CLPlacemark - 州缩写？
Posted
技术标签:
【中文标题】CLPlacemark - 州缩写？【英文标题】：CLPlacemark - State Abbreviations? 【发布时间】：2012-06-19 19:57:18 【问题描述】：我想知道是否可以从 CLPlacemark 获得州缩写？
在 Apple 的 CLPlacemark Reference 中指出：
行政区 与地标关联的州或省。 （只读） @property(nonatomic, readonly) NSString *administrativeArea 讨论 例如，如果地标位置是 Apple 的总部，则此属性的值将是字符串“CA”或“California”。
但每当我使用它时，我只会得到完整的州（即加利福尼亚州）而不是缩写（即 CA）。有人可以帮我吗？
【问题讨论】：
CLPlacemark-StateAbbreviation 正是这样做的。 【参考方案1】：对于其他需要解决方案的人，我为 CLPlacemark 创建了一个类别类，它返回短状态字符串。您只需拨打myPlacemark shortState
CLPlacemark+ShortState.h
#import <CoreLocation/CoreLocation.h>
#import <Foundation/Foundation.h>
@interface CLPlacemark (ShortState)
- (NSString *)shortState;
@end
CLPlacemark+ShortState.m
#import "CLPlacemark+ShortState.h"
@interface CLPlacemark (ShortStatePrivate)
- (NSDictionary *)nameAbbreviations;
@end
@implementation CLPlacemark (ShortState)
- (NSString *)shortState
NSString *state = [self.administrativeArea lowercaseString];
if (state.length==0)
return nil;
return [[self nameAbbreviations] objectForKey:state];
- (NSDictionary *)nameAbbreviations
static NSDictionary *nameAbbreviations = nil;
static dispatch_once_t onceToken;
dispatch_once(&onceToken, ^
nameAbbreviations = [NSDictionary dictionaryWithObjectsAndKeys:
@"AL",@"alabama",
@"AK",@"alaska",
@"AZ",@"arizona",
@"AR",@"arkansas",
@"CA",@"california",
@"CO",@"colorado",
@"CT",@"connecticut",
@"DE",@"delaware",
@"DC",@"district of columbia",
@"FL",@"florida",
@"GA",@"georgia",
@"HI",@"hawaii",
@"ID",@"idaho",
@"IL",@"illinois",
@"IN",@"indiana",
@"IA",@"iowa",
@"KS",@"kansas",
@"KY",@"kentucky",
@"LA",@"louisiana",
@"ME",@"maine",
@"MD",@"maryland",
@"MA",@"massachusetts",
@"MI",@"michigan",
@"MN",@"minnesota",
@"MS",@"mississippi",
@"MO",@"missouri",
@"MT",@"montana",
@"NE",@"nebraska",
@"NV",@"nevada",
@"NH",@"new hampshire",
@"NJ",@"new jersey",
@"NM",@"new mexico",
@"NY",@"new york",
@"NC",@"north carolina",
@"ND",@"north dakota",
@"OH",@"ohio",
@"OK",@"oklahoma",
@"OR",@"oregon",
@"PA",@"pennsylvania",
@"RI",@"rhode island",
@"SC",@"south carolina",
@"SD",@"south dakota",
@"TN",@"tennessee",
@"TX",@"texas",
@"UT",@"utah",
@"VT",@"vermont",
@"VA",@"virginia",
@"WA",@"washington",
@"WV",@"west virginia",
@"WI",@"wisconsin",
@"WY",@"wyoming",
nil];
);
return nameAbbreviations;
@end
【讨论】：
谢谢！我认为这对于看到此内容的未来用途/访问者可能非常有用！ 感谢您在最初的问题之后这么长时间接受这个......希望是的，它确实对某人有用。 嘿，我把它变成了一个plist，其中还包括加拿大各省。更简洁+您可以添加更多内容（只需从文件中加载字典，而不是用代码编写）。这个方法虽然很棒。 cl.ly/3c0W1s3E1J1Y @shim - 很好。谢谢！【参考方案2】：我认为您无法获得各州的缩写，但您可以为此创建自己的课程..
列出所有状态（状态是标准） 比较这些状态并返回缩写代码..
