Convert dictionary to query string in swift?

Question

I have a dictionary as [String:Any].Now i want to convert this dictionary keys & value as key=value&key=value.I have created below extensio

Answer 1

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

Hope it helps. Happy Coding!!

Answer 2

Use NSURLQueryItem.

An NSURLQueryItem object represents a single name/value pair for an item in the query portion of a URL. You use query items with the queryItems property of an NSURLComponents object.

To create one use the designated initializer queryItemWithName:value: and then add them to NSURLComponents to generate an NSURL. For example:

OBJECTIVE-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

Swift:

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

Answer 3

Add this function to your controller

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
    }

Example

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

Answer 4

Compact version of @luckyShubhra's answer

Swift 5.0

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

Usage

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

Answer 5

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
        }
    }

That extension will allow you to parse URL where you have both encoded + sign and space with plus, for example:

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

That extension will parse "q" as "First question" and "email" as "mail+10@mail.com"