解析NSURL查询属性

时间:2010-10-22 14:48:49

标签: iphone objective-c nsurl

我有一个像myApp://action/1?parameter=2&secondparameter=3

这样的网址

通过属性查询,我得到了URL

的部分内容
parameter=2&secondparameter=3

有什么方法可以轻松地将其放在NSDictionaryArray中吗?

很多

17 个答案:

答案 0 :(得分:142)

您可以在URLComponents中使用queryItems

  

当你获得这个属性的值时,NSURLComponents类会解析查询字符串并返回一个NSURLQueryItem对象数组,每个对象代表一个键值对,按照它们在原始查询字符串中出现的顺序。 / p>

夫特

let url = "http://example.com?param1=value1&param2=param2"
let queryItems = URLComponents(string: url)?.queryItems
let param1 = queryItems?.filter({$0.name == "param1"}).first
print(param1?.value)

目标C

使用NSURLComponents

NSURLComponents *urlComponents = [NSURLComponents componentsWithURL:url 
                                            resolvingAgainstBaseURL:NO];
NSArray *queryItems = urlComponents.queryItems;
NSString *param1 = [self valueForKey:@"param1" 
                  fromQueryItems:queryItems];
NSLog(@"%@", param1);
…

- (NSString *)valueForKey:(NSString *)key
           fromQueryItems:(NSArray *)queryItems
{
    NSPredicate *predicate = [NSPredicate predicateWithFormat:@"name=%@", key];
    NSURLQueryItem *queryItem = [[queryItems 
                                  filteredArrayUsingPredicate:predicate]
                                 firstObject];
    return queryItem.value;
}

答案 1 :(得分:55)

我有理由为这种可能派上用场的行为编写一些扩展。首先是标题:

#import <Foundation/Foundation.h>

@interface NSString (XQueryComponents)
- (NSString *)stringByDecodingURLFormat;
- (NSString *)stringByEncodingURLFormat;
- (NSMutableDictionary *)dictionaryFromQueryComponents;
@end

@interface NSURL (XQueryComponents)
- (NSMutableDictionary *)queryComponents;
@end

@interface NSDictionary (XQueryComponents)
- (NSString *)stringFromQueryComponents;
@end

这些方法扩展了NSString,NSURL和NSDictionary,允许您在包含结果的查询组件字符串和字典对象之间进行转换。

现在相关的.m代码:

#import "XQueryComponents.h"

@implementation NSString (XQueryComponents)
- (NSString *)stringByDecodingURLFormat
{
    NSString *result = [self stringByReplacingOccurrencesOfString:@"+" withString:@" "];
    result = [result stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
    return result;
}

- (NSString *)stringByEncodingURLFormat
{
    NSString *result = [self stringByReplacingOccurrencesOfString:@" " withString:@"+"];
    result = [result stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
    return result;
}

- (NSMutableDictionary *)dictionaryFromQueryComponents
{
    NSMutableDictionary *queryComponents = [NSMutableDictionary dictionary];
    for(NSString *keyValuePairString in [self componentsSeparatedByString:@"&"])
    {
        NSArray *keyValuePairArray = [keyValuePairString componentsSeparatedByString:@"="];
        if ([keyValuePairArray count] < 2) continue; // Verify that there is at least one key, and at least one value.  Ignore extra = signs
        NSString *key = [[keyValuePairArray objectAtIndex:0] stringByDecodingURLFormat];
        NSString *value = [[keyValuePairArray objectAtIndex:1] stringByDecodingURLFormat];
        NSMutableArray *results = [queryComponents objectForKey:key]; // URL spec says that multiple values are allowed per key
        if(!results) // First object
        {
            results = [NSMutableArray arrayWithCapacity:1];
            [queryComponents setObject:results forKey:key];
        }
        [results addObject:value];
    }
    return queryComponents;
}
@end

@implementation NSURL (XQueryComponents)
- (NSMutableDictionary *)queryComponents
{
    return [[self query] dictionaryFromQueryComponents];
}
@end

@implementation NSDictionary (XQueryComponents)
- (NSString *)stringFromQueryComponents
{
    NSString *result = nil;
    for(__strong NSString *key in [self allKeys])
    {
        key = [key stringByEncodingURLFormat];
        NSArray *allValues = [self objectForKey:key];
        if([allValues isKindOfClass:[NSArray class]])
            for(__strong NSString *value in allValues)
            {
                value = [[value description] stringByEncodingURLFormat];
                if(!result)
                    result = [NSString stringWithFormat:@"%@=%@",key,value];
                else 
                    result = [result stringByAppendingFormat:@"&%@=%@",key,value];
            }
        else {
            NSString *value = [[allValues description] stringByEncodingURLFormat];
            if(!result)
                result = [NSString stringWithFormat:@"%@=%@",key,value];
            else 
                result = [result stringByAppendingFormat:@"&%@=%@",key,value];
        }
    }
    return result;
}
@end

答案 2 :(得分:54)

类似的东西:

NSMutableDictionary *params = [[NSMutableDictionary alloc] init];
for (NSString *param in [url componentsSeparatedByString:@"&"]) {
  NSArray *elts = [param componentsSeparatedByString:@"="];
  if([elts count] < 2) continue;
  [params setObject:[elts lastObject] forKey:[elts firstObject]];
}

注意:这是示例代码。不管理所有错误情况。

答案 3 :(得分:13)

试试这个;)!