类状态缩写
StateAbbreviation.h
@interface StateAbbreviation : NSString
+ (NSString *)allStates:(int)index;
+ (NSString *)abbreviatedState:(NSString *)strState;
@end
StateAbbreviation.m
@implementation StateAbbreviation
+ (NSString *)allStates:(NSString *)strState
// Remove all space on the string
strState = [strState stringByReplacingOccurrencesOfString:@" " withString:@""];
//Sample states
NSArray *states = [[NSArray alloc] initWithObjects:
@"ALABAMA",
@"ALASKA", //AK
@"AMERICANSAMOA", //AS
@"ARIZONA", //AZ
@"ARKANSAS", //AR
@"CALIFORNIA", //CA
nil];
NSUInteger n = [states indexOfObject:strState];
if (n > [states count] - 1)
strAbbreviation = @"NOSTATE";
else
strAbbreviation =[self abbreviatedState:n];
[states release];
return strAbbreviation;
+ (NSString *)abbreviatedState:(int)index
NSArray *states = [[NSArray alloc] initWithObjects:
@"AL",
@"AK",
@"AS",
@"AZ",
@"AR",
@"CA",
nil];
NSString *strAbbreviation = [states objectAtIndex:index];
[states release];
return strAbbreviation;
@end
当你调用这个类时，它应该是这样的
NSString *upperCase = [@"California" uppercaseString]; // California could be from (NSString *)placemark.administrativeArea;
NSString *abbr = [StateAbbreviation allStates:upperCase];
NSLog(@"%@", abbr); // Result should be CA
这只是您可以研究所有状态的示例，states and their abbreviations 也像这样states and their abbreviations
【讨论】：
谢谢！但是您能否更新代码以显示如何使用管理区域来实现这一点？我只是把它放在 objectForKey 之类的东西中吗？我对如何做到这一点有点困惑。感谢您的回答！ 感谢更新，但执行此操作时出现 2 个错误。 1. 在类方法中访问的实例变量“strAbbreviation”。 2. NSUInteger 到 NSString * 的隐式转换在 ARC 中是不允许的。我拿出了发布声明以使其在 ARC 中工作，但我不知道如何解决这两个问题。再次感谢。对于帮助，我很感激。 没问题，你能帮我解决这个问题吗：p？【参考方案3】：我认为文档不正确。行政区域总是要返回美国地方的全州名称。要获得州缩写，您很可能必须创建一个字典查找表，以便搜索键“California”将返回值“CA”。
【讨论】：
【参考方案4】：这是另一个使用 FormattedAddressLines 的类别，它返回类似加利福尼亚州的结果
-(NSString *) stateWithAbbreviation
if ([[self.addressDictionary objectForKey:@"CountryCode"] isEqualToString:@"US"] && self.addressDictionary)
NSDictionary *addressLines = [self.addressDictionary objectForKey:@"FormattedAddressLines"];
for (NSString* addressLine in addressLines)
NSRange stateRange = [addressLine rangeOfString:self.postalCode options:NSCaseInsensitiveSearch];
if (stateRange.length > 0)
NSRange lastSpace = [addressLine rangeOfString:@" " options:NSBackwardsSearch];
if (lastSpace.length > 0)
NSString *state = [[addressLine substringToIndex:lastSpace.location] stringByTrimmingCharactersInSet:[NSCharacterSet whitespaceCharacterSet]];
lastSpace = [state rangeOfString:@" " options:NSBackwardsSearch];
if (lastSpace.length > 0)
NSString *abbr = [[state substringFromIndex:lastSpace.location] stringByTrimmingCharactersInSet:[NSCharacterSet whitespaceCharacterSet]];
return [NSString stringWithFormat:@"%@, %@", self.administrativeArea, abbr];
return self.administrativeArea;
