我有一个PHP web api,它以下列格式返回json:
{"users":[
{"user": {id:"1","name":"ahmad"}},
...
]}
在我的Swift 2代码中,我能够检索上面的数据,将其存储在名为NSArray
的{{1}}
现在,我需要迭代抛出每个users
以将其转换为对象:
user
输出类似的东西:
for user in users {
print("found: \(user)")
}
但是当我尝试访问该对象的任何元素时,我收到错误:
found: {
user = {
id = 1;
name = ahmad;
};
}
然后我试了一下:
let id = user["user"]["id"] //does not work: Xcode wont compile
let id2 = user["user"]!["id"]! //does not work: Xcode wont compile
let id3 = user!["user"]!["id"]! //does not work: Xcode wont compile
我在if let u=user["user"] { //does not work: Xcode wont compile
// do somthing
}
处设置了一个断点,看看发生了什么,这就是我发现的:
当我打印每个人print("\(user)")
的描述时,我得到:
如何在Swift 2中访问此JSON数据的元素?
答案 0 :(得分:2)
NSArray
只保留AnyObject
,因此您必须将其投放到Array<Dictionary<String, Dictionary<String, String>>>
。下方您会看到速记:
// this is a forced cast and you probably get runtime errors if users cannot be casted
for user in users as! [[String : [String : String]]] {
print("found: \(user)")
}