Swift
easy value JSON
, Support all systems: macOS
iOS
watchOS
tvOS
linux
... As long as you can run Swift
一个Swift
语言简易的JSON
取值,支持 macOS
iOS
watchOS
tvOS
linux
... 所有系统,只要能运行 Swift
File -> Swift Packages -> Add Package dependency
import JSONValue
let jsonData = ["key": "value"]
/*
/// or
let jsonData = ["key", "value"]
/// or
let jsonData = "{\"key\":\"value\"}"
/// or
let jsonData = "{\"key\":
\"value\"}".data(using: .utf8)!
jsonData格式
Dictionary<String,Any>,Array<Any>,String,Number,Int,UInt,Double,Float,Bool,nil
*/
let json = JSONValue(jsonData)
/// or
let json = jsonData.json
/// Array
let item = json[0]
let number = json[0].number
let string = json[0].number.string
let bool = json[0].number.bool
let item = json.0
let number = json.0.number
let string = json.0.number.string
let bool = json.0.number.bool
json[0] = "string".json
json[0] = 123.json
/// Dictionary
let item = json["key"]
let number = json["key"].number
let string = json["key"].number.string
let bool = json["key"].number.bool
let item = json.key
let number = json.key.number
let string = json.key.number.string
let bool = json.key.number.bool
json.key = "string".json
json.key = 123.json
/// path
let item = json["key1"]["key2"]
/// or
let item = json["key"][0]
/// or
let item = json[["key", 0]]
json["key1"]["key2"] = "string".json
json["key"][0] = "string".json
json[["key", 0]] = "string".json
let item = json.key1.key2
let item = json.key.0
json.key1.key2 = "string".json
json.key.0 = true.json
let dictionary = json.dictionary
let array = json.array
let number = json.number
let string = json.number.string
let bool = json.number.bool
let int = json.number.int
let double = json.number.double
let float = json.number.float
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Last commit: 2 years ago |
Swift JSON Value Simplify JSON Value
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