不完美，但只要 Apple 改变我认为的地址行格式，它就可以工作。
【讨论】：
【参考方案5】：对于需要交换对象和键的状态列表的人（例如，在 ios 7 上，我从 placemark.administrativeArea 获得“CA”）：
NSDictionary *nameAbbreviations = [NSDictionary dictionaryWithObjectsAndKeys:
@"alabama",@"AL",
@"alaska",@"AK",
@"arizona",@"AZ",
@"arkansas",@"AR",
@"california",@"CA",
@"colorado",@"CO",
@"connecticut",@"CT",
@"delaware",@"DE",
@"district of columbia",@"DC",
@"florida",@"FL",
@"georgia",@"GA",
@"hawaii",@"HI",
@"idaho",@"ID",
@"illinois",@"IL",
@"indiana",@"IN",
@"iowa",@"IA",
@"kansas",@"KS",
@"kentucky",@"KY",
@"louisiana",@"LA",
@"maine",@"ME",
@"maryland",@"MD",
@"massachusetts",@"MA",
@"michigan",@"MI",
@"minnesota",@"MN",
@"mississippi",@"MS",
@"missouri",@"MO",
@"montana",@"MT",
@"nebraska",@"NE",
@"nevada",@"NV",
@"new hampshire",@"NH",
@"new jersey",@"NJ",
@"new mexico",@"NM",
@"new york",@"NY",
@"north carolina",@"NC",
@"north dakota",@"ND",
@"ohio",@"OH",
@"oklahoma",@"OK",
@"oregon",@"OR",
@"pennsylvania",@"PA",
@"rhode island",@"RI",
@"south carolina",@"SC",
@"south dakota",@"SD",
@"tennessee",@"TN",
@"texas",@"TX",
@"utah",@"UT",
@"vermont",@"VT",
@"virginia",@"VA",
@"washington",@"WA",
@"west virginia",@"WV",
@"wisconsin",@"WI",
@"wyoming",@"WY",
nil];
【讨论】：
【参考方案6】：至少从 iOS 8 开始，CLPlacemark 的 administrativeArea 返回美国各州的两个字母缩写。
只要您的目标是 iOS 8 和更高版本（您现在应该是），您就不需要使用已接受答案中的类别来扩展 CLPlacemark。
CLGeocoder *geocoder = [[CLGeocoder alloc] init];
[geocoder geocodeAddressString:@"1 Infinite Loop, Cupertino, CA" completionHandler:^(NSArray *placemarks, NSError *error)
CLPlacemark *placemark = [placemarks firstObject];
NSLog(@"State: %@", placemark.administrativeArea);
];
运行这个，你会得到：
State: CA
【讨论】：
至少根据 Apple 提供的最新文档，这不是一个真实的陈述：此属性中的字符串可以是行政区域的拼写名称或其指定的缩写，如果有的话存在。例如，如果地标位置是 Apple 的总部，则此属性的值将是字符串“CA”或“California”。【参考方案7】：字典的 SWIFT 变体
let states = [
"AL":"alabama",
"AK":"alaska",
"AZ":"arizona",
"AR":"arkansas",
"CA":"california",
"CO":"colorado",
"CT":"connecticut",
"DE":"delaware",
"DC":"district of columbia",
"FL":"florida",
"GA":"georgia",
"HI":"hawaii",
"ID":"idaho",
"IL":"illinois",
"IN":"indiana",
"IA":"iowa",
"KS":"kansas",
"KY":"kentucky",
"LA":"louisiana",
"ME":"maine",
"MD":"maryland",
"MA":"massachusetts",
"MI":"michigan",
"MN":"minnesota",
"MS":"mississippi",
"MO":"missouri",
"MT":"montana",
"NE":"nebraska",
"NV":"nevada",
"NH":"new hampshire",
"NJ":"new jersey",
"NM":"new mexico",
"NY":"new york",
"NC":"north carolina",
"ND":"north dakota",
"OH":"ohio",
"OK":"oklahoma",
"OR":"oregon",
"PA":"pennsylvania",
"RI":"rhode island",
"SC":"south carolina",
"SD":"south dakota",
"TN":"tennessee",
"TX":"texas",
"UT":"utah",
"VT":"vermont",
"VA":"virginia",
"WA":"washington",
"WV":"west virginia",
"WI":"wisconsin",
"WY":"wyoming"
]
【讨论】：
以上是关于CLPlacemark - 州缩写？的主要内容，如果未能解决你的问题，请参考以下文章