Question

我有一个像

这样的字符串

NSString* str = @"[90, 5, 6]";

我需要将其转换为类似

的数组

NSArray * numbers = [90, 5 , 6];

我做了很长的路：


+ (NSArray*) stringToArray:(NSString*)str
{
    NSString *sep = @"[,";
    NSCharacterSet *set = [NSCharacterSet characterSetWithCharactersInString:sep];
    NSArray *temp=[str componentsSeparatedByCharactersInSet:set];

    NSMutableArray* numbers = [[NSMutableArray alloc] init];
    for (NSString* s in temp) {
        NSNumber *n = [NSNumber numberWithInteger:[s integerValue]];
        [numbers addObject:n];
    }

    return numbers;
}

有没有任何简洁快捷的方式进行此类转换？

由于

Answer 1

首先从字符串中删除不需要的字符，如空格和大括号：

NSString* str = @"[90, 5, 6]";
NSCharacterSet* characterSet = [[NSCharacterSet
                                 characterSetWithCharactersInString:@"0123456789,"] invertedSet];
NSString* newString = [[str componentsSeparatedByCharactersInSet:characterSet]
                       componentsJoinedByString:@""];

您将拥有这样的字符串：90,5,6。然后只需使用逗号分割并转换为NSNumber：

NSArray* arrayOfStrings = [newString componentsSeparatedByString:@","];
NSMutableArray* arrayOfNumbers = [NSMutableArray arrayWithCapacity:arrayOfStrings.count];
for (NSString* string in arrayOfStrings) {
    [arrayOfNumbers addObject:[NSDecimalNumber decimalNumberWithString:string]];
}

使用this response中的NSString类别，可以简化为：

NSArray* arrayOfStrings = [newString componentsSeparatedByString:@","];
NSArray* arrayOfNumbers = [arrayOfStrings valueForKey: @"decimalNumberValue"];

Answer 2

NSString* str = @"[90, 5, 6]";
NSCharacterSet *characterSet = [NSCharacterSet characterSetWithCharactersInString:@"[] "];
NSArray *array = [[[str componentsSeparatedByCharactersInSet:characterSet]
                        componentsJoinedByString:@""]     
                        componentsSeparatedByString:@","];

Answer 3

NSString *newSTR = [str stringByReplacingOccurrencesOfString:@"[" withString:@""];

newSTR = [newSTR stringByReplacingOccurrencesOfString:@"]" withString:@""];

NSArray *items = [newSTR componentsSeparatedByString:@","];

Answer 4

试试这个

NSArray *arr = [string componentsSeparatedByString:@","];

Answer 5

您可以使用正则表达式

来实现

（[0-9] +）

NSError* error = nil;
NSString* str = @"[90, 5, 6]";
NSRegularExpression* regex = [NSRegularExpression regularExpressionWithPattern:@"([0-9]+)" options:0 error:&error];
NSArray* matches = [regex matchesInString:str options:0 range:NSMakeRange(0, [str length])];

然后你有一个字符串的NSArray，你只需要迭代它并将字符串转换为数字并将它们插入数组。

如何将NSString转换为NSArray

5 个答案: