Question

我希望在Swift（2.0）中反序列化JSON时不传递任何选项。我最初尝试过：

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery-validate/1.14.0/jquery.validate.min.js" type="text/javascript"></script>

但那不能编译，我收到错误：

类型NSJSONReadingOptions不符合协议NilLiteralConvertible

枚举NSJSONReadingOptions没有任何“无”选项，那么如果我不想要这些选项，我该怎么办？

Answer 1

在swift 2中，你应该使用空数组[]来表示no options：

NSJSONSerialization.JSONObjectWithData(data, options: [])

Answer 2

tldr; Swift 3：跳过选项参数，一切都会好的。

JSONSerialization.jsonObject(with: data)

说明：

在swift 3中，函数调用是

class func jsonObject(with data: Data, options opt: JSONSerialization.ReadingOptions = []) throws -> AnyObject

ReadingOptions是一个选项集，Option Set协议的标题有

/// When you need to create an instance of an option set, assign one of the
/// type's static members to your variable or constant. Alternately, to create
/// an option set instance with multiple members, assign an array literal with
/// multiple static members of the option set. To create an empty instance,
/// assign an empty array literal to your variable.
///
///     let singleOption: ShippingOptions = .priority
///     let multipleOptions: ShippingOptions = [.nextDay, .secondDay, .priority]
///     let noOptions: ShippingOptions = []

option set docs are here

表示您可以致电

JSONSerialization.jsonObject(with: data, options: [])

但是，选项已经在函数定义中定义了默认[]，因此您可以完全跳过它并调用

JSONSerialization.jsonObject(with: data)

Answer 3

您还可以为NSJSONReadingOptions对象使用空构造函数。

NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions())

如何在Swift中向NSJSONSerialization.JSONObjectWithData传递任何选项

3 个答案: