URL Encoding a string with special characters

Question

Please consider below code that I implemented in my iOS App:

let kBaseURL = "https://www.sampledomain.com/"
let method = "api/MasterData/GetCompanyList?jsonArgs={\"AccountID\":%d}"

private func prepareURL(api: String) -> String {
    return (kBaseURL + api).addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)!
}

In above code, "method" has "%d" which will be replaced by account id (at run time) that is an integer.

let accountID = 20
let methodToCall = String(format: method, accountID)
var apiRequest = URLRequest(url: URL(string: prepareURL(api: methodToCall))!)

Above code is crashing as URL is nil which is due to "method" that has special characters in it.

Crash reason given by Xcode is: Thread 1: Fatal error: Unexpectedly found nil while unwrapping an Optional value

where are you calling the method prepareUrl in the viewController? — Karthick Ramesh, May 24 '18 at 07:35
@KarthickRamesh, Its mentioned in the description. var apiRequest = URLRequest(url: URL(string: prepareURL(api: method))!) — Satyam, May 24 '18 at 07:36
@KarthickRamesh, I updated my description to understand the code better. Can you please check it. — Satyam, May 24 '18 at 07:41

beshio · Answer 1 · 2018-05-24T14:11:58.403

As found in Apple's doc, if you have some characters in the string which are illegal in a URL, URL(string:) returns nil. At this case, you need to % encode as follows.

let url = URL(string: "123\"456") // this returns nil due to double quote

So you need to mofify to

let url = URL(string: "123\"456".addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)!)

I came across the same issue when I was handling non-English letters; for my case they are "Kanji". So, I always use percent encoding w/o checking what's in the string when calling URL(string:). Actually, there are so many illegal URL letters w/ "Kanji" and it's next to impossible to check what's legal and what's not manually. URL(string:) w/ the percent encoded string by addingPercentEncoding(withAllowedCharacters: .urlHostAllowed) always works nicely for me.

life saver. I was having issues with one of my query param. i resolved like : URL(string: str.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)!) — Vikas saini, Jul 15 '22 at 11:54

Matic Oblak · Answer 2 · 2018-05-24T07:56:15.090

You need to encode those characters as well. You can check what percent escaped values are for { and } and put them in manually. But I rather suggest you have the ability to create a query string from dictionary. Something like the following should fix your issue:

func createQueryString(from parameters: [String: String]) -> String {
    let components: [String] = parameters.compactMap { item in
        guard let encoded = item.value.addingPercentEncoding(withAllowedCharacters: CharacterSet(charactersIn: "!*'\"(){};:@&=+$,/?%#[]% ").inverted) else { return nil }
        return item.key + "=" + encoded
    }
    return components.joined(separator: "&")
}

Now all you need to do is use it and ensure jsonArgs={\"AccountID\":%d} is used in a dictionary as well. Try something like:

func generateAccountsRequest(accountID: Int) -> URLRequest {
    let kBaseURL = "https://www.sampledomain.com"
    let endpoint = "api/MasterData/GetCompanyList"

    func prepareURL(endpoint: String, query: [String: String]) -> String {
        return kBaseURL + "/" + endpoint + "?" + createQueryString(from: query)
    }

    return URLRequest(url: URL(string: prepareURL(endpoint: endpoint, query: ["jsonArgs": "{\"AccountID\":\(accountID)}"]))!)
}

I intentionally didn't add it in here but you should probably check if query is empty in which case you should not append `?` and just return `base + endpoint`. — Matic Oblak, May 24 '18 at 07:43

URL Encoding a string with special characters

2 Answers2