在ObjectiveC中从NSDictionary对象创buildURL查询参数
所有的URL处理对象都位于标准的Cocoa库（NSURL，NSMutableURL，NSMutableURLRequest等）中，我知道我必须忽略一个简单的方式来编程地编写一个GET请求。
目前我手动追加“？” 后跟由“＆”连接的名称值对，但是我所有的名称和值对都需要手动编码，所以NSMutableURLRequest在尝试连接到URL时不会完全失败。
这感觉就像我应该能够使用一个预先烘焙的API …有没有什么开箱即可追加查询参数的NSDictionary到NSURL？ 还有另一种方法我应该这样做？
在iOS8和OS X 10.10中NSURLQueryItem是NSURLQueryItem ，它可以用来构build查询。 从NSURLQueryItem上的文档：
NSURLQueryItem对象表示URL的查询部分中的项目的单个名称/值对。 您使用具有NSURLComponents对象的queryItems属性的查询项目。
要创build一个使用指定的初始化程序queryItemWithName:value:然后将它们添加到NSURLComponents以生成一个NSURL 。 例如：
NSURLComponents *components = [NSURLComponents componentsWithString:@"http://stackoverflow.com"]; NSURLQueryItem *search = [NSURLQueryItem queryItemWithName:@"q" value:@"ios"]; NSURLQueryItem *count = [NSURLQueryItem queryItemWithName:@"count" value:@"10"]; components.queryItems = @[ search, count ]; NSURL *url = components.URL; // http://stackoverflow.com?q=ios&count=10
请注意，问号和＆符号会自动处理。 从参数字典中创build一个NSURL非常简单：
NSDictionary *queryDictionary = @{ @"q": @"ios", @"count": @"10" }; NSMutableArray *queryItems = [NSMutableArray array]; for (NSString *key in queryDictionary) { [queryItems addObject:[NSURLQueryItem queryItemWithName:key value:queryDictionary[key]]]; } components.queryItems = queryItems;
我还写了一篇关于如何使用NSURLComponents和NSURLQueryItems构buildURL的NSURLComponents NSURLQueryItems 。
你可以为NSDictionary创build一个类别来做到这一点 – 我可以findCocoa库中的标准方式。 我使用的代码如下所示：
// file "NSDictionary+UrlEncoding.h" #import <cocoa/cocoa.h> @interface NSDictionary (UrlEncoding) -(NSString*) urlEncodedString; @end
有了这个实现：
// file "NSDictionary+UrlEncoding.m" #import "NSDictionary+UrlEncoding.h" // helper function: get the string form of any object static NSString *toString(id object) { return [NSString stringWithFormat: @"%@", object]; } // helper function: get the url encoded string form of any object static NSString *urlEncode(id object) { NSString *string = toString(object); return [string stringByAddingPercentEscapesUsingEncoding: NSUTF8StringEncoding]; } @implementation NSDictionary (UrlEncoding) -(NSString*) urlEncodedString { NSMutableArray *parts = [NSMutableArray array]; for (id key in self) { id value = [self objectForKey: key]; NSString *part = [NSString stringWithFormat: @"%@=%@", urlEncode(key), urlEncode(value)]; [parts addObject: part]; } return [parts componentsJoinedByString: @"&"]; } @end
我认为这段代码非常简单，但是我在http://blog.ablepear.com/2008/12/urlencoding-category-for-nsdictionary.html上详细讨论。;
我想用克里斯的答案，但它不是自动引用计数 （ARC），所以我更新了它。 我想我会粘贴我的解决scheme，以防其他人有这个相同的问题。 注意：在适当情况下用self的实例或类名replaceself 。
+(NSString*)urlEscapeString:(NSString *)unencodedString { CFStringRef originalStringRef = (__bridge_retained CFStringRef)unencodedString; NSString *s = (__bridge_transfer NSString *)CFURLCreateStringByAddingPercentEscapes(NULL,originalStringRef, NULL, (CFStringRef)@"!*'\"();:@&=+$,/?%#[]% ", kCFStringEncodingUTF8); CFRelease(originalStringRef); return s; } +(NSString*)addQueryStringToUrlString:(NSString *)urlString withDictionary:(NSDictionary *)dictionary { NSMutableString *urlWithQuerystring = [[NSMutableString alloc] initWithString:urlString]; for (id key in dictionary) { NSString *keyString = [key description]; NSString *valueString = [[dictionary objectForKey:key] description]; if ([urlWithQuerystring rangeOfString:@"?"].location == NSNotFound) { [urlWithQuerystring appendFormat:@"?%@=%@", [self urlEscapeString:keyString], [self urlEscapeString:valueString]]; } else { [urlWithQuerystring appendFormat:@"&%@=%@", [self urlEscapeString:keyString], [self urlEscapeString:valueString]]; } } return urlWithQuerystring; }
