Question

当我运行下面的代码时，它会向我显示如下错误

警告：'NSMutableDictionary'可能无法响应'-setObject：forKey：forKey：'

我的代码中存在什么问题。

myNumberOfPlayers = players;

myDifficultyLevel = difficulty;

myDuration = duration;

myExcerciseId = excerciseId;

NSMutableDictionary *myDictionary = [[ NSMutableDictionary alloc] init];
    [ myDictionary setObject:players forKey:players  forKey:@"no_of_players"];
    [ myDictionary setObject:difficulty forKey:@"difficulty_level"];
    [ myDictionary setObject:duration forKey:@"duration_excercise"];
    [ myDictionary setObject:excerciseId forKey:@"Excercise_id"];

Answer 1

应该是

[myDictionary setObject:players forKey:@"no_of_players"];

而不是

[myDictionary setObject:players forKey:players  forKey:@"no_of_players"];

Answer 2

NSMutableDictionary * myDictionary = [[NSMutableDictionary alloc] init];

[ myDictionary setObject:players forKey:@"no_of_players"];
[ myDictionary setObject:difficulty forKey:@"difficulty_level"];
[ myDictionary setObject:duration forKey:@"duration_excercise"];
[ myDictionary setObject:excerciseId forKey:@"Excercise_id"];

Answer 3

您正在调用不存在的方法。你应该试试这个：

[ myDictionary setObject:players forKey:players ]

或者

[ myDictionary setObject:players forKey:@"no_of_players"]

我敢打赌，第二种选择是正确的选择。您的代码中可能存在拼写错误，并且您意外地重复了players变量。

字典可以为确定的键设置对象。在同一个调用中设置一个给出两个不同键的对象是没有意义的。

Answer 4

此代码必须类似于

 NSMutableDictionary *myDictionary = [[ NSMutableDictionary alloc] init];
[myDictionary setObject:players forKey:@"no_of_players"];//you cant call for key two time.
[myDictionary setObject:difficulty forKey:@"difficulty_level"];
[myDictionary setObject:duration forKey:@"duration_excercise"];
[myDictionary setObject:excerciseId forKey:@"Excercise_id"];

帮我解决nsdictionary问题

4 个答案: