解析NSURL查询属性

Question

我有一个像这样的网址 myApp://action/1?parameter=2&secondparameter=3

通过属性查询,我得到了我的部分内容 URL

parameter=2&secondparameter=3

有没有什么方法可以把它放在一个NSDictionary或一个Array？

多谢

Answer 1

您可以使用queryItems在URLComponents.

当您获得此属性的值时,NSURLComponents类将解析查询字符串并返回NSURLQueryItem对象的数组,每个对象表示一个键值对,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)

目标C.

使用queryItems.

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
    }
}

Answer 2

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,允许您转换查询组件字符串和包含结果的字典对象.

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

Answer 3

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.

Answer 4

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]];
}

Answer 5

以前的所有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];
}

Answer 6

这是swift中的扩展:

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
        }
    }

Answer 7

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,你可以在这里得到一个查询参数:

例如,URL包含param = some_value对

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

  }
}

Answer 8

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

如果您想要 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;
}

注意：URL 可以有重复的参数，但字典将只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。

Answer 9

对于使用Bolts Framework的用户,您可以使用:

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

记得导入:

#import <Bolts/BFURL.h>

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