85

我有一个类似的网址myApp://action/1?parameter=2&secondparameter=3

通过属性查询，我得到以下部分URL

parameter=2&secondparameter=3

有什么方法可以很容易地将它放入 aNSDictionary或 an中Array吗？

多谢

4

17 回答 17

151

您可以queryItems在URLComponents.

当你得到这个属性的值时，NSURLComponents 类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。

迅速

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)

或者，您可以在 URL 上添加扩展名以使事情变得更容易。 

extension URL {
    var queryParameters: QueryParameters { return QueryParameters(url: self) }
}

class QueryParameters {
    let queryItems: [URLQueryItem]
    init(url: URL?) {
        queryItems = URLComponents(string: url?.absoluteString ?? "")?.queryItems ?? []
        print(queryItems)
    }
    subscript(name: String) -> String? {
        return queryItems.first(where: { $0.name == name })?.value
    }
}

然后，您可以通过其名称访问参数。

let url = "http://example.com?param1=value1&param2=param2"
print(url.queryParameters["param1"])
于 2014-10-16T14:05:12.083 回答
55

我有理由为此行为编写一些可能派上用场的扩展。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，允许您在查询组件字符串和包含结果的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
于 2011-03-23T19:19:34.483 回答
54

像这样的东西：

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。

于 2010-10-22T15:03:42.640 回答
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]];
}
于 2010-10-22T14:57:47.480 回答
9

以前的所有帖子都没有正确地进行 url 编码。我建议以下方法：

+(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];
}
于 2012-10-18T12:15:46.533 回答
8

根据Onato 已经非常干净的答案，我在 Swift 中为 NSURL 编写了一个扩展，您可以在其中获得如下查询参数：

例如 URL 包含对 param=some_value

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

扩展看起来像：

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

  }
}
于 2015-02-17T09:03:36.167 回答
7

这是swift的扩展：

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
        }
    }
于 2014-10-13T18:34:03.843 回答
5

对于那些使用Bolts 框架的人，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，那么您也有Bolts。Facebook 正在使用这个框架作为依赖项。

于 2015-02-11T13:20:11.813 回答
5

现在处理 URL 的首选方法是NSURLComponents. 特别是queryItems返回NSArray参数的属性。

如果你想要 a 中的参数NSDictionary，这里有一个方法：

+(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;
}

警告：URL 可以有重复的参数，但字典只会包含任何重复参数的最后一个值。如果不希望这样做，请queryItems直接使用数组。

于 2015-09-14T03:20:13.393 回答
4

斯威夫特 2.1

单线：

"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：

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 [:]
            }
        }
    }

}

例子：

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NSURLComponents和属性并直接queryItems从创建字典。NSURLQueryItems

这是一个NSURLComponents基于NSURL扩展的解决方案：

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
        }
    }

}

NSURL 扩展的一个脚注是，在 Swift 中实际上可以为属性赋予与现有字符串属性相同的名称——<code>query。直到我尝试过才知道，但是 Swift 中的多态性让你只在返回类型上有所不同。因此，如果扩展的 NSURL 属性public var query: [String:String]有效。我没有在示例中使用它，因为我觉得它有点疯狂，但它确实有效......

于 2015-03-17T11:18:51.900 回答
2

我在麻省理工学院发布了一个简单的课程：

https://github.com/anegmawad/URLQueryToCocoa

有了它，您可以在查询中拥有数组和对象，它们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

会变成：

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

您可以将其视为NSJSONSerializer.

于 2014-07-15T14:03:51.517 回答
2

看起来您正在使用它来处理来自另一个 iOS 应用程序的传入数据。如果是这样，这就是我用于相同目的的东西。

初始调用（例如在外部应用程序中）：

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];
于 2015-08-18T18:49:05.293 回答
0

这个类是一个很好的 url 解析解决方案。

.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
于 2014-07-10T11:38:59.467 回答
0

Hendrik在这个问题中写了一个很好的扩展示例，但是我不得不重写它以不使用任何objective-c 库方法。快速使用NSArray不是正确的方法。

这就是结果，一切都迅速且更安全。Swift 1.2 的使用示例将减少代码行数。

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)")
        }
    }
}
于 2015-03-16T15:47:49.860 回答
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;
}
于 2017-06-15T12:39:41.443 回答
0

相当紧凑的方法：

    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
    }
于 2020-05-14T02:41:41.620 回答
-6

如果您使用 URL 将数据从 Web 应用程序传递到手机，并且您想要传递数组、数字、字符串......

JSON 在 PHP 中对您的对象进行编码

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

和 JSON 解码 iOS 中的结果

NSData *data = [[[request URL] host] dataUsingEncoding:NSUTF8StringEncoding];
NSDictionary *packed = [NSJSONSerialization JSONObjectWithData:data options:0 error:nil];
于 2012-07-20T01:39:38.947 回答