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ios - 参数类型 'AnyObject' 不符合预期类型 NSCopying

转载 作者:行者123 更新时间:2023-11-30 13:30:58 24 4
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我尝试在 Swift 中使用 NSDictionary ,但遇到了上述问题。我有以下格式的字典:

let xyz: NSMutableDictionary = ["1":[1,2,3,4,"1","n","1","2"],"2":[1,2,3,4,"+","o","6","2"]]

我想迭代字典中的键并提取数组的第 6 个元素。我尝试了以下方法;但没有遇到任何运气:

for keys in dictKeyMutableDict {
let xCentVal = xyz[keys as! [NSCopying]][6]
}

我不断收到下标错误,如果我删除 as! [NSCopying],我收到上述错误。有谁知道如何处理这种情况吗?

最佳答案

删除 NSMutableDictionary 并通过将其设为 var 使其可变。现在您可以删除 as! [NS复制]

var xyz = ["1":[1,2,3,4,"1","n","1","2"],"2":[1,2,3,4,"+","o","6","2"]]

for keys in dictKeyMutableDict {
let xCentVal = xyz[keys]![6]
}

或者更好地选择打开它:

for keys in dictKeyMutableDict {
if let v = xyz[keys] {
let xCentVal = v[6]
}
}

关于ios - 参数类型 'AnyObject' 不符合预期类型 NSCopying,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36583232/

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