SwiftyJSON 打印字典数组中的所有值

Posted 2023-03-15

【中文标题】SwiftyJSON 打印字典数组中的所有值

【英文标题】：SwiftyJSON print all values in an array of dictionaries

【发布时间】：2017-10-29 21:21:11

【问题描述】：

我有一个字典数组：

var dictArray = [[String:String]]()

我附加了一些字典并将dictArray 转换为 JSON：

let json = JSON(dictArray)

然后我在尝试使用此循环以 JSON 格式实际打印这些值时遇到了麻烦：

for (index,subJson):(String, JSON) in json 
    print(index)

上面的循环取自 SwiftyJSON 文档，但它似乎对我不起作用。我希望输出看起来像这样：

 
"dictArray": [
  
  "a": "a stuff 1",
  "b": "b stuff 1",
  "c": "c stuff 1",
  "d": "d stuff 1"
,

  "a": "a stuff 2",
  "b": "b stuff 2",
  "c": "c stuff 2",
  "d": "d stuff 2"
,

  "a": "a stuff 3",
  "b": "b stuff 3",
  "c": "c stuff 3",
  "d": "d stuff 3"
    
  ]

非常感谢所有帮助。

【参考方案1】：

您不需要迭代字典来打印它。你可以简单地用print(json)打印它

如果你想循环字典，你应该使用这样的东西：

for dictionaries in json 
    for (key, value) in dictionaries 
        print("\(key): \(value)")
    

【讨论】：

这段代码给了我以下错误：Error: type 'JSONGenerator.Element' (aka '(String, JSON)') does not conform to protocol 'Sequence' 和 type 'JSONGenerator.Element' (aka '(String, JSON)') does not conform to protocol 'Sequence'

【参考方案2】：

你可以这样使用：

var array = [String:[[String:String]]]()

var dictArray = [[String:String]]()

var stuffArray1 = [String: String]()
stuffArray1["a"] = "a stuff 1"
stuffArray1["b"] = "b stuff 1"
stuffArray1["c"] = "c stuff 1"
stuffArray1["d"] = "d stuff 1"
dictArray.append(stuffArray1)

var stuffArray2 = [String: String]()
stuffArray2["a"] = "a stuff 2"
stuffArray2["b"] = "b stuff 2"
stuffArray2["c"] = "c stuff 2"
stuffArray2["d"] = "d stuff 2"
dictArray.append(stuffArray2)

var stuffArray3 = [String: String]()
stuffArray3["a"] = "a stuff 3"
stuffArray3["b"] = "b stuff 3"
stuffArray3["c"] = "c stuff 3"
stuffArray3["d"] = "d stuff 3"
dictArray.append(stuffArray3)

array["dictArray"] = dictArray

let json = JSON(array)
print(json)

【参考方案3】：

试试这个：

for (_, object) in json ?? [] 
  let dict = NSMutableDictionary()
  for(key, value) in object 
      dict.setValue(value.stringValue, forKey: key)
     // all the values must be *String*, otherwise this fails.
  

  print(dict)