如何在Swift 3.1中从jsonObject获取JSONArray

Question

{
"status": true,
"status_code": 1,
"content": [
    {
        "cat_id": "3",
        "cat_name": "Food",
        "cat_parentid": "2"
    },
    {
        "cat_id": "4",
        "cat_name": "Entertainment",
        "cat_parentid": "2"
    },
    {
        "cat_id": "5",
        "cat_name": "Cars",
        "cat_parentid": "2"
    },
    {
        "cat_id": "12",
        "cat_name": "Personal Care",
        "cat_parentid": "2"
    }
],
"message": "Success"
}

UPDATE

do {
        //create json object from data
        if let json = try JSONSerialization.jsonObject(with: data, options: .mutableContainers) as? [String: Any] {
            completion((json as? AnyObject)!) //here completion callback will return the jsonObject to my UIViewController.
        }

    } catch let error {
        print(error.localizedDescription)
    }

this is my JSONObject. I am very new to the swift. how to get the content JSONArray and further process in swift.? Anybody can help me? Help will be appreciated.

Answer 1

This code checks if the status is true , gets the array for key content and prints all values in the array.

The array is clearly [[String:String]] so cast the object to this specific type.

do {
    //create json object from data
    if let json = try JSONSerialization.jsonObject(with: data) as? [String: Any] {
        if let status = json["status"] as? Bool, status == true {
            if let content = json["content"] as? [[String:String]] {
                for category in content {
                    let id = category["cat_id"]
                    let name =  category["cat_name"]
                    let parentId =  category["cat_parentid"]
                    print(id , name, parentId)
                }
            }
        }
    }

} catch let error {
    print(error.localizedDescription)
}

PS: As always, never use .mutableContainers in Swift. It's meaningless

Answer 2

Check whether your json has content array

if let content = json["content"] as? [Dictionary<String, AnyObject>] {
      print(content) // it will give you content array
}

Answer 3

you can extract you data by providing key

if let array = result["content"] as? Array<AnyObject> {
      print(arry)  
    }

Answer 4

let content = dict.objectForKey("content")! as NSArray

Then you can get json of single object for parsing by

for var cat in content
{
    print(cat) 
}

Answer 5

Another alternative way, by using the library.

First, import JSON library for Swift - SwiftyJSON and use the code:

import SwiftyJSON

let json = JSON(<jsonObject data>)

let contentArray: Array<JSON> = json["content"].arrayValue

Library Integration
If you're using cocoapods then use this pod:

pod 'SwiftyJSON'

OR else just drag SwiftyJSON.swift to the project tree.

Answer 6

You can access like below

    if  let filePath = Bundle.main.path(forResource: "sample", ofType: "json"), let data = FileManager().contents(atPath: filePath) {
      do {
             let dicRes =  try JSONSerialization.jsonObject(with: data, options: .allowFragments) as? [String: Any]
             let contentArray = dicRes?["content"]
              print("contentArray == \(contentArray)")
         } catch {

         }
    }

Answer 7

Get content array like this:

let allContent = json["content"] as? [[String: Any]]

Full sample:

do {
        //create json object from data
        if let json = try JSONSerialization.jsonObject(with: data, options: .mutableContainers) as? [String: Any] {
            if let allContent = json["content"] as? [[String: Any]] {
                for content in allContent {
                    let catId = content["cat_id"] as? String
                    let catName = content["cat_name"] as? String
                    let catParentId = content["cat_parentid"] as? String
                    print(">> catid=" + catId!)
                    print(">> catName=" + catName!)
                    print(">> catparentID=" + catParentId!)
                }
            }
        }

    } catch let error {
        print(error.localizedDescription)
    }