NSString *query = @"parameter=2&secondparameter=3"; // replace this with [url query];
NSArray *components = [query componentsSeparatedByString:@"&"];
NSMutableDictionary *parameters = [[NSMutableDictionary alloc] init];
for (NSString *component in components) {
    NSArray *subcomponents = [component componentsSeparatedByString:@"="];
    [parameters setObject:[[subcomponents objectAtIndex:1] stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding]
                   forKey:[[subcomponents objectAtIndex:0] stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];
}

答案 4 :(得分:9)

所有以前的帖子都没有正确进行网址编码。我会建议以下方法:

+(NSString*)concatenateQuery:(NSDictionary*)parameters {
    if([parameters count]==0) return nil;
    NSMutableString* query = [NSMutableString string];
    for(NSString* parameter in [parameters allKeys])
        [query appendFormat:@"&%@=%@",[parameter stringByAddingPercentEncodingWithAllowedCharacters:NSCharacterSet.URLQueryAllowedCharacterSet],[[parameters objectForKey:parameter] stringByAddingPercentEncodingWithAllowedCharacters:NSCharacterSet.URLQueryAllowedCharacterSet]];
    return [[query substringFromIndex:1] copy];
}
+(NSDictionary*)splitQuery:(NSString*)query {
    if([query length]==0) return nil;
    NSMutableDictionary* parameters = [NSMutableDictionary dictionary];
    for(NSString* parameter in [query componentsSeparatedByString:@"&"]) {
        NSRange range = [parameter rangeOfString:@"="];
        if(range.location!=NSNotFound)
            [parameters setObject:[[parameter substringFromIndex:range.location+range.length] stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding] forKey:[[parameter substringToIndex:range.location] stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];
        else [parameters setObject:[[NSString alloc] init] forKey:[parameter stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];
    }
    return [parameters copy];
}

答案 5 :(得分:7)

这是swift中的扩展名:

extension NSURL{
        func queryParams() -> [String:AnyObject] {
            var info : [String:AnyObject] = [String:AnyObject]()
            if let queryString = self.query{
                for parameter in queryString.componentsSeparatedByString("&"){
                    let parts = parameter.componentsSeparatedByString("=")
                    if parts.count > 1{
                        let key = (parts[0] as String).stringByReplacingPercentEscapesUsingEncoding(NSUTF8StringEncoding)
                        let value = (parts[1] as String).stringByReplacingPercentEscapesUsingEncoding(NSUTF8StringEncoding)
                        if key != nil && value != nil{
                            info[key!] = value
                        }
                    }
                }
            }
            return info
        }
    }

答案 6 :(得分:7)

根据已经very clean answer of Onato,我在Swift中为NSURL写了一个扩展,你可以在这里获得一个查询参数:

e.g。该URL包含param = some_value

let queryItem = url.queryItemForKey("param")
let value = queryItem.value // would get String "someValue"

扩展名如下:

extension NSURL {

  var allQueryItems: [NSURLQueryItem] {
      get {
          let components = NSURLComponents(URL: self, resolvingAgainstBaseURL: false)!
          let allQueryItems = components.queryItems!
          return allQueryItems as [NSURLQueryItem]
      }
  }

  func queryItemForKey(key: String) -> NSURLQueryItem? {

      let predicate = NSPredicate(format: "name=%@", key)!
      return (allQueryItems as NSArray).filteredArrayUsingPredicate(predicate).first as? NSURLQueryItem

  }
}

答案 7 :(得分:5)

对于那些使用 Bolts Framework 的人,您可以使用:

NSDictionary *parameters = [BFURL URLWithURL:yourURL].inputQueryParameters;

记得导入:

#import <Bolts/BFURL.h>

如果您的项目中碰巧有 Facebook SDK ,那么您还有螺栓。 Facebook正在使用这个框架作为依赖。

答案 8 :(得分:3)

Swift 2.1

Oneliner:

"p1=v1&p2=v2".componentsSeparatedByString("&").map {
    $0.componentsSeparatedByString("=") 
}.reduce([:]) {
    (var dict: [String:String], p) in
    dict[p[0]] = p[1]
    return dict
}

