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ios - 解析 JSON 并将其变为变量

转载 作者:行者123 更新时间:2023-11-29 11:30:39 24 4
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SWIFT4 JSON 解析 Firebase我有一个已解析的 JSON,如何从中获取电子邮件和用户名到变量?以及如何将 citiesName 从它获取到 citiesName 数组?

struct User: Codable {

let email: String
let username: String
let cities: [city]
}
struct city: Codable {
let cityname: String
}
if snapshot.exists()
{
print(snapshot)
guard let value = snapshot.value else { return }
do {
let parsedData = try FirebaseDecoder().decode([User].self, from: value)
print(parsedData)
} catch let error {
print(error)
}
}


JSON: Snap (users) {
0 = {
cinemas = {
0 = {
cityname = Kazan;
};
1 = {
cityname = Kursk;
};
2 = {
cityname = Volgograd;
};
};
email = "placeholder@gmail.com";
username = “Alex”;
};
}

最佳答案

你可以试试循环

parsedData.forEach {

print($0.email)

}

如果是1个用户

if ler user = parsedData.first  {  
print(user.email)
let cities = user.cities.map { $0.cityname }
print(cities)
}

关于ios - 解析 JSON 并将其变为变量,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53923049/

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