ios - 如何将 CLLCoordinate2D 纬度/经度转换为 NSNumber
<p><p>我正在尝试使用 Yelp API，并且我正在尝试将 lat/long 作为 API 搜索的参数。但是，它不采用 <code>double</code> 类型，它只接受 Objective-C 对象。不了解 Objective-C，您建议 lat 和 long be 参数的类型是什么？我尝试了 NSNumber，但是当我尝试将类型为 <code>CLLocationCoordinate2D</code> 的经纬度坐标转换为接受 double 的 NSNumber 时，其值为 nil</p>
<p>这是我正在使用的 Yelp API:</p>
<pre><code>- (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 = ;
NSURLSession *session = ;
[[session dataTaskWithRequest:searchRequest completionHandler:^(NSData *data, NSURLResponse *response, NSError *error) {
NSHTTPURLResponse *httpResponse = (NSHTTPURLResponse *)response;
if (!error && httpResponse.statusCode == 200) {
NSDictionary *searchResponseJSON = ;
NSArray *businessArray = searchResponseJSON[@"businesses"];
if ( > 0) {
NSDictionary *firstBusiness = ;
NSString *firstBusinessID = firstBusiness[@"id"];
NSLog(@"%lu businesses found, querying business info for the top result: %@", (unsigned long), firstBusinessID);
;
} 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];
}
</code></pre>
<p>这是参数</p>
<pre><code>- (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 ;
}
</code></pre>
<p>这是我当前从 YelpAPi 调用查询的 Swift 方法:</p>
<pre><code>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: !, 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);
}
</code></pre></p>
<br><hr><h1><strong>Best Answer-推荐答案</ strong></h1><br>
<p><p>要将 <code>CLLocationCoordinate2D</code> 转换为 <code>NSNumber</code>，您不能将其包含在单个 <code>NSNumber</code> 中。您可以转换为两个 <code>NSNumber</code> 对象，如下所示:</p>
<pre><code> CLLocationCoordinate2D location; // This is the location you have.
NSNumber *latitude = ;
NSNumber *longitude = ;
</code></pre>
<p>使用 Modern ObjC，您可以转换为:</p>
<pre><code> CLLocationCoordinate2D location; // This is the location you have.
NSNumber *latitude = @(location.latitude);
NSNumber *longitude = @(location.longitude);
</code></pre>
<p>调用你的</p>
<pre><code>NSURLRequest *searchRequest = ;
</code></pre>
<p>不要与我刚刚命名的类似 <code>location</code> 的变量混淆。因为您的函数 <code>searchRequestWithTerm:location:latitude:longitude:</code> 有一个名为 <code>location</code> 的参数，它接受 <code>NSString</code>。</p>
<p>这可能会对你有所帮助。</p></p>
<p style="font-size: 20px;">关于ios - 如何将 CLLCoordinate2D 纬度/经度转换为 NSNumber，我们在Stack Overflow上找到一个类似的问题：
<a href="https://stackoverflow.com/questions/31616043/" rel="noreferrer noopener nofollow" style="color: red;">
https://stackoverflow.com/questions/31616043/
</a>
</p>
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