如何解析特定的谷歌地图 API 网络服务 JSON 响应

Posted 2023-03-15

【中文标题】如何解析特定的谷歌地图 API 网络服务 JSON 响应

【英文标题】：How to parse specific googlemaps API webservices JSON repsonse

【发布时间】：2011-12-26 10:19:18

【问题描述】：

google Maps API JSON 响应返回具有相同名词的不同参数：

  "status": "OK",
  "results": [ 
    "types": [ "street_address" ],
    "formatted_address": "1600 Amphitheatre Pkwy, Mountain View, CA 94043, USA",
    "address_components": [ 
      "long_name": "1600",
      "short_name": "1600",
      "types": [ "street_number" ]
    , 
      "long_name": "Amphitheatre Pkwy",
      "short_name": "Amphitheatre Pkwy",
      "types": [ "route" ]
    , 
      "long_name": "Mountain View",
      "short_name": "Mountain View",
      "types": [ "locality", "political" ]
    , 
      "long_name": "California",
      "short_name": "CA",
      "types": [ "administrative_area_level_1", "political" ]
    , 
      "long_name": "United States",
      "short_name": "US",
      "types": [ "country", "political" ]
    , 
      "long_name": "94043",
      "short_name": "94043",
      "types": [ "postal_code" ]
     ],
    "geometry": 
      "location": 
        "lat": 37.4219720,
        "lng": -122.0841430
      ,
      "location_type": "ROOFTOP",
      "viewport": 
        "southwest": 
          "lat": 37.4188244,
          "lng": -122.0872906
        ,
        "northeast": 
          "lat": 37.4251196,
          "lng": -122.0809954
        
      
    
   ]

参数lat和lng在响应中存在多次，例如假设我需要获取location参数的lat/lng：

"location": 
            "lat": 37.4219720,
            "lng": -122.0841430
          ,

我的 JSON 解析代码应该怎么做：

NSDictionary *responseDict = [responseString JSONValue];
    double latitude = [responseDict objectForKey:@"lat"];
    double longitude = [responseDict objectForKey:@"lng"];

这是我写的，解析器如何知道我是明确表示位置参数的纬度/经度还是另一个？

【参考方案1】：

为确保获得所需的正确 lat 和 lon 值，您需要获取 results 数组，然后是 results 字典，然后是 geometry 字典，然后是 location字典，像这样：

NSDictionary *responseDict = [responseString JSONValue];

// The "results" object is an array that contains a single dictionary: 
// "results": [...]
// So first lets get the results array
NSArray *resultsArray = [responseDict objectForKey:@"results"];

// Then get the results dictionary
NSDictionary *resultsDict = [resultsArray objectAtIndex:0];

// Once we have the results dictionary, we can get the geometry
NSDictionary *geometryDict = [resultsDict objectForKey:@"geometry"];

// Then we get the location
NSDictionary *locationDict = [geometryDict objectForKey:@"location"];

// Now we can get the latitude and longitude
double latitude = [locationDict objectForKey:@"lat"];
double longitude = [locationDict objectForKey:@"lng"];

【讨论】：

Mutix，谢谢你的回答，你能澄清objectAtIndex:3 的3 指的是什么吗？因为，它与上例中的 Geometry 参数不匹配。

对不起，索引 3 应该是几何字典，但我没有很好地阅读 JSON。我已经相应地更新了我的答案。

哦，是的，非常感谢 Mutix，请原谅我没有将其标记为已接受，我忘记了，您的回答正是我所需要的 :)))))

【参考方案2】：

可能没有必要提及，但为了初学者，最后两行应该是：

double latitude = [[locationDict objectForKey:@"lat"] doubleValue];
double longitude = [[locationDict objectForKey:@"lng"] doubleValue];

【参考方案3】：

这个函数会给你预期的结果

- (CLLocationCoordinate2D) geoCodeUsingAddress:(NSString *)address

  double latitude = 0, longitude = 0;
  NSString *straddr =  [address stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
  NSString *req = [NSString stringWithFormat:@"http://maps.google.com/maps/api/geocode/json?sensor=false&address=%@", straddr];
  NSString *result = [NSString stringWithContentsOfURL:[NSURL URLWithString:req] encoding:NSUTF8StringEncoding error:NULL];
 if (result) 
 
    NSScanner *scanner = [NSScanner scannerWithString:result];
    if ([scanner scanUpToString:@"\"lat\" :" intoString:nil] && [scanner scanString:@"\"lat\" :" intoString:nil])
    
        [scanner scanDouble:&latitude];
        if ([scanner scanUpToString:@"\"lng\" :" intoString:nil] && [scanner scanString:@"\"lng\" :" intoString:nil])
        
            [scanner scanDouble:&longitude];
        
    

CLLocationCoordinate2D center;
center.latitude = latitude;
center.longitude = longitude;

return center;

希望这会对某人有所帮助。