如果值是string，其他的答案工作很好，但如果值是字典或数组，那么这段代码将处理。
值得注意的是，没有通过查询string传递数组/字典的标准方式，但PHP处理这个输出就好了
-(NSString *)serializeParams:(NSDictionary *)params { /* Convert an NSDictionary to a query string */ NSMutableArray* pairs = [NSMutableArray array]; for (NSString* key in [params keyEnumerator]) { id value = [params objectForKey:key]; if ([value isKindOfClass:[NSDictionary class]]) { for (NSString *subKey in value) { NSString* escaped_value = (NSString *)CFURLCreateStringByAddingPercentEscapes(NULL, (CFStringRef)[value objectForKey:subKey], NULL, (CFStringRef)@"!*'();:@&=+$,/?%#[]", kCFStringEncodingUTF8); [pairs addObject:[NSString stringWithFormat:@"%@[%@]=%@", key, subKey, escaped_value]]; } } else if ([value isKindOfClass:[NSArray class]]) { for (NSString *subValue in value) { NSString* escaped_value = (NSString *)CFURLCreateStringByAddingPercentEscapes(NULL, (CFStringRef)subValue, NULL, (CFStringRef)@"!*'();:@&=+$,/?%#[]", kCFStringEncodingUTF8); [pairs addObject:[NSString stringWithFormat:@"%@[]=%@", key, escaped_value]]; } } else { NSString* escaped_value = (NSString *)CFURLCreateStringByAddingPercentEscapes(NULL, (CFStringRef)[params objectForKey:key], NULL, (CFStringRef)@"!*'();:@&=+$,/?%#[]", kCFStringEncodingUTF8); [pairs addObject:[NSString stringWithFormat:@"%@=%@", key, escaped_value]]; [escaped_value release]; } } return [pairs componentsJoinedByString:@"&"]; }
例子
[foo] => bar [translations] => { [one] => uno [two] => dos [three] => tres }
富=酒吧和翻译[酮] = UNO＆翻译[2] =＆DOS翻译[3] = TRES
[foo] => bar [translations] => { uno dos tres }
富=酒吧和翻译[] = UNO＆翻译[] = DOS＆翻译[] = TRES
我重构并转换为AlBeebe的ARC答案
- (NSString *)serializeParams:(NSDictionary *)params { NSMutableArray *pairs = NSMutableArray.array; for (NSString *key in params.keyEnumerator) { id value = params[key]; if ([value isKindOfClass:[NSDictionary class]]) for (NSString *subKey in value) [pairs addObject:[NSString stringWithFormat:@"%@[%@]=%@", key, subKey, [self escapeValueForURLParameter:[value objectForKey:subKey]]]]; else if ([value isKindOfClass:[NSArray class]]) for (NSString *subValue in value) [pairs addObject:[NSString stringWithFormat:@"%@[]=%@", key, [self escapeValueForURLParameter:subValue]]]; else [pairs addObject:[NSString stringWithFormat:@"%@=%@", key, [self escapeValueForURLParameter:value]]]; } return [pairs componentsJoinedByString:@"&"];
}
- (NSString *)escapeValueForURLParameter:(NSString *)valueToEscape { return (__bridge_transfer NSString *) CFURLCreateStringByAddingPercentEscapes(NULL, (__bridge CFStringRef) valueToEscape, NULL, (CFStringRef) @"!*'();:@&=+$,/?%#[]", kCFStringEncodingUTF8); }
如果您已经在使用AFNetworking （就像我一样），那么您可以使用它的类AFHTTPRequestSerializer来创build所需的NSURLRequest 。
[[AFHTTPRequestSerializer serializer] requestWithMethod:@"GET" URLString:@"YOUR_URL" parameters:@{PARAMS} error:nil];
如果您只需要工作的URL，请使用NSURLRequest.URL 。
这是Swift （iOS8 +）中的一个简单例子：
