xcode Swift 4：无法将“__NSCFNumber”类型的值转换为“NSString”
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Swift 4: Could not cast value of type '__NSCFNumber' to 'NSString'
提问by jgravois
I have tried
我试过了
self.adc_role_id = String(res["adc_role_id"])
self.adc_role_id = "\(res["adc_role_id']"
self.adc_role_id = (\(res["adc_role_id"] as? String)!
but still get
但仍然得到
Could not cast value of type '__NSCFNumber' to 'NSString'
无法将“__NSCFNumber”类型的值转换为“NSString”
I added the dump of res[4] below
我在下面添加了 res[4] 的转储
As new as I am to Swift, I don't know anything else to try
就像我对 Swift 的新手一样，我不知道还有什么可以尝试的
回答by Mike Taverne
In Swift 4, the String initializer requires the describing:argument label.
在 Swift 4 中，字符串初始值设定项需要describing:参数标签。
I don't know if this will solve your problem, but your first line of code should be written:
我不知道这是否能解决您的问题，但您的第一行代码应该是这样写的：
self.adc_role_id = String(describing: res["adc_role_id"])
回答by Eric Aya
In your screenshot we can see that res["adc_role_id"]is an NSNumber.
在您的屏幕截图中，我们可以看到这res["adc_role_id"]是一个 NSNumber。
To transform an NSNumber to a String you should use its stringValueproperty.
要将 NSNumber 转换为 String，您应该使用其stringValue属性。
And since a dictionary gives an Optional, you should use optional binding to safely unwrap it.
并且由于字典给出了 Optional，您应该使用 optional 绑定来安全地解包它。
Example:
例子：
if let val = res["adc_role_id"] {
self.adc_role_id = val.stringValue
}
You could also, if you want, use string interpolation instead of the property:
如果需要，您也可以使用字符串插值而不是属性：
if let val = res["adc_role_id"] {
self.adc_role_id = "\(val)"
}
but I think using the property is more relevant.
但我认为使用该属性更相关。
If for some reason the compiler complains about the type of the content, cast it:
如果由于某种原因编译器抱怨内容的类型，请将其转换为：
if let val = res["adc_role_id"] as? NSNumber {
self.adc_role_id = val.stringValue
}
Note that you should notuse String(describing:)because this initializer will try to represent the string in many ways, and some of them will give inaccurate and unexpected results (for example, if String(describing:)resolves to use the debugDescriptionproperty, as explained in the documentation, you may get a totally different string than the one you want).
请注意，您应不使用String(describing:)，因为这个初始化会尽量表现在许多方面的字符串，其中一些将给予不准确和意想不到的结果（例如，如果String(describing:)解析为使用debugDescription属性，如文档中解释的，你可能会得到一个与您想要的字符串完全不同）。
It's also worth noting that using String(describing:)with an optional value such as your dictionary will resolve to a wrong string: String(describing: res["adc_role_id"])will give Optional(yourNumber)! This is why Mike's answer is wrong. Be careful about this. My advice is to avoid using String(describing:)altogether unless for debugging purposes.
还值得注意的是，使用String(describing:)可选值（例如您的字典）将解析为错误的字符串：String(describing: res["adc_role_id"])将给出Optional(yourNumber)! 这就是为什么迈克的答案是错误的。小心这一点。我的建议是避免String(describing:)完全使用，除非出于调试目的。
回答by vadian
The error message is clear and the dump is clear, too.
错误消息很清楚，转储也很清楚。
The value is not a String, it's an Int(64)wrapped in NSNumber
值不是 a String，它是一个Int(64)包裹在NSNumber
Optional bind the value directly to Int(NSNumberis implicit bridged to Int) and use the Stringinitializer.
可选地将值直接绑定到Int（NSNumber隐式桥接到Int）并使用String初始值设定项。
if let roleID = res["adc_role_id"] as? Int {
self.adc_role_id = String(roleID)
}
Please conform to the naming convention that variable names are camelCasedrather than snake_cased
请遵守命名约定，即变量名是驼峰式的而不是蛇形的
