如何解析 url 以获取键和值-解网

问：

我引用这个答案来解析 url，但是当解析这个 url

NSString *urlString = @"https://www.example.com/product-detail?journey_id=123456&iswa=1";

我正在得到我的第一把钥匙，但我只需要作为我的钥匙。https://www.example.com/product-detail?journey_idjourney_id

这是我在编码中所做的：

NSString *urlString = @"https://www.example.com/product-detail?journey_id=123456&iswa=1";
        
NSMutableDictionary *waLoginDictionary = [[NSMutableDictionary alloc] init];
NSArray *urlComponents = [urlString componentsSeparatedByString:@"&"];
                            
for (NSString *keyValuePair in urlComponents) {
NSArray *pairComponents = [keyValuePair componentsSeparatedByString:@"="];
NSString *key = [[pairComponents firstObject] stringByRemovingPercentEncoding];
NSString *value = [[pairComponents lastObject] stringByRemovingPercentEncoding];
[waLoginDictionary setObject:value forKey:key];

}
                            
NSLog(@"%@", waLoginDictionary);

我得到这个输出：

{
"https://www.example.com/product-detail?journey_id" = 123456;
iswa = 1;
}

iOS Swift Objective-C 解析 Cocoa-Touch

阅读问题的所有答案。有时有旧的，从那时起就有了 iOS/SDK/Lib 的改进，如答案的第一行所述（即：“编辑（2018 年 6 月）：”这个答案更好”。苹果在iOS 7中添加了“），并且由于解决方案由作者自行决定，因此他们并不总是选择”最好的”。NSURLComponents

答：

3赞 Prasad Parab 4/25/2022 #1

你所指的答案已经过时了，作者本人已经进行了相应的更新。Apple 在对象中添加了。[URLQueryItem]URLComponent

试试这个。

迅速

    let urlString = "https://www.example.com/product-detail?journey_id=123456&iswa=1"
    var dict: [String : String] = [:]
    if let urlComponents = URLComponents(string: urlString), let queryItems = urlComponents.queryItems {
        for item in queryItems {
            dict[item.name] = item.value
        }
    }
    print("dict : \(dict)")

目标 - C

NSString *urlString = @"https://www.example.com/product-detail?journey_id=123456&iswa=1";
NSMutableDictionary *dict = [NSMutableDictionary dictionary];

NSURLComponents *urlComponents = [NSURLComponents componentsWithString:urlString];
NSArray *queryItems = [urlComponents queryItems];

for (NSURLQueryItem *item in queryItems) {
    [dict setValue:item.value forKey:item.name];
}

NSLog(@"dict %@", dict);

如何解析 url 以获取键和值

How to parse a url to get key and value

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