iphone - 如何让sortedArrayUsingSelector使用整数而不是字符串进行排序？

Question

我有一些这样的数据:

1、111、2、333、45、67、322、4445

NSArray *array = [[myData allKeys]sortedArrayUsingSelector: @selector(compare:)];

如果我运行这段代码，它会像这样排序:

1, 111, 2,322, 333, 4445, 45, 67,

但我实际上想要这个:

1, 2, 45, 67, 111, 322, 333, 4445

我该如何实现它？谢谢你。

Answer 1

扩展 Paul Lynch 的答案，下面是我使用比较方法作为 NSString 上的类别来执行此操作的示例。此代码仅处理数字后跟可选非数字限定符的情况，但如果需要，您可以将其扩展以处理“1a10”等情况。

创建类别方法后，您只需执行以下操作

[[myData allKeys]sortedArrayUsingSelector:@selector(psuedoNumericCompare:)];

@interface NSString (Support) 
- (NSComparisonResult) psuedoNumericCompare:(NSString *)otherString;
@end

@implementation NSString (Support) 

// "psuedo-numeric" comparison
//   -- if both strings begin with digits, numeric comparison on the digits
//   -- if numbers equal (or non-numeric), caseInsensitiveCompare on the remainder

- (NSComparisonResult) psuedoNumericCompare:(NSString *)otherString {

    NSString *left  = self;
    NSString *right = otherString;
    NSInteger leftNumber, rightNumber;


    NSScanner *leftScanner = [NSScanner scannerWithString:left];
    NSScanner *rightScanner = [NSScanner scannerWithString:right];

    // if both begin with numbers, numeric comparison takes precedence
    if ([leftScanner scanInteger:&leftNumber] && [rightScanner scanInteger:&rightNumber]) {
        if (leftNumber < rightNumber)
            return NSOrderedAscending;
        if (leftNumber > rightNumber)
            return NSOrderedDescending;

        // if numeric values tied, compare the rest 
        left = [left substringFromIndex:[leftScanner scanLocation]];
        right = [right substringFromIndex:[rightScanner scanLocation]];
    }

    return [left caseInsensitiveCompare:right];
}