获取复杂请求核心数据的谓词和表达式

Posted 2023-03-12

【中文标题】获取复杂请求核心数据的谓词和表达式

【英文标题】：Predicate and expression to fetch complex request core data

【发布时间】：2013-08-31 16:32:31

【问题描述】：

我必须提出一个复杂的核心数据获取请求，但我不知道是否可以提出。 这是我的场景：只有一个具有这些属性的实体（费用）：

成本

(NSDecimalNumber)

存款

(NSDecimalNumber)

类别

（NSString）

付费

（布尔值）

请求应该返回 3 个最昂贵的类别，但这些是必须遵守的规则：

如果

Paid == YES

，应将费用成本添加到费用类别总计中 如果

Paid == NO && Deposit > 0

，费用押金应添加到费用类别总额中 如果

Paid == NO

，则不应将任何内容添加到费用类别总计中

使用 NSExpression，我可以计算每个类别的每个总数，但它还包括未支付的费用成本。 有没有办法做到这一点？ 非常感谢！

【参考方案1】：

例如，您可以使用NSFetchRequest：

// Build the fetch request
NSString *entityName = NSStringFromClass([Expense class]);
NSFetchRequest *request = [[NSFetchRequest alloc] init];
request.entity = entity;

仅过滤相关费用：

NSPredicate *predicate = [NSPredicate predicateWithFormat:@"(paid == YES) OR ((paid == NO) AND (deposit > 0))"];
request.predicate = predicate;

并总结cost和depost属性：

NSExpressionDescription *(^makeExpressionDescription)(NSString *, NSString *) = ^(NSString *keyPath, NSString *name)

    // Create an expression for the key path.
    NSExpression *keyPathExpression = [NSExpression expressionForKeyPath:keyPath];

    // Create an expression to represent the function you want to apply
    NSExpression *totalExpression = [NSExpression expressionForFunction: @"sum:" arguments: @[keyPathExpression]];

    NSExpressionDescription *expressionDescription = [[NSExpressionDescription alloc] init];

    // The name is the key that will be used in the dictionary for the return value
    expressionDescription.name = name;
    expressionDescription.expression = totalExpression;
    expressionDescription.expressionResultType = NSDecimalAttributeType;

    return expressionDescription;
;

NSExpressionDescription *totalCostDescription = makeExpressionDescription(@"cost", @"totalCost");
NSExpressionDescription *totalDepositDescription = makeExpressionDescription(@"deposit", @"totalDeposit");

// Specify that the request should return dictionaries.
request.resultType = NSDictionaryResultType;

request.propertiesToFetch = @[categoryDescription,
                              paidDescription,
                              totalCostDescription,
                              totalDepositDescription];

并按类别和付费状态对结果进行分组：

// Get 'category' and 'paid' attribute descriptions
NSEntityDescription *entity = [NSEntityDescription entityForName:entityName
                                              inManagedObjectContext:context];
NSDictionary *attributes = [entity attributesByName];
NSAttributeDescription *categoryDescription = attributes[@"category"];
NSAttributeDescription *paidDescription = attributes[@"paid"];

// Group by 'category' and 'paid' attributes
request.propertiesToGroupBy = @[categoryDescription, paidDescription];

您将获得汇总的已付和未付费用

NSError *error = nil;
NSArray *results = [context executeFetchRequest:request error:&error];

您需要做的就是合并（和排序）然后：

if (results) 
    NSMutableDictionary *combined = [NSMutableDictionary dictionary];

    for (NSDictionary *result in results) 
        NSString *category = result[@"category"];

        BOOL paid = [result[@"paid"] boolValue];

        NSDecimalNumber *total = result[paid ? @"totalCost" : @"totalDeposit"];

        NSDecimalNumber *sum = combined[category];

        if (sum) 
            total = [total decimalNumberByAdding:sum];
        

        combined[category] = total;
    

    NSArray *sortedCategories = [combined keysSortedByValueUsingSelector:@selector(compare:)];

    [sortedCategories enumerateObjectsWithOptions:NSEnumerationReverse usingBlock:^(id obj, NSUInteger idx, BOOL *stop) 
        NSLog(@"Category %@: %@", obj, combined[obj]);
    ];

else 
    NSLog(@"Error: %@", error);

【讨论】：

eik，非常感谢！这正是我一直在寻找的东西，我挽救了我的一天！ ;) 它就像一个魅力！