Question

如何使用消息中的数百个键转换此JSON响应，以便将其分配给变量类型public void actionPerformed(ActionEvent evt) { if (evt.getSource().equals(addButton)) { if (source.getSelectedValue() != null) { String str = (String) source.getSelectedValue(); if(str != null) { destModel.addElement(str); dest.setSelectedIndex(0); sourceModel.removeElement(str); source.setSelectedIndex(0); } } } }

JSON

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尝试

解析JSON之后，我无法将[AnyHashable: Any]分配给接受{ "content": { "message": { "greet": "Hello world", "error": "There is an error", "thanks": "Thank you", ... } } }的变量。我试图将jsonData的类型更改为[AnyHashable: Any]，但是它不符合message。

[AnyHashable: Any]

Answer 1

你们不能同时拥有。

要么想用Decodable解析JSON，要么必须将testVariable声明为

private var testVariable : TestCase?

或者您想将testVariable保留为[AnyHashable: Any]，然后删除结构并使用传统的JSONSerialization解码JSON

~~//print(s) struct TestCase: Decodable { let content: Content } struct Content: Decodable { let message: [String: String] }~~

// ViewController.swift

private var testVariable = [AnyHashable:Any]()

private func loadJson(filename: String) {
    if let url = Bundle.main.url(forResource: filename, withExtension: "json") {
        do {
            let data = try Data(contentsOf: url)
            testVariable = try JSONSerialization.jsonObject(with: data) as? [AnyHashable: Any] ?? [:]

        } catch {
            print("JsonError: ", error)
        }
    }
}

如何在Swift中将JSON响应转换为字典格式

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