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?

12 Solutions collect form web for “在ObjectiveC中从NSDictionary对象创buildURL查询参数”

在iOS8和OS X 10.10中NSURLQueryItemNSURLQueryItem ,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NSURLComponentsNSURLQueryItems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&#x3002;

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); 

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