Question

我使用以下格式创建了一个具有JSON响应的API：

[{"name":"xxx","direct_link":"http:\/\/domain.com\/images\/xxx.png","image":"http:\/\/domain.com\/images\/xxx.png"},{"name":"yyy","direct_link":"http:\/\/domain.com\/images\/yyy.png","image":"http:\/\/domain.com\/images\/yyy.png"}]

注意JSON响应如何没有数组标题。

我的Swift代码如下所示：

       do {
            //converting resonse to NSDictionary
            var teamJSON: NSDictionary!
            teamJSON =  try NSJSONSerialization.JSONObjectWithData(data!, options: .MutableContainers) as? NSDictionary


            //getting the JSON array teams from the response
            let teams: NSArray = teamJSON["teams"] as! NSArray


            //looping through all the json objects in the array teams
            for i in 0 ..< teams.count{

                //getting the data at each index
                let teamId:Int = teams[i]["name"] as! String!
                let teamName:String = teams[i]["direct_link"] as! String!
                let teamMember:Int = teams[i]["image"] as! Int!

                //displaying the data
                print("name -> ", teamId)
                print("direct_link -> ", teamName)
                print("image -> ", teamMember)
                print("===================")
                print("")

            }

注意数组如何寻找标题＆＃34;团队＆＃34;。如何确保正确解析JSON并显示JSON响应所需的3个值？我是应用程序编码的新手，并且拥有网络背景，仍在努力解决这个问题。

当我尝试构建并运行时，我遇到以下错误：fatal error unexpectedly found nil while unwrapping an optional value

Answer 1

问题在于这一行：

teamJSON =  try NSJSONSerialization.JSONObjectWithData(data!, options: .MutableContainers) as? NSDictionary

因为您正在尝试将JSON数组转换为字典。

只需使用

teamJSON =  try NSJSONSerialization.JSONObjectWithData(data!, options: .MutableContainers) as? NSArray

然后你也不需要这样做：

//getting the JSON array teams from the response
let teams: NSArray = teamJSON["teams"] as! NSArray

因为teamJSON已经是数组。

我认为您必须注意的关键问题是NSJSONSerialization.JSONObjectWithData()可以返回NSDictionary或NSArray，具体取决于JSON中的根级别数据是数组还是字典。

Answer 2

试试这个：

do {
    guard let teams = try NSJSONSerialization.JSONObjectWithData(data!, options: .MutableContainers) as? NSArray else {
        //Doesn't exist, or isn't an NSArray
        return
    }

    for team in teams {
        //getting the data at each index
        let teamId = team["name"] as! String
        let teamName = team["direct_link"] as! String
        let teamMember = team["image"] as! Int

        //displaying the data
        print("name -> ", teamId)
        print("direct_link -> ", teamName)
        print("image -> ", teamMember)
        print("===================")
        print()
    }
}
//...

一些注意事项：

请勿投放到隐式展开的选项（例如String!）
不要添加不必要的类型注释
使用guard let强制执行前提条件（例如，JSON不是nil，可以转换为NSArray）。
首选迭代范围（for team in teams）迭代数组元素（for i in 0..<teams.count）
- 如果您只需要索引，请使用for i in teams.indices
- 如果您需要索引和元素，请使用for (index, team) in teams.enumerate()

Answer 3

<强> ISSUE：

您的NSJSONSerialization.JSONObjectWithData会返回NSArray而非NSDictionary

解决方案：

更新代码如下：

var teamJSON =  try NSJSONSerialization.JSONObjectWithData(data!, options: .MutableContainers) as? NSArray

//getting the JSON array teams from the response

let teams: NSArray = teamJSON

同时

请勿使用 !
强制投射值
始终使用?
将其强制转换为可选项

Swift解析JSON格式

3 个答案: