Question

我有两个实体

书 - 属性：name，desc - 关系：书籍类型为To Many to BookLang（逆：bookLang）

BookLang - 属性：nameLang，descLang，lang - 关系：bookLang，类型为To One to Book（反向：书）

在BookLang中，fr，de语言有两个对象如果我使用

let lang = "fr"
let pre = NSPredicate(format: "book.lang == %@", lang);
Then I have to loop through the fetchResult and then loop through the book NSSet

for b in bookSet {

    let lan = b.valueForKey("lang") as! String
        if(lan == lang){
        let name = b.valueForKey("nameLang") as! String
        println(name) // the book title in french
        break;
    }                  

}

我如何才能在NSSet中找到我想要的对象

我尝试

let subPre = NSPredicate(format: "SUBQUERY(BookLang, $s, $s.lang == %@)", lang);

有错误

Unable to parse the format string "SUBQUERY(BookLang, $s, $s.lang == %@)"'

Answer 1

保持简单。关系，循环，子查询 - 都是不必要的！

简单的字符串属性lang怎么样？

let frenchBooks = allBooks.filteredSetUsingPredicate(NSPredicate(format: "lang = %@", "fr"))

或者，以真正的Swift方式

let frenchBooks = allBooks.filter() { $0.lang = "fr" }

Answer 2

为什么不使用？来创建谓词？要获取所有符合条件的书籍，您可以对核心数据上下文使用获取请求。

    let fetchRequest = NSFetchRequest(entityName: "Book") //supposing that you managed object subclass is named Book
    fetchRequest.predicate = NSPredicate(format: "lang == %@", lang);
    if let fetchResults = managedObjectContext!.executeFetchRequest(fetchRequest, error: nil) as? [Book] {
//DO what you want
    }

这只是一个你应该完成错误和else语句的片段

是否可以仅返回Core Data中的所需对象

2 个答案: