快速将字典转换为查询字符串？

Question

我有一个字典为[String:Any]。现在我想将此字典键和值转换为key=value&key=value。我创建了下面的扩展来工作，但它对我不起作用。

extension Dictionary {

    var queryString: String? {
        var output: String = ""
        for (key,value) in self {
            output +=  "\(key)=\(value)\(&)"
        }
        return output
    }
}

Answer 1

使用NSURLQueryItem。

一个 NSURLQueryItem 对象代表一个单一的名称/值对 URL 的查询部分中的项目。您将查询项与 NSURLComponents 对象的 queryItems 属性。

要创建一个使用指定的初始化器queryItemWithName:value:，然后将它们添加到NSURLComponents 以生成NSURL。例如：

目标-C：

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;
NSURL *url = components.URL; // http://stackoverflow.com?q=ios&count=10

斯威夫特：

let queryDictionary = [ "q": "ios", "count": "10" ]
var components = URLComponents()
components.queryItems = queryDictionary.map {
     URLQueryItem(name: $0, value: $1)
}
let URL = components.url

Answer 2

另一种 Swift 风格的方法：

let params = [
    "id": 2,
    "name": "Test"
]

let urlParams = params.flatMap({ (key, value) -> String in
    return "\(key)=\(value)"
}).joined(separator: "&")

Answer 3

var populatedDictionary = ["key1": "value1", "key2": "value2"]

extension Dictionary {
    var queryString: String {
        var output: String = ""
        for (key,value) in self {
            output +=  "\(key)=\(value)&"
        }
        output = String(output.characters.dropLast())
        return output
    }
}

print(populatedDictionary.queryString)

// Output : key1=value1&key2=value2

希望对您有所帮助。快乐编码！

Answer 4

extension Dictionary {
    var queryString: String? {
        return self.reduce("") { "\($0!)\($1.0)=\($1.1)&" }
    }
}

Answer 5

与@KKRocks 相同，更新为 swift 4.1.2

func queryItems(dictionary: [String:Any]) -> String {
    var components = URLComponents()
    print(components.url!)
    components.queryItems = dictionary.map {
        URLQueryItem(name: $0, value: String(describing: $1))
    }
   return (components.url?.absoluteString)!
}

Answer 6

@luckyShubhra 答案的精简版

Swift 5.0

extension Dictionary {
    var queryString: String {
        var output: String = ""
        forEach({ output += "\($0.key)=\($0.value)&" })
        output = String(output.dropLast())
        return output
    }
}

用法

let populatedDictionary = ["key1": "value1", "key2": "value2"]
let urlQuery = populatedDictionary.queryString
print(urlQuery)

Answer 7

试试这个：

func queryItems(dictionary: [String:Any]) -> String {
        var components = URLComponents()
        print(components.url!)
        components.queryItems = dictionary.map {
            URLQueryItem(name: $0, value: $1)
        }
       return (components.url?.absoluteString)!
    }

Answer 8

import Foundation

    extension URL {
        var queryItemsDictionary: [String: String] {
            var queryItemsDictionary = [String: String]()

            // we replace the "+" to space and then encode space to "%20" otherwise after creating URLComponents object
            // it's not possible to distinguish the real percent from the space in the original URL
            let plusEncodedString = self.absoluteString.replacingOccurrences(of: "+", with: "%20")

            if let queryItems = URLComponents(string: plusEncodedString)?.queryItems {
                queryItems.forEach { queryItemsDictionary[$0.name] = $0.value }
            }
            return queryItemsDictionary
        }
    }

该扩展将允许您解析同时编码 + 符号和带有加号的空格的 URL，例如：

https://stackoverflow.com/?q=First+question&email=mail%2B10@mail.com

该扩展程序会将“q”解析为“第一个问题”，将“email”解析为“mail+10@mail.com”

Answer 9

protocol ParametersConvertible {
    func asParameters() -> [String:Any] 
}

protocol QueryStringConvertible {
    func asQuery() -> String
}

extension QueryStringConvertible where Self: ParametersConvertible {
    func asQuery() -> String {
        var queries: [URLQueryItem] = []
        for (key, value) in self.asParameters() {
            queries.append(.init(name: key, value: "\(value)"))
        }

        guard var components = URLComponents(string: "") else {
            return ""
        }

        components.queryItems = queries
        return components.percentEncodedQuery ?? ""
    }
}

Answer 10

将此功能添加到您的控制器

func getQueryString(params : [String : Any])-> String{

        let urlParams = params.compactMap({ (key, value) -> String in
            return "\(key)=\(value)"
        }).joined(separator: "&")
        var urlString = "?" + urlParams
        if let url = urlString.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed){
            urlString = url
        }
          return urlString
    }

示例

self.getQueryString(params: ["name" : "deep ios developer" , "age" :22])

Answer 11

适合任何想要“短、短”版本的人。

使用 map 进行压缩，您不需要 forEach 或 for 循环。对字典的类型强制也有协议约束。

extension Dictionary where Key : StringProtocol, Value : StringProtocol {
    var queryString: String {
        self.map { "\($0)=\($1)" }.joined(separator: "&")
    }
}

Answer 12

我写了这个helper fn，我认为它很干净：

private static func queryStringParamsToString(_ dictionary: [String: Any]) -> String {
    return dictionary
        .map({(key, value) in "\(key)=\(value)"})
        .joined(separator: "&")
        .addingPercentEncoding(withAllowedCharacters: .urlPathAllowed)!

}