private let kSNStockInfoFetchRequestPath: String = "http://dev.markitondemand.com/Api/v2/Quote/json" private func SNStockInfoFetchRequestURL(symbol:String) -> NSURL? { if let components = NSURLComponents(string:kSNStockInfoFetchRequestPath) { components.queryItems = [NSURLQueryItem(name:"symbol", value:symbol)] return components.URL } return nil }
我有另一个解决scheme：
http://splinter.com.au/build-a-url-query-string-in-obj-c-from-a-dict
+(NSString*)urlEscape:(NSString *)unencodedString { NSString *s = (NSString *)CFURLCreateStringByAddingPercentEscapes(NULL, (CFStringRef)unencodedString, NULL, (CFStringRef)@"!*'\"();:@&=+$,/?%#[]% ", kCFStringEncodingUTF8); return [s autorelease]; // Due to the 'create rule' we own the above and must autorelease it } // Put a query string onto the end of a url +(NSString*)addQueryStringToUrl:(NSString *)url params:(NSDictionary *)params { NSMutableString *urlWithQuerystring = [[[NSMutableString alloc] initWithString:url] autorelease]; // Convert the params into a query string if (params) { for(id key in params) { NSString *sKey = [key description]; NSString *sVal = [[params objectForKey:key] description]; // Do we need to add ?k=v or &k=v ? if ([urlWithQuerystring rangeOfString:@"?"].location==NSNotFound) { [urlWithQuerystring appendFormat:@"?%@=%@", [Http urlEscape:sKey], [Http urlEscape:sVal]]; } else { [urlWithQuerystring appendFormat:@"&%@=%@", [Http urlEscape:sKey], [Http urlEscape:sVal]]; } } } return urlWithQuerystring; }
你可以像这样使用它：
NSDictionary *params = @{@"username":@"jim", @"password":@"abc123"}; NSString *urlWithQuerystring = [self addQueryStringToUrl:@"https://myapp.com/login" params:params];
-(NSString*)encodeDictionary:(NSDictionary*)dictionary{ NSMutableString *bodyData = [[NSMutableString alloc]init]; int i = 0; for (NSString *key in dictionary.allKeys) { i++; [bodyData appendFormat:@"%@=",key]; NSString *value = [dictionary valueForKey:key]; NSString *newString = [value stringByReplacingOccurrencesOfString:@" " withString:@"+"]; [bodyData appendString:newString]; if (i < dictionary.allKeys.count) { [bodyData appendString:@"&"]; } } return bodyData; }
我采取乔尔的使用URLQueryItems的build议，并成为一个Swift扩展（Swift 3）
extension URL { /// Creates an NSURL with url-encoded parameters. init?(string : String, parameters : [String : String]) { guard var components = URLComponents(string: string) else { return nil } components.queryItems = parameters.map { return URLQueryItem(name: $0, value: $1) } guard let url = components.url else { return nil } // Kinda redundant, but we need to call init. self.init(string: url.absoluteString) } }
（ self.init方法有点俗气，但没有组件的NSURL init）
可以用作
URL(string: "http://www.google.com/", parameters: ["q" : "search me"])
还有一种解决scheme，如果你使用RestKit，RKURLEncodedSerialization中有一个名为RKURLEncodedStringFromDictionaryWithEncoding的函数，它正是你想要的。
Objective-c中将NSDictionary转换为url查询string的简单方法
例如：first_name = Steve＆middle_name =盖茨＆姓氏=工作＆地址=加利福尼亚州帕洛阿尔托
NSDictionary *sampleDictionary = @{@"first_name" : @"Steve", @"middle_name" : @"Gates", @"last_name" : @"Jobs", @"address" : @"Palo Alto, California"}; NSMutableString *resultString = [NSMutableString string]; for (NSString* key in [sampleDictionary allKeys]){ if ([resultString length]>0) [resultString appendString:@"&"]; [resultString appendFormat:@"%@=%@", key, [sampleDictionary objectForKey:key]]; } NSLog(@"QueryString: %@", resultString);
希望能帮到:)
