如何在Swift(iOS,Xcode)中将NSDictionary转换为Json String?

时间:2015-12-17 04:47:02

标签: ios json swift nsdictionary

我遇到以下问题.... 我的NSDictionary就是这样:

var dic : NSDictionary = [ "level" :
    [
        ["column" : 0,"down" : 0,"left" : 0,"right" : 0,"row" : 0,"up" : 0],
        ["column" : 1,"down" : 0,"left" : 0,"right" : 0,"row" : 0,"up" : 0],
        ["column" : 2,"down" : 0,"left" : 0,"right" : 0,"row" : 0,"up" : 0],
        ["column" : 0,"down" : 0,"left" : 0,"right" : 0,"row" : 1,"up" : 0],
        ["column" : 1,"down" : 0,"left" : 0,"right" : 0,"row" : 1,"up" : 0],
        ["column" : 2,"down" : 0,"left" : 0,"right" : 0,"row" : 1,"up" : 0]
    ]
]

但如果我打印出来,

print(dic);  or print(“\(dic)”);

Out out就是这样:

{
    level =     (
                {
            column = 0;
            down = 0;
            left = 0;
            right = 0;
            row = 0;
            up = 0;
        },
                {
            column = 1;
            down = 0;
            left = 0;
            right = 0;
            row = 0;
            up = 0;
        },
                {
            column = 2;
            down = 0;
            left = 0;
            right = 0;
            row = 0;
            up = 0;
        },
                {
            column = 0;
            down = 0;
            left = 0;
            right = 0;
            row = 1;
            up = 0;
        },
                {
            column = 1;
            down = 0;
            left = 0;
            right = 0;
            row = 1;
            up = 0;
        },
                {
            column = 2;
            down = 0;
            left = 0;
            right = 0;
            row = 1;
            up = 0;
        }
    ); }

如何获得精确的Json字符串? 在swift中,xcode?

1 个答案:

答案 0 :(得分:28)

无需实现这种复杂的逻辑,

你可以简单地这样做

var jsonData: NSData = NSJSONSerialization.dataWithJSONObject(dictionary, options: NSJSONWritingOptions.PrettyPrinted, error: &error)!
    if error == nil {
        return NSString(data: jsonData, encoding: NSUTF8StringEncoding)! as String
    }

如果您想将API与服务器一起发送,则无需将其转换为String