps.Map(document.getElementById('map'), { 
      center: new google.maps.LatLng(queryLat, queryLong), 
      zoom: 15 
     }); 

     var request ={ 
      location: new google.maps.LatLng(queryLat, queryLong) 
      radius:'500', 
      query:'mechanic' 
     }; 

     service = new google.maps.places.PlacesService(map); 
     service.textSearch(request, callback); 
}); 

Answer 1

三个问题：

1. 告诉$.ajax()期待jsonp当你location: new google.maps.LatLng(queryLat, queryLong)
后调用的JSON API返回
2. var place = mapResultsi];应该var place = mapResults[i];
3. 缺少一个逗号

Answer 2

function callback(mapsResults, status){ 
      if(status == google.maps.places.PlacesServiceStatus.OK){ 
       for(var i = 0; i < mapResults.length; i++){ 
        **var place = mapResultsi];** 
        createMarker(mapResults[i]); 
       } 
      } 
     }