// ["p1": "v1", "p2": "v2"]

用作NSURL的扩展程序:

extension NSURL {

    /**
     * URL query string as dictionary. Empty dictionary if query string is nil.
     */
    public var queryValues : [String:String] {
        get {
            if let q = self.query {
                return q.componentsSeparatedByString("&").map {
                    $0.componentsSeparatedByString("=") 
                }.reduce([:]) {
                    (var dict: [String:String], p) in
                    dict[p[0]] = p[1]
                    return dict
                }
            } else {
                return [:]
            }
        }
    }

}

示例:

let url = NSURL(string: "http://example.com?p1=v1&p2=v2")!
let queryDict = url.queryValues

// ["p1": "v1", "p2": "v2"]

请注意,如果使用OS X 10.10或iOS 8(或更高版本),最好使用NSURLComponentsqueryItems属性,并直接从NSURLQueryItems创建字典。

以下是基于NSURLComponents的{​​{1}}扩展解决方案:

NSURL

NSURL扩展的脚注是,Swift实际上可以为属性提供与现有字符串属性相同的名称 - extension NSURL { /// URL query string as a dictionary. Empty dictionary if query string is nil. public var queryValues : [String:String] { get { guard let components = NSURLComponents(URL: self, resolvingAgainstBaseURL: false) else { return [:] } guard let queryItems = components.queryItems else { return [:] } var result:[String:String] = [:] for q in queryItems { result[q.name] = q.value } return result } } } 。直到我尝试它才知道,但是Swift中的多态性只允许你在返回类型上有所不同。因此,如果扩展的NSURL属性为query则可行。我没有在示例中使用它,因为我发现它有点疯狂,但它确实有用......

答案 9 :(得分:3)

目前处理网址的首选方式是NSURLComponents。特别是queryItems属性,它返回NSDictionary个参数。

如果你想要+(NSDictionary<NSString *, NSString *>*)queryParamsFromURL:(NSURL*)url { NSURLComponents* urlComponents = [NSURLComponents componentsWithURL:url resolvingAgainstBaseURL:NO]; NSMutableDictionary<NSString *, NSString *>* queryParams = [NSMutableDictionary<NSString *, NSString *> new]; for (NSURLQueryItem* queryItem in [urlComponents queryItems]) { if (queryItem.value == nil) { continue; } [queryParams setObject:queryItem.value forKey:queryItem.name]; } return queryParams; } 中的参数,这里有一个方法:

queryItems

警告:URL可以有重复的参数,但字典只包含任何重复参数的最后一个值。如果不合需要,请直接使用{ "query": { "bool": { "must": [ { "terms": { "field1": [ "a", "b", "c" ] } }, { "terms": { "field2": [ "d", "e", "f" ] } }, { "terms": { "field3": [ "g", "h", "i" ] } } ] } } } 数组。

答案 10 :(得分:2)

我在麻省理工学院发表了一个简单的课程:

https://github.com/anegmawad/URLQueryToCocoa

有了它,你可以在查询中拥有数组和对象,这些数组和对象被收集并粘合在一起

例如

users[0][firstName]=Amin&users[0][lastName]=Negm&name=Devs&users[1][lastName]=Kienle&users[1][firstName]=Christian

将成为:

@{
   name : @"Devs",
   users :
   @[
      @{
         firstName = @"Amin",
         lastName = @"Negm"
      },
      @{
         firstName = @"Christian",
         lastName = @"Kienle"
      }
   ]
 }

您可以将其视为NSJSONSerializer的网址查询对象。

答案 11 :(得分:1)

看起来您正在使用它来处理来自其他iOS应用程序的传入数据。如果是这样,这就是我用于同一目的的。

初始通话(例如在外部申请中):

UIApplication *application = [UIApplication sharedApplication];
NSURL *url = [NSURL URLWithString:@"myApp://action/1?parameter=2&secondparameter=3"];
if ([application canOpenURL:url]) {
    [application openURL:url];
    NSLog(@"myApp is installed");
} else {
    NSLog(@"myApp is not installed");
}

从NSURL中提取QueryString数据并保存为NSDictionary的方法:

-(NSDictionary *) getNSDictionaryFromQueryString:(NSURL *)url {
   NSMutableDictionary *result = [[NSMutableDictionary alloc] init];
   NSRange needle = [url.absoluteString rangeOfString:@"?" options:NSCaseInsensitiveSearch];
   NSString *data = nil;

   if(needle.location != NSNotFound) {
       NSUInteger start = needle.location + 1;
       NSUInteger end = [url.absoluteString length] - start;
       data = [url.absoluteString substringWithRange:NSMakeRange(start, end)];
   }

   for (NSString *param in [data componentsSeparatedByString:@"&"]) {
       NSArray *keyvalue = [param componentsSeparatedByString:@"="];
       if([keyvalue count] == 2){
           [result setObject:[keyvalue objectAtIndex:1] forKey:[keyvalue objectAtIndex:0]];
       }
   }

  return result;
}

用法:

