JSON object to String in Swift

Question

I have spend a lot of time googling and I could not get a solution to work so I hope someone can help me.

I have a string called user which is of type: < hit>JSON. I can get information out of it by using:

user.object["displayname"] ?? ""

This works fine however it gives me the result between "". So if someones displayname is stackoverflowuser it will give me: "stackoverflowuser". Is there a way to get this as a string without the ""?

Edit:

I use the following API: https://www.algolia.com/doc/api-reference/api-methods/search/#examples.

This is my code to get the user object:

index.search(query: "\(searchtext)") { result in
   if case .success(let response) = result {
      for user in response.hits {
         let resultDisplayname = "\(user.object["displayname"] ?? "")"

      }
   }
 }

The resultDisplayname is an example of a variable that I want to turn into a String.

Is there something wrong with, for example, "chrispsv"? Cause it's swift syntax. Check this documentation: developer.apple.com/documentation/swift/string#2919514 — sonle
– sonle, Commented May 4, 2022 at 8:45
Well the problem is that if I want to show someones username it is between "". And that just doesn't look good. Also if I want to show someones profilepicture. The link to the url is also between "". This way the url doesn't work and I can't show the picture. — chrispsv
– chrispsv, Commented May 4, 2022 at 9:26
So what I can do is remove the first and last character of the string (because those are the "") but that is an ugly solution. — chrispsv
– chrispsv, Commented May 4, 2022 at 9:41
What do you mean show someones username? If you got the string from displayname and set to label, like, nameLabel.text = user.object["displayname"] ?? "". It will work perfectly. — sonle
– sonle, Commented May 4, 2022 at 9:52
If I add .text behind it I get the following error in xcode: "Value of type 'JSON' has no member 'text'" — chrispsv
– chrispsv, Commented May 4, 2022 at 10:10

sonle · Accepted Answer · 2022-05-04 12:38:48Z

1

let's assume your JSON is like this format

{
    "displayname": "chrispv",
    "age": 30
}

Create a swift object to represent it

struct Person: Decodable {
    let displayname: String
    let age: Int
}

Then execute your search query using object above

index.search(query: "\(searchtext)") { result in
    if case .success(let response) = result {
        do {
            let persons: [Person] = try response.extractHits()
            for person in persons {
                print(person.displayname)
            }
        } catch let error {
            print("Decoding error")
        }
    }
}

edited May 4, 2022 at 12:38

answered May 4, 2022 at 12:32

sonle

10.4k3 gold badges17 silver badges31 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

chrispsv Over a year ago

Yes I understand this is the best solution. However it is very hard to create a struct of the query result since the user returns a very complicated JSON format. Is there a way to create a struct that is flexible?

sonle Over a year ago

I think it's better if the data model mapping 1-1 with JSON structure. However, you can just take what you really need. In this scenario, remove age and keep displayname it's ok. Make sense?

chrispsv Over a year ago

Thank you very much! This fixed my issue!! I did not know that it was possible to decode JSON in a struct that wasn't mapped 1-1!

sonle Over a year ago

It's no big deal.

Collectives™ on Stack Overflow

JSON object to String in Swift

1 Answer 1

4 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

4 Comments

Your Answer

Sign up or log in

Post as a guest

Related