How to parse JSON object without JSONdecoder in iOS?

Question

Okay, so I received this werid JSON format couple days back, I don't know if it is possible to decode it but I would like to learn if we could.

{
"status" : true,
"orderHistory": {
    "5" : {
        "productId" : 0,
        "productName" : "Anchovy",
        "quantity" : 1,
        "price" : "199"
        "variation": ""
    },
    
    "17" : {
        "productId" : 0,
        "productName" : "Anchovy",
        "quantity" : 1,
        "price" : "199"
        "variation": ""
    },
    
    "98" : {
        "productId" : 0,
        "productName" : "Anchovy",
        "quantity" : 1,
        "price" : "199"
        "variation": ""
    }
}

}

Order History is supposed to be an Array, but in this case backend is sending these Objects with random string keys, Reason we can't create a Decodable Model is that the keys are random,

Any ideas on how to decode this data and use it into our iOS app?

There is no need to throw the baby out with the bathwater. You can still use `JSONDecoder` here. — Sweeper, Aug 27 '21 at 06:52
`JSONSerialization`, but you can still use `JSONDecoder` with a property at some point: `let orderHistory: [String: [OrderHistoryDecodableStruct]]` — Larme, Aug 27 '21 at 07:01
Your JSON dataset is invalid since there's no comma after the price key. — El Tomato, Aug 28 '21 at 00:08

score 0 · Answer 1 · answered Aug 27 '21 at 07:00

0

It is best to continue using the JSON decoder.

The root model matching the JSON would look like this:

struct Body: Decodable {
    let status: Bool
    let orderHistory: [String: Order]
}

The Order Model will be simple.

answered Aug 27 '21 at 07:00

Johannes

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How to parse JSON object without JSONdecoder in iOS?

1 Answers1