我店的名字作为键和分数作为值到NSDictionary用于节省NSUserDefaults 。然后，我想回去按分数排序的钥匙，但我似乎无法给他们只能通过串数字进行排序。得分100，50，300，名单200，500，例如，给了我100，200，300，50，500。

可以这样做或做我需要去了解这个不同？

NSString *defaultNames[] = {@"Matt", @"Terry",@"Jessica",@"Sean",nil};
NSNumber *defaultScores[] = {@"600", @"500",@"100",@"50", nil};

NSDictionary *newScoreDict =  [NSDictionary dictionaryWithObjects:(id *)defaultScores forKeys:(id *)defaultNames count:7];

NSArray *currScores = [scoreDict keysSortedByValueUsingSelector:@selector(compare:)];

Answer 1:

有关使用keysSortedByValueUsingSelector如何（NSDictionary的）

似乎是你所需要的按在Xcode中的文档

Answer 2:

NSString *defaultNames[] = {@"Matt", @"Terry",@"Jessica",@"Sean",nil};
NSNumber *defaultScores[] = {@"600", @"500",@"100",@"50", nil};
NSDictionary *newScoreDict =  [NSDictionary dictionaryWithObjects:(id *)defaultScores forKeys:(id *)defaultNames count:7];
NSArray *currScores = [scoreDict keysSortedByValueUsingSelector:@selector(localizedStandardCompare:)];

Answer 3:

@implementation NSString (numericComparison)

- (NSComparisonResult) floatCompare:(NSString *) other
{
    float myValue = [self floatValue];
    float otherValue = [other floatValue];
    if (myValue == otherValue) return NSOrderedSame;
    return (myValue < otherValue ? NSOrderedAscending : NSOrderedDescending);
}

- (NSComparisonResult) intCompare:(NSString *) other
{
    int myValue = [self intValue];
    int otherValue = [other intValue];
    if (myValue == otherValue) return NSOrderedSame;
    return (myValue < otherValue ? NSOrderedAscending : NSOrderedDescending);
}

@end

NSString *defaultNames[] = {@"Matt", @"Terry",@"Jessica",@"Sean",nil};
// NSNumber *defaultScores[] = {@"600", @"500",@"100",@"50", nil};

NSNumber *defaultScores[] = {                                                                   
            [NSNumber  numberWithInt:600],
            [NSNumber  numberWithInt:500],
            [NSNumber  numberWithInt:100],
            [NSNumber  numberWithInt:50],
            nil 
    };

NSDictionary *newScoreDict =  [NSDictionary dictionaryWithObjects:(id *)defaultScores forKeys:(id *)defaultNames count:4];

NSArray *currScores = [newScoreDict keysSortedByValueUsingSelector:@selector(intCompare:NotSureWhatGoesHere:)];

我仍然与前行困惑？

难道我只是用

//
NSArray *currScores = [newScoreDict keysSortedByValueUsingSelector:@selector(intCompare:other:)];
//

是一个数字数组OK，或者是有一个更简单的方法？

非常感谢你...

Answer 4:

-compare：是一个字符串比较。通过用于比较，例如一个不同的方法：

@implementation NSString (numericComparison)

- (NSComparisonResult) compareNumerically:(NSString *) other
{
float myValue = [self floatValue];
float otherValue = [other floatValue];
if (myValue == otherValue) return NSOrderedSame;
return (myValue < otherValue ? NSOrderedAscending : NSOrderedDescending);
}

@end

在特定情况下，你可以使用-intValue代替。

Answer 5:

不知道它会帮助，但你也可以保存一个NSArray中的plist; 不像一个NSDictionary（返回基本上随机顺序键），你让他们回来，你把他们进来。

Answer 6:

我认为这个问题的最简单的方法是使用比较..

NSString *defaultNames[] = {@"Matt", @"Terry",@"Jessica",@"Sean",nil};
NSNumber *defaultScores[] = {@(600), @(500),@(400),@(50), nil};
NSDictionary *newScoreDict =  [NSDictionary dictionaryWithObjects:defaultNames       forKeys:defaultScores count:4];
NSArray *currScores = [newScoreDict keysSortedByValueUsingComparator:^NSComparisonResult(id obj1, id obj2) {
    if ([obj1 integerValue] > [obj2 integerValue]) {
        return NSOrderedAscending;
    }else{
        return NSOrderedDescending;
    }}];

for (NSString *string in currScores) {
    NSLog(@"%@",string);
}

试试这个..我注意到，我不能达到的值与使用NSNumber的对象，所以如果你想达到目标值比我改变你的NSNumber分数的NSString解决，并将其转换为数字，而订货。您可以使用如下..

NSString *defaultNames[] = {@"Matt", @"Terry",@"Jessica",@"Sean",nil};
NSString *defaultScores[] = {@"600", @"500",@"400",@"50", nil};
NSMutableDictionary *newScoreDict =  [NSMutableDictionary   dictionaryWithObjects:defaultScores  forKeys:defaultNames count:4];

NSArray *currScores = [newScoreDict keysSortedByValueUsingComparator:^NSComparisonResult(id obj1, id obj2) {
NSNumberFormatter *formatter = [[NSNumberFormatter alloc] init];
[formatter setNumberStyle:NSNumberFormatterDecimalStyle];
NSNumber *number1 = [formatter numberFromString:obj1];
NSNumber *number2 = [formatter numberFromString:obj2];
    if (number1.intValue > number2.intValue) {
        return NSOrderedDescending;
    }else{
        return NSOrderedAscending;
    }}];

for (NSString *name in currScores) {
    NSLog(@"key %@ value %@",name,[newScoreDict valueForKey:name]);
}

希望能帮助到你..

文章来源: NSDictionary keys sorted by value numerically