2016-09-21 21 views
-3
NSMutableArray *array = [NSMutableArray arrayWithObjects:@"22,343", @"44,323",@"34,5678",@"22,725", nil]; 

我有这个数组,我要总结两个值这个数组。 这样array[0] + array [1]如何求和数组的两个元素?

如何总结这个元素? 谢谢!!

+0

你有你自己尝试新鲜事物? – Tavo

+0

是的,我试图增加一倍mydouble1 =阵列[0]; double mudouble2 = array [1];但是这种方法不正确 –

+2

'array [anIndex]'是你的示例代码中的'NSString'对象,而不是'int','NSInteger'或其他原语。所以'NSInteger sum = [array [0] integerValue] + [array [1] integerValue];' – Larme

回答

0

解决方案是

NSString *strVaue1 = [array[0] stringByReplacingOccurrencesOfString:@"," withString:@""]; 
NSString *strVaue2 = [array[1] stringByReplacingOccurrencesOfString:@"," withString:@""]; 
NSUInteger arrValue1 = [strVaue1 integerValue]; 
NSUInteger arrValue2 = [strVaue2 integerValue]; 
NSUInteger sum = arrValue1 + arrValue2; 
NSNumberFormatter *fmt = [[NSNumberFormatter alloc] init]; 
[fmt setNumberStyle:NSNumberFormatterDecimalStyle]; // to get commas (or locale equivalent) 
[fmt setMaximumFractionDigits:0]; // to avoid any decimal 
NSString *result = [fmt stringFromNumber:@(sum)]; 
NSLog(@"The result is - %@",result); 

打印的结果是

The result is - 66,666 
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