Question

我的模型有字段日期。我需要获取一年中所有日期的所有记录数，或者如果今天没有记录提取为0.现在我获取一年中的所有记录并手动制作。我可以使用核心数据查询吗？

Answer 1

通常在Core Data中，您需要预先完成所有提取操作，而不是将其分解为多次提取。

我试图通过两种方式计算一年中的记录：

获取一年的所有记录。查找这些记录的不同日期。循环日期并计算记录。
在一年中的每一天循环。对具有该日期的记录执行仅计数提取。

使用小型数据库（48条记录），第一种方法的速度提高了约90倍。我想第一种方法的表现会随着更多记录的增加而变得更糟，第二种方法会保持不变。

以下是代码：

- (void)doFullLoadTestWithYear:(int)year {
 NSFetchRequest *fetchRequest = [[NSFetchRequest alloc] init];

 NSEntityDescription *entity = [NSEntityDescription entityForName:@"Record" inManagedObjectContext:self.managedObjectContext];
 [fetchRequest setEntity:entity];

 NSCalendar *gregorian = [[[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar] autorelease];
 NSDateComponents *comps = [[[NSDateComponents alloc] init] autorelease];
 [comps setDay:1];
 [comps setMonth:1];
 [comps setYear:year];
 NSDate *start = [gregorian dateFromComponents:comps];
 [comps setDay:31];
 [comps setMonth:12];
 NSDate *end = [gregorian dateFromComponents:comps];

 NSPredicate *yearPred = [NSPredicate predicateWithFormat:@"date >= %@ && date <= %@",start,end];
 [fetchRequest setPredicate:yearPred];

 NSError *error = nil;
 NSArray *array = [self.managedObjectContext executeFetchRequest:fetchRequest error:&error];
 [fetchRequest release];
 if (error) {
  NSLog(@"Error during fetch: %@",[error localizedDescription]);
  abort();
 }
 int dateCount = 0, recordCount = 0;

 NSArray *dates = [array valueForKeyPath:@"@distinctUnionOfObjects.date"];

 for (NSDate *date in dates) {
  NSPredicate *pred = [NSPredicate predicateWithFormat:@"date == %@",date];

  NSArray *records = [array filteredArrayUsingPredicate:pred];

  dateCount++;
  recordCount += [records count];

  NSLog(@"%d record(s) with date %@",[records count],date);
 }

 NSLog(@"Record count for year is %d",recordCount);
 NSLog(@"Distinct dates count is %d",dateCount);
}

- (void)doSeparateTestWithYear:(int)year {
 NSFetchRequest *fetchRequest = [[NSFetchRequest alloc] init];

 NSEntityDescription *entity = [NSEntityDescription entityForName:@"Record" inManagedObjectContext:self.managedObjectContext];
 [fetchRequest setEntity:entity];

 NSCalendar *gregorian = [[[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar] autorelease];
 NSDateComponents *comps = [[[NSDateComponents alloc] init] autorelease];
 [comps setDay:1];
 [comps setMonth:1];
 [comps setYear:year];
 NSDate *start = [gregorian dateFromComponents:comps];
 [comps setYear:year+1];
 NSDate *end = [gregorian dateFromComponents:comps];

 NSDateComponents *oneDay = [[[NSDateComponents alloc] init] autorelease];
 [oneDay setDay:1];

 int dateCount = 0, recordCount = 0;

 NSDate *date = start;
 while (![date isEqual:end]) {
  NSPredicate *dayPred = [NSPredicate predicateWithFormat:@"date == %@",date];
  [fetchRequest setPredicate:dayPred];

  NSError *error = nil;
  int count = [self.managedObjectContext countForFetchRequest:fetchRequest error:&error];
  if (error) {
   NSLog(@"Error during fetch: %@",[error localizedDescription]);
   abort();
  }

  if (count > 0) {
   NSLog(@"%d record(s) with date %@",count,date);
   dateCount++;
   recordCount += count;
  }

  date = [gregorian dateByAddingComponents:oneDay toDate:date options:0];
 }

 NSLog(@"Record count for year is %d",recordCount);
 NSLog(@"Distinct dates count is %d",dateCount);

 [fetchRequest release];
}

核心数据复杂查询

1 个答案: