我正在尝试使用Yelp API,并且我试图将lat / long作为API搜索的参数。但是,它不采用类型double
,它只接受Objective-C对象。不了解Objective-C,你对lat和long参数的类型有什么建议?我尝试过NSNumber,但当我尝试将CLLocationCoordinate2D
类型的纬度/经度坐标转换为接受双精度的NSNumber时,其值为nil
这是我正在使用的Yelp API:
- (void)queryTopBusinessInfoForTerm:(NSString *)term location:(NSString *)location latitude:(NSNumber *)latitude longitude:(NSNumber *)longitude completionHandler:(void (^)(NSDictionary *topBusinessJSON, NSError *error))completionHandler {
NSLog(@"Querying the Search API with term \'%@\' and location \'%@'", term, location);
//Make a first request to get the search results with the passed term and location
NSURLRequest *searchRequest = [self _searchRequestWithTerm:term location:location latitude:latitude longitude:longitude];
NSURLSession *session = [NSURLSession sharedSession];
[[session dataTaskWithRequest:searchRequest completionHandler:^(NSData *data, NSURLResponse *response, NSError *error) {
NSHTTPURLResponse *httpResponse = (NSHTTPURLResponse *)response;
if (!error && httpResponse.statusCode == 200) {
NSDictionary *searchResponseJSON = [NSJSONSerialization JSONObjectWithData:data options:0 error:&error];
NSArray *businessArray = searchResponseJSON[@"businesses"];
if ([businessArray count] > 0) {
NSDictionary *firstBusiness = [businessArray firstObject];
NSString *firstBusinessID = firstBusiness[@"id"];
NSLog(@"%lu businesses found, querying business info for the top result: %@", (unsigned long)[businessArray count], firstBusinessID);
[self queryBusinessInfoForBusinessId:firstBusinessID completionHandler:completionHandler];
} else {
completionHandler(nil, error); // No business was found
}
} else {
completionHandler(nil, error); // An error happened or the HTTP response is not a 200 OK
}
}] resume];
}
这是params
- (NSURLRequest *)_searchRequestWithTerm:(NSString *)term location:(NSString *)location latitude:(NSNumber *) latitude longitude:(NSNumber *)longitude {
NSDictionary *params = @{
@"term": term,
@"location": location,
@"cll": latitude,
@"cll": longitude,
@"limit": kSearchLimit
};
return [NSURLRequest requestWithHost:kAPIHost path:kSearchPath params:params];
}
这是我当前的Swift方法从YelpAPi调用查询:
func yelpApi() {
var latitude = NSNumber(double: businessStreetAddress.latitude)
var longitude = NSNumber(double: businessStreetAddress.longitude)
var searchTerm: NSString = "Asian Food";
var defaultLocation: NSString = "New York"
var APISample:YPAPISample = YPAPISample();
var requestGroup:dispatch_group_t = dispatch_group_create();
APISample.queryTopBusinessInfoForTerm(searchTerm as String, location: defaultLocation as String, latitude: latitude, longitude: longitude) { (topBusinessJSON: [NSObject: AnyObject]!, error: NSError!) -> Void in
if((error) != nil) {
println("Error happened during the request" + error.localizedDescription);
} else if((topBusinessJSON) != nil) {
println("Top business info",topBusinessJSON);
} else {
println("No business was found");
}
dispatch_group_leave(requestGroup);
}
dispatch_group_wait(requestGroup, DISPATCH_TIME_FOREVER);
}
答案 0 :(得分:2)
要将CLLocationCoordinate2D
转换为NSNumber
,您无法将其NSNumber
转换为NSNumber
。您可以转换为两个 CLLocationCoordinate2D location; // This is the location you have.
NSNumber *latitude = [NSNumber numberWithDouble:location.latitude];
NSNumber *longitude = [NSNumber numberWithDouble:location.longitude];
对象,如下所示:
CLLocationCoordinate2D location; // This is the location you have.
NSNumber *latitude = @(location.latitude);
NSNumber *longitude = @(location.longitude);
使用Modern ObjC,您可以隐藏为:
NSURLRequest *searchRequest = [self _searchRequestWithTerm:term location:location latitude:latitude longitude:longitude];
并致电您的
location
不要与名为searchRequestWithTerm:location:latitude:longitude:
的变量混淆,我只是命名它。因为您的函数location
有一个名为NSString
的参数,它接受[Flags]
enum People { Adam = 0x1, Barry = 0x2, Chris = 0x4, David = 0x8,
Eric = 0x10, Frank = 0x20, George = 0x40, Harold = 0x80 };
Dictionary<string, People> EUInterest = new Dictionary<string, People>() {
{ "athletes", People.Adam | People.Barry },
{ "artists", People.Frank | People.Harold } };
Dictionary<string, People> USInterest = new Dictionary<string, People>() {
{ "athletes", People.Chris | People.Harold },
{ "artists", People.Eric } };
var result = from interest in EUInterest.Keys
where USInterest.ContainsKey(interest)
let v1 = EUInterest[interest]
let v2 = USInterest[interest]
select new Dictionary<string, People>() { {interest, v1 | v2 } };
。
这可能会对你有帮助。