How can I build a URL with query parameters containing multiple values for the same key in Swift?

IosSwiftSwift2Nsurl

Ios Problem Overview


I am using AFNetworking in my iOS app and for all the GET requests it makes, I build the url from a base URL and than add parameters using NSDictionary Key-Value pairs.

The problem is that I need same key for different values.

Here is an example of what I need the finally URL to look like -

http://example.com/.....&id=21212&id=21212&id=33232

It's not possible in NSDictionary to have different values in same keys. So I tried NSSet but did not work.

let productIDSet: Set = [prodIDArray]
let paramDict = NSMutableDictionary()
paramDict.setObject(productIDSet, forKey: "id")

Ios Solutions


Solution 1 - Ios

All you need is URLComponents (or NSURLComponents in Obj-C). The basic idea is to create a bunch of query items for your id's. Here's code you can paste into a playground:

import Foundation
import XCPlayground

let queryItems = [URLQueryItem(name: "id", value: "1"), URLQueryItem(name: "id", value: "2")]
var urlComps = URLComponents(string: "www.apple.com/help")!
urlComps.queryItems = queryItems
let result = urlComps.url!
print(result)

You should see an output of

> www.apple.com/help?id=1&id=2

Solution 2 - Ios

Method 1

It can add the QueryItem to your existing URL.

extension URL {
    
    func appending(_ queryItem: String, value: String?) -> URL {
                                
        guard var urlComponents = URLComponents(string: absoluteString) else { return absoluteURL }
        
        // Create array of existing query items
        var queryItems: [URLQueryItem] = urlComponents.queryItems ??  []
        
        // Create query item
        let queryItem = URLQueryItem(name: queryItem, value: value)
        
        // Append the new query item in the existing query items array
        queryItems.append(queryItem)
        
        // Append updated query items array in the url component object
        urlComponents.queryItems = queryItems
        
        // Returns the url from new url components
        return urlComponents.url!
    }
}

How to use

var url = URL(string: "https://www.example.com")!
let finalURL = url.appending("test", value: "123")
                  .appending("test2", value: nil)

Method 2

In this method, the URL will be updated automatically.

extension URL {

    mutating func appendQueryItem(name: String, value: String?) {

        guard var urlComponents = URLComponents(string: absoluteString) else { return }

        // Create array of existing query items
        var queryItems: [URLQueryItem] = urlComponents.queryItems ??  []

        // Create query item
        let queryItem = URLQueryItem(name: name, value: value)

        // Append the new query item in the existing query items array
        queryItems.append(queryItem)

        // Append updated query items array in the url component object
        urlComponents.queryItems = queryItems

        // Returns the url from new url components
        self = urlComponents.url!
    }
}

// How to use
var url = URL(string: "https://www.example.com")!
url.appendQueryItem(name: "name", value: "bhuvan")

Solution 3 - Ios

func queryString(_ value: String, params: [String: String]) -> String? {    
    var components = URLComponents(string: value)
    components?.queryItems = params.map { element in URLQueryItem(name: element.key, value: element.value) }
        
    return components?.url?.absoluteString
}

Solution 4 - Ios

An URL extension to append query items, similar to Bhuvan Bhatt idea, but with a different signature:

  • it can detect failures (by returning nil instead of self), thus allowing custom handling of cases where the URL is not RFC 3986 compliant for instance.
  • it allows nil values, by actually passing any query items as parameters.
  • for performance, it allows passing multiple query items at a time.
extension URL {
    /// Returns a new URL by adding the query items, or nil if the URL doesn't support it.
    /// URL must conform to RFC 3986.
    func appending(_ queryItems: [URLQueryItem]) -> URL? {
        guard var urlComponents = URLComponents(url: self, resolvingAgainstBaseURL: true) else {
            // URL is not conforming to RFC 3986 (maybe it is only conforming to RFC 1808, RFC 1738, and RFC 2732)
            return nil
        }
        // append the query items to the existing ones
        urlComponents.queryItems = (urlComponents.queryItems ?? []) + queryItems
        
        // return the url from new url components
        return urlComponents.url
    }
}

###Usage

let url = URL(string: "https://example.com/...")!
let queryItems = [URLQueryItem(name: "id", value: nil),
                  URLQueryItem(name: "id", value: "22"),
                  URLQueryItem(name: "id", value: "33")]
let newUrl = url.appending(queryItems)!
print(newUrl)

Output: >https://example.com/...?id&id=22&id=33

Solution 5 - Ios

2019

private func tellServerSomething(_ d: String, _ s: String) {
    
    var c = URLComponents(string: "https://you.com/info")
    c?.queryItems = [
        URLQueryItem(name: "description", value: d),
        URLQueryItem(name: "summary", value: s)
    ]
    guard let u = c?.url else { return print("url fail") }
    do {
        let r = try String(contentsOf: u)
        print("Server response \(r)")
    }
    catch { return print("comms fail") }
}

Percent-encoding and everything else is handled.

Solution 6 - Ios

In Swift Forming URL with multiple params

func rateConversionURL(with array: [String]) -> URL? {
            var components = URLComponents()
            components.scheme = "https"
            components.host = "example.com"
            components.path = "/hello/"
            components.queryItems = array.map { URLQueryItem(name: "value", value: $0)}
            
        return components.url
    }

Solution 7 - Ios

I guess u just have to do something like this:

let params = ["id" : [1, 2, 3, 4], ...];

which will be encoded into: ....id%5B%5D=1&id%5B%5D=2&id%5B%5D=3&id%5B%5D=4....

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionAbhishekDwivediView Question on Stackoverflow
Solution 1 - IosDaniel GalaskoView Answer on Stackoverflow
Solution 2 - IosBhuvan BhattView Answer on Stackoverflow
Solution 3 - IosKirit VaghelaView Answer on Stackoverflow
Solution 4 - IosCœurView Answer on Stackoverflow
Solution 5 - IosFattieView Answer on Stackoverflow
Solution 6 - IosWasimView Answer on Stackoverflow
Solution 7 - IosSerg DortView Answer on Stackoverflow