對象排序與對象

我有一個列表視圖，並希望對數組進行排序，以便每個名稱都在正確的UILocalizedIndexCollation部分下。以下是正在傳入的部分片段。我想通過name進行排序，但我也希望保持對象完整無缺，以便可以使用其他數據。對象排序與對象

2012-03-23 12:10:14.083 MyApp[61241:14803] Section: (
     { 
     "c_id" = 261; 
     "customer_id" = 178664; 
     "first_name" = My; 
     "last_name" = Test; 
     name = "Test, My"; 
    }, 
     { 
     "c_id" = 261; 
     "customer_id" = 185182; 
     "first_name" = valid; 
     "last_name" = Test; 
     name = "Test, valid"; 
    }, 
     { 
     "c_id" = 261; 
     "customer_id" = 178729; 
     "first_name" = Test; 
     "last_name" = Three; 
     name = "Three, Test"; 
    }, 
     { 
     "c_id" = 261; 
     "customer_id" = 178727; 
     "first_name" = Test; 
     "last_name" = Two; 
     name = "Two, Test"; 
    }, 
     { 
     "c_id" = 261; 
     "customer_id" = 178728; 
     "first_name" = Test; 
     "last_name" = Two; 
     name = "Two, Test"; 
    } 
)

分區

-(NSArray *)partitionObjects:(NSArray *)array collationStringSelector:(SEL)selector 
{ 
    UILocalizedIndexedCollation *collation = [UILocalizedIndexedCollation currentCollation]; 
    NSInteger sectionCount = [[collation sectionTitles] count]; 
    NSMutableArray *unsortedSections = [NSMutableArray arrayWithCapacity:sectionCount]; 

    for (int i = 0; i < sectionCount; i++) { 
     [unsortedSections addObject:[NSMutableArray array]]; 
    } 

    for (id object in array) { 
     NSInteger index = [collation sectionForObject:[object objectForKey:@"name"] collationStringSelector:selector]; 
     [[unsortedSections objectAtIndex:index] addObject:object]; 
    } 

    NSMutableArray *sections = [NSMutableArray arrayWithCapacity:sectionCount]; 

    for (NSMutableArray *section in unsortedSections) { 
     [sections addObject:[collation sortedArrayFromArray:section collationStringSelector:selector]]; 
    } 

    return sections; 
}

調用分區

NSDictionary *json = [NSJSONSerialization JSONObjectWithData:data options:kNilOptions error:&error]; 
     NSMutableArray *tempCustomers = [[NSMutableArray alloc] init]; 
     NSString *name; 
     for (NSDictionary *dict in [json objectForKey:@"data"]) { 
      name = [NSString stringWithFormat:@"%@, %@",[dict objectForKey:@"last_name"],[dict objectForKey:@"first_name"]]; 
      NSMutableDictionary *tempDict = [[NSMutableDictionary alloc] initWithDictionary:dict]; 
      [tempDict setValue:name forKey:@"name"]; 
      [tempCustomers addObject:tempDict]; 

     } 

     self.customers = tempCustomers; 
     self.customerData = [self partitionObjects:[self customers] collationStringSelector:@selector(self)];

我需要保持對象完好，但按名稱排序，並讓每個對象被放置到它們與相應章節UILocalizedIndexCollation

ANSWER使用代碼：

-(NSArray *)partitionObjects:(NSArray *)array collationStringSelector:(SEL)selector 
{ 
    UILocalizedIndexedCollation *collation = [UILocalizedIndexedCollation currentCollation]; 
    NSInteger sectionCount = [[collation sectionTitles] count]; 
    NSMutableArray *unsortedSections = [NSMutableArray arrayWithCapacity:sectionCount]; 

    for (int i = 0; i < sectionCount; i++) { 
     [unsortedSections addObject:[NSMutableArray array]]; 
    } 

    for (id object in array) { 
     NSInteger index = [collation sectionForObject:[object objectForKey:@"name"] collationStringSelector:selector]; 
     [[unsortedSections objectAtIndex:index] addObject:object]; 
    } 

    NSMutableArray *sections = [NSMutableArray arrayWithCapacity:sectionCount]; 
    NSSortDescriptor *sortDescriptor = [[NSSortDescriptor alloc] initWithKey:@"name" ascending:YES selector:@selector(localizedCaseInsensitiveCompare:)]; 
    NSArray *sortDescriptors = [NSArray arrayWithObject:sortDescriptor]; 


    for (NSMutableArray *section in unsortedSections) { 
     NSArray *sortedArray = [section sortedArrayUsingDescriptors:sortDescriptors]; 
     collationStringSelector:selector]]; 
     [sections addObject:sortedArray]; 
    } 

    return sections; 
}

來源

2012-03-23 Bot

這可能會有所幫助： http://stackoverflow.com/questions/805547/how-to-sort-an-nsmutablearray-with-custom- objects-in-it – benuuu 2012-03-23 19:31:18

你能把代碼放在這裏嗎？ – Shmidt 2012-12-06 11:33:51

@Flink你是什麼意思？代碼在上面發佈。 – Bot 2012-12-06 17:46:33

你應該可以用比較方法來做到這一點。

的一個很好的例子在這裏提供：https://stackoverflow.com/a/805589/580291

來源

2012-03-23 20:24:18 benuuu

這工作完美。我認爲我將不得不使用它與整理排序，但NOPE！謝謝！ – Bot 2012-03-23 20:29:00

對象排序與對象

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