NSDictionary *result = [self getNSDictionaryFromQueryString:url];

答案 12 :(得分:0)

这个类是解析url的好方法。

.h文件

@interface URLParser : NSObject {
    NSArray *variables;
}

@property (nonatomic, retain) NSArray *variables;

- (id)initWithURLString:(NSString *)url;
- (NSString *)valueForVariable:(NSString *)varName;

@end

.m文件

#import "URLParser.h"

@implementation URLParser
@synthesize variables;

- (id) initWithURLString:(NSString *)url{
    self = [super init];
    if (self != nil) {
        NSString *string = url;
        NSScanner *scanner = [NSScanner scannerWithString:string];
        [scanner setCharactersToBeSkipped:[NSCharacterSet characterSetWithCharactersInString:@"&?"]];
        NSString *tempString;
        NSMutableArray *vars = [NSMutableArray new];
        [scanner scanUpToString:@"?" intoString:nil];       //ignore the beginning of the string and skip to the vars
        while ([scanner scanUpToString:@"&" intoString:&tempString]) {
            [vars addObject:[tempString copy]];
        }
        self.variables = vars;
    }
    return self;
}

- (NSString *)valueForVariable:(NSString *)varName {
    for (NSString *var in self.variables) {
        if ([var length] > [varName length]+1 && [[var substringWithRange:NSMakeRange(0, [varName length]+1)] isEqualToString:[varName stringByAppendingString:@"="]]) {
            NSString *varValue = [var substringFromIndex:[varName length]+1];
            return varValue;
        }
    }
    return nil;
}

@end

答案 13 :(得分:0)

Hendrik在这个问题中为扩展写了一个很好的例子,但是我不得不重新编写它以不使用任何objective-c库方法。在swift中使用NSArray不是正确的方法。

这是结果,所有快速,更安全。使用示例将使用Swift 1.2减少代码行。

public extension NSURL {
    /*
    Set an array with all the query items
    */
    var allQueryItems: [NSURLQueryItem] {
        get {
            let components = NSURLComponents(URL: self, resolvingAgainstBaseURL: false)!
            if let allQueryItems = components.queryItems {
                return allQueryItems as [NSURLQueryItem]
            } else {
                return []
            }
        }
    }

    /**
    Get a query item form the URL query

    :param: key The parameter to fetch from the URL query

    :returns: `NSURLQueryItem` the query item
    */
    public func queryItemForKey(key: String) -> NSURLQueryItem? {
        let filteredArray = filter(allQueryItems) { $0.name == key }

        if filteredArray.count > 0 {
            return filteredArray.first
        } else {
            return nil
        }
    }
}

用法:

let queryItem = url.queryItemForKey("myItem")

或者,更详细的用法:

if let url = NSURL(string: "http://www.domain.com/?myItem=something") {
    if let queryItem = url.queryItemForKey("myItem") {
        if let value = queryItem.value {
            println("The value of 'myItem' is: \(value)")
        }
    }
}

答案 14 :(得分:0)

试试这个:

-(NSDictionary *)getUrlParameters:(NSString *)url{
    NSArray *justParamsArr = [url componentsSeparatedByString:@"?"];
    url = [justParamsArr lastObject];
    NSMutableDictionary *params = [[NSMutableDictionary alloc] init];
    for (NSString *param in [url componentsSeparatedByString:@"&"]) {
        NSArray *elts = [param componentsSeparatedByString:@"="];
        if([elts count] < 2) continue;
        [params setObject:[elts lastObject] forKey:[elts firstObject]];
    }
    return params;
}

答案 15 :(得分:0)

完全紧凑的方法:

    func stringParamsToDict(query: String) -> [String: String] {
        let params = query.components(separatedBy: "&").map {
            $0.components(separatedBy: "=")
        }.reduce(into: [String: String]()) { dict, pair in
            if pair.count == 2 {
                dict[pair[0]] = pair[1]
            }
        }
        return params
    }

答案 16 :(得分:-5)

最强大的解决方案,如果您使用网址将数据从网络应用程序传递到手机,并且您想传递数组,数字,字符串,......

JSON使用PHP编码您的对象

header("Location: myAppAction://".urlencode(json_encode($YOUROBJECT)));

JSON解码iOS中的结果

NSData *data = [[[request URL] host] dataUsingEncoding:NSUTF8StringEncoding];
NSDictionary *packed = [NSJSONSerialization JSONObjectWithData:data options:0 error:nil];