iOS 根據時間排序

這個是回答這篇文章的問題

建立個date模型

#import 

@inte***ce
date : nsobject
@property (nonatomic, copy) nsstring *time;
@end

date *d1 = [[date alloc] init];

d1.time = @"2012-12-02";
date *d2 = [[date alloc] init];
d2.time = @"2412-12-02";
date *d3 = [[date alloc] init];
d3.time = @"2312-12-02";
date *d4 = [[date alloc] init];
d4.time = @"2312-12-02";
date *d5 = [[date alloc] init];
d5.time = @"2812-08-02";
date *d6 = [[date alloc] init];
d6.time = @"0000-09-02";
_dataarray = [nsmutablearray array];
[_dataarray addobject:d1];
[_dataarray addobject:d2];
[_dataarray addobject:d3];
[_dataarray addobject:d4];
[_dataarray addobject:d5];
[_dataarray addobject:d6];
for (int i = 0; i < _dataarray.count; i ++) 
nslog(@"\n");
nssortdescriptor *sorter = [[nssortdescriptor alloc] initwithkey:@"time" ascending:yes];
nsmutablearray *sortdescriptors = [[nsmutablearray alloc] initwithobjects:&sorter count:1];
nsarray *sortarray = [_dataarray sortedarrayusingdescriptors:sortdescriptors];
for (int j = 0; j < sortarray.count; j ++)

2017-06-10

19:59:35.046
timesort[4740:678220] 排序前 ==
2012-12-02
2017-06-10
19:59:35.047
timesort[4740:678220] 排序前 ==
2412-12-02
2017-06-10
19:59:35.047
timesort[4740:678220] 排序前 ==
2312-12-02
2017-06-10
19:59:35.047
timesort[4740:678220] 排序前 ==
2312-12-02
2017-06-10
19:59:35.047
timesort[4740:678220] 排序前 ==
2812-08-02
2017-06-10
19:59:35.047
timesort[4740:678220] 排序前 ==
0000-09-02
2017-06-10
19:59:35.048
timesort[4740:678220] 
2017-06-10
19:59:35.048
timesort[4740:678220] 排序後 ==
0000-09-02
2017-06-10
19:59:35.048
timesort[4740:678220] 排序後 ==
2012-12-02
2017-06-10
19:59:35.048
timesort[4740:678220] 排序後 ==
2312-12-02
2017-06-10
19:59:35.048
timesort[4740:678220] 排序後 ==
2312-12-02
2017-06-10
19:59:35.048
timesort[4740:678220] 排序後 ==
2412-12-02
2017-06-10
19:59:35.049
timesort[4740:678220] 排序後 ==
2812-08-02

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