我无法想出以优雅的方式解析这个回应。我正在尝试将不同的字段类型匹配到包含以下属性的对象。
@property (nonatomic) NSString *name;
@property (nonatomic) NSString *address;
@property (nonatomic) NSString *city;
@property (nonatomic) NSString *postalcode;
@property (nonatomic) NSString *district;
@property (nonatomic) NSString *country;
过去我经历了字典,子字段与我搜索的字段匹配。
{
"results" : [
{
"address_components" : [
{
"long_name" : "10",
"short_name" : "10",
"types" : [ "street_number" ]
},
{
"long_name" : "Warschauer Straße",
"short_name" : "Warschauer Straße",
"types" : [ "route" ]
},
{
"long_name" : "Friedrichshain",
"short_name" : "Friedrichshain",
"types" : [ "sublocality", "political" ]
},
{
"long_name" : "Friedrichshain-Kreuzberg",
"short_name" : "Friedrichshain-Kreuzberg",
"types" : [ "sublocality", "political" ]
},
{
"long_name" : "Berlin",
"short_name" : "Berlin",
"types" : [ "locality", "political" ]
},
{
"long_name" : "Berlin",
"short_name" : "Berlin",
"types" : [ "administrative_area_level_1", "political" ]
},
{
"long_name" : "Germany",
"short_name" : "DE",
"types" : [ "country", "political" ]
},
{
"long_name" : "10243",
"short_name" : "10243",
"types" : [ "postal_code" ]
}
],
"formatted_address" : "Warschauer Straße 10, 10243 Berlin, Germany",
"geometry" : {
"location" : {
"lat" : 52.51329350,
"lng" : 13.45246930
},
"location_type" : "ROOFTOP",
"viewport" : {
"northeast" : {
"lat" : 52.51464248029150,
"lng" : 13.45381828029150
},
"southwest" : {
"lat" : 52.51194451970850,
"lng" : 13.45112031970850
}
}
},
"types" : [ "street_address" ]
}
],
"status" : "OK"
}
答案 0 :(得分:1)
确定。我想出了以下解决方案。我已经定义了一个Dictionary并覆盖了Setter来将传入的字典解析为我自己的解析器:
-(void)addressByGoogle: (NSDictionary*)newAddressComponents{
for (NSArray *googleResult in newAddressComponents) {
for (NSArray *fields in googleResult) {
if ([[fields valueForKey:@"types"] count] >= 1) {
NSString *googleType = [[fields valueForKey:@"types"] objectAtIndex:0];
if ( [googleType isEqualToString:@"sublocality"]) {
self.district = [fields valueForKey:@"long_name"];
}
if ( [googleType isEqualToString:@"country"]) {
self.country = [fields valueForKey:@"long_name"];
}
if ( [googleType isEqualToString:@"locality"]) {
self.city = [fields valueForKey:@"long_name"];
}
if ( [googleType isEqualToString:@"street_number"]) {
self.streetnumber = [fields valueForKey:@"long_name"];
}
if ( [googleType isEqualToString:@"route"]) {
self.address = [fields valueForKey:@"long_name"];
}
if ( [googleType isEqualToString:@"postal_code"]) {
self.postalcode = [fields valueForKey:@"long_name"];
}
if ( [googleType isEqualToString:@"administrative_area_level_1"]) {
self.adminAreaLevel1 = [fields valueForKey:@"long_name"];
}
if ( [googleType isEqualToString:@"administrative_area_level_2"]) {
self.adminAreaLevel2 = [fields valueForKey:@"long_name"];
}
}
}
}
}
- (void) setAddressComponents:(NSDictionary *)newAddressComponents{
[self addressByGoogle: newAddressComponents];
}
答案 1 :(得分:0)
正如Jim所说,没有优雅的方法来使用Restkit映射这个对象,你可以做的是将这种格式直接映射到一个“丑陋的对象”并将该对象与另一个具有该接口的对象包装起来如你所愿。
类似的东西:
@interface MYMappedAddress : NSObject
@property (nonatomic) NSArray *addressComponents;
@property (nonatomic) NSArray *types;
@property (nonatomic) MYGeometry *geometry;
@property (nonatomic) NSString * formattedAddress;
@end
@interface MYAddress : MYMappedAddress
@property (nonatomic) NSString *name;
@property (nonatomic) NSString *address;
@property (nonatomic) NSString *city;
@property (nonatomic) NSString *postalcode;
@property (nonatomic) NSString *district;
@property (nonatomic) NSString *country;
@end