Question

我正在开发一个appplication，其中我需要时间戳。我有两个包含小时，分钟和秒的字符串对象的数组，我需要减去它们并将它们作为小时，分钟和秒作为三个对象放在第三个数组中。 / p>

示例：

arr1=[9,10,22]
arr2=[10,12,42];

所以答案应该是

arr3=[1,2,22]

我需要将这个数组对象放在由' :'

分隔的字符串中

请帮助。

Answer 1

尝试这个

NSMutableArray * arr3=[[NSMutableArray alloc]init];

[arr3 addObject:[NSString stringWithFormat:@"%d",[[arr2 objectAtIndex:0]integerValue]-[[arr1 objectAtIndex:0]integerValue]]];

[arr3 addObject:[NSString stringWithFormat:@"%d",[[arr2 objectAtIndex:1]integerValue]-[[arr1 objectAtIndex:1]integerValue]]];

[arr3 addObject:[NSString stringWithFormat:@"%d",[[arr2 objectAtIndex:2]integerValue]-[[arr1 objectAtIndex:2]integerValue]]];

Answer 2

你也可以试试这个：

NSMutableArray *arr3=[[NSMutableArray alloc]init];

    [arr3 addObject:[NSString stringWithFormat:@"%d",[[arr2 objectAtIndex:0]intValue]-[[arr1 objectAtIndex:0]intValue]]];
    [arr3 addObject:[NSString stringWithFormat:@"%d",[[arr2 objectAtIndex:1]intValue]-[[arr1 objectAtIndex:1]intValue]]];
    [arr3 addObject:[NSString stringWithFormat:@"%d",[[arr2 objectAtIndex:2]intValue]-[[arr1 objectAtIndex:2]intValue]]];

Answer 3

试试这个：

NSMutableArray *arr3 = [NSMutableArray alloc]init];
for(int i=0;i<[arr1 count];i++)
{
  [arr3 addObject:[NSString stringWithFormat:@"%d",[[arr2 objectAtIndex:i]integerValue]-[[arr1 objectAtIndex:i]integerValue]]];
}

NSString *combinedStr = [arr3 componentsJoinedByString:@":"];

如何减去两个数组对象？

3 个答案: