Swift NSPredicate 在复合语句时抛出 EXC_BAD_ACCESS(Code=1, address=0x1)

Question

我正在尝试在 Swift 中使用 NSPredicate 来查询核心数据，但在尝试运行时会抛出 EXC_BAD_ACCESS(Code=1, address=0x1) 错误，我做错了什么？

这是发生错误的文件

class LevelsScreenModel : UIViewController {

func getWord(level: Int, section: Int) -> String
{
    let fetchRequest = NSFetchRequest(entityName: "Words")

    //This is the line where the error happens
    fetchRequest.predicate = NSPredicate(format: "level = %@", level)
    fetchRequest.predicate = NSPredicate(format: "section = %@", section)

    let word = AppDelegate().managedObjectContext!.executeFetchRequest(fetchRequest, error: nil) as [Words]

    if(word.count > 1)
    {
        for words in word
        {
            println(words.word)
            return words.word
        }
    }

    return "ERROR"
  }
}

Answer 1

谓词格式字符串中的 %@ 占位符用于 Objective-C 对象，所以你必须将整数包装成 NSNumber:

fetchRequest.predicate = NSPredicate(format: "level = %@", NSNumber(integer: level))

或使用 ld 来格式化一个（长）整数：

fetchRequest.predicate = NSPredicate(format: "level = %ld", level)

另请注意

fetchRequest.predicate = NSPredicate(format: ...)
fetchRequest.predicate = NSPredicate(format: ...)

不创建复合谓词，简单的秒赋值覆盖第一个。您可以使用 NSCompoundPredicate:

let p1 = NSPredicate(format: "level = %ld", level)!
let p2 = NSPredicate(format: "section = %ld", section)!
fetchRequest.predicate = NSCompoundPredicate.andPredicateWithSubpredicates([p1, p2])

或者简单地将谓词与 "AND":

结合起来

fetchRequest.predicate = NSPredicate(format: "level = %ld AND section = %ld", level, section)

Answer 2

您可以使用 PredicatePal 框架，而不是忙于格式转换和 AND 子谓词：

fetchRequest.predicate = *(Key("level") == level && Key("section") == section)

请注意，您需要使用 == 而不是 = 进行相等比较。

Swift NSPredicate 在复合语句时抛出 EXC_BAD_ACCESS(Code=1, address=0x1)

Swift NSPredicate throwing EXC_BAD_ACCESS(Code=1, address=0x1) when compounding statements

swift

core-data

nspredicate