Question

我有实体：语言，正确和答案。

模型看起来像Language{A:name(NSString), R:propers(NSSet)} --->> Proper{A:name(NSString), R:answer(Answer)} ---> Answer{A:answer(NSString)}

所以，我得到了带有参数的NSDictionary：{@"key1", @"value1"}, {@"key2", @"value2"}... i

我需要从此词典中创建 NSPredicate ，以便从我的NSDictionary中获取所有语言，其中propers.name = key[i]和propers.answer.answer = value[i]。

示例：

C++

level : high

try/catch : yes

typization : static

Java

level : high

try/catch : yes

typization : dynamic

NSDictionary : {level : hight}, {try/catch : yes}, {typization : dynamic}

//制作NSPredicate并将其设置为数组控制器 //数组控制器arrangeObjects将返回Java

抱歉语法不好：/

更新 * 经过2个星期的不眠之夜和专家系统老师的工作，没有检查它就去了实验室。杀了我算了。非常感谢大家。 *

Answer 1

您需要来自SUBQUERY。类似的东西：

[NSPredicate predicateWithFormat:@"B.key = %@ AND SUBQUERY(B, $B, $B.C.value = %@).@count > 0", key, value];

SUBQUERY遍历对象，如果你有多个多对多的关系，你也可以拥有嵌套的SUBQUERY。

您可以使用和“ANY ...”但它并不适用于所有情况。

Answer 2

我只想猜你想做什么，所以这里是一个代码：

- (NSPredicate *)constructPredicateWithDictionary:(NSDictionary *)dictionary
{
    NSArray *allKeys = [dictionary allKeys];
    NSMutableArray *predicates = [NSMutableArray array];

    for (NSString *key in allKeys) {
        NSPredicate *predicate = [NSPredicate predicateWithFormat:@"SUBQUERY(B, $B, $B.key = %@ && $B.C.value = %@).@count > 0", key, [dictionary valueForKey:key]];
        [predicates addObject:predicate];
    }


    //not quite sure what you need so I am guessing

    NSPredicate *finalAndPredicate = [NSCompoundPredicate andPredicateWithSubpredicates:predicates]; //if you want all the predicates to be concatenated with and '&&' - logical expression - so all of the subqueries have to be correct
    NSPredicate *finalOrPredicate = [NSCompoundPredicate orPredicateWithSubpredicates:predicates]; //if you want all the predicates to be concatenated with or '||' - logical expression - so any of the subqueries has to be correct

    return finalOrPredicate; //return the needed predicate
}

NSPredicate对许多人来说像＆lt; -----＆gt;＆gt; b -----＆以及c

2 个答案: