解析Core Data中按获取请求分组的NSDictionary结果

Question

我需要使用 CoreData 从表中获取一些统计信息。我已经使用本手册使用“group by”表达式提取数据。

这是我的源代码：

var popularGuestNames: [String] = []
        let keypathExp = NSExpression(forKeyPath: "guestName")
        let expression = NSExpression(forFunction: "count:", arguments: [keypathExp])

        let countDesc = NSExpressionDescription()
        countDesc.expression = expression
        countDesc.name = "count"
        countDesc.expressionResultType = .integer64AttributeType

        let request = NSFetchRequest<NSFetchRequestResult>(entityName: "GuestsTable")
        request.returnsObjectsAsFaults = false
        request.propertiesToGroupBy = ["guestName"]
        request.propertiesToFetch = ["guestName", countDesc]
        request.resultType = .dictionaryResultType

        let matchedGuests = try? context.fetch(request)
        guard matchedGuests != nil else {return popularGuestNames}

但作为结果，我得到了类型为 [Any] 的matchedGuests。而我卡在这里。我可以使用 print 看到有 JSON 对象：

Optional(
[{
    count = 1;
    guestName = "\U0412\U0430\U043d\U044f";
}, {
    count = 13;
    guestName = "\U0413\U043e\U0441\U0442\U044c 1";
}, {
    count = 9;
    guestName = "\U0413\U043e\U0441\U0442\U044c 2";
}, {
    count = 3;
    guestName = "\U0413\U043e\U0441\U0442\U044c 3";
}, {
    count = 1;
    guestName = "\U0413\U043e\U0441\U0442\U044c 4";
}, {
    count = 1;
    guestName = "\U0413\U043e\U0441\U0442\U044c 5";
}, {
    count = 1;
    guestName = "\U041d\U043d\U043d";
}, {
    count = 1;
    guestName = "\U0422\U0435\U0441\U0442";
}, {
    count = 1;
    guestName = "\U0422\U0435\U0441\U0442 \U0442\U0435\U0441\U0442\U043e\U0432\U0438\U0447";
}, {
    count = 1;
    guestName = "\U0423\U0410\U0443\U043f\U0430\U0438\U0443\U0430\U0433\U0443";
}])

但我找不到解析它的方法。我知道我应该使用 JSONSerialization.jsonObject，但是这个方法需要数据类型的输入变量。而且我无法理解如何将 Any 转换为 Data。

Answer 1

首先，Core Data fetch 的结果绝对不是JSON 对象

其次要更具体！

由于获取请求的返回类型显然是字典使用泛型类型传递 NSDictionary

let request = NSFetchRequest<NSDictionary>(entityName: "GuestsTable")

然后将类型相应地转换为实际类型[[String:Any]]。

var matchedGuests = [[String:Any]]()
...
if let result = try? context.fetch(request) as! [[String:Any]] {
   matchedGuests = result
} 

即使try?发生错误，假设提取请求有效，强制解包的返回类型也永远不会导致崩溃。

如果你只想要客人命名map数组

let popularGuestNames = matchedGuests.flatMap { $0["guestName"] as? String }