Swift NSPredicate 在复合语句时抛出 EXC_BAD_ACCESS(Code=1, address=0x1) [英] Swift NSPredicate throwing EXC_BAD_ACCESS(Code=1, address=0x1) when compounding statements

问题描述

我试图在 Swift 中使用 NSPredicate 来查询核心数据，但在尝试运行它时抛出 EXC_BAD_ACCESS(Code=1, address=0x1) 错误，我做错了什么?

I am trying to use NSPredicate in Swift to query Core Data but it throws an EXC_BAD_ACCESS(Code=1, address=0x1) error when trying to run it, what am I doing wrong?

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"
  }
}

推荐答案

谓词格式字符串中的 %@ 占位符用于 Objective-C对象，所以你必须把整数包装成一个 NSNumber:

The %@ placeholder in predicate format strings is for Objective-C objects, so you have to wrap the integer into an NSNumber:

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

或使用 ld 来格式化(长)整数:

or use ld instead to format a (long) integer:

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

还要注意

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

不创建复合谓词，秒赋值简单覆盖第一个.您可以使用 NSCompoundPredicate:

does not create a compound predicate, the seconds assignment simply overwrites the first. You can use an 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:

or simply combine the predicates with "AND":

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

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