NSString 到 NSArray

Question

我想将NSString 拆分为NSArray。例如，给定：

NSString *myString=@"ABCDEF";

我想要一个NSArray 喜欢：

NSArray *myArray={A,B,C,D,E,F};

如何用 Objective-C 和 Cocoa 做到这一点？

Answer 1

NSMutableArray *letterArray = [NSMutableArray array];
NSString *letters = @"ABCDEF?क्";
[letters enumerateSubstringsInRange:NSMakeRange(0, [letters length]) 
                            options:(NSStringEnumerationByComposedCharacterSequences) 
                         usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) {
    [letterArray addObject:substring];
}];

for (NSString *i in letterArray){
    NSLog(@"%@",i);
}

结果

A
B
C
D
E
F
?
क्

enumerateSubstringsInRange:options:usingBlock: _{适用于 iOS 4+} 可以枚举不同样式的字符串。一种称为NSStringEnumerationByComposedCharacterSequences，它将逐个字母枚举，但对代理对、基本字符加组合标记、Hangul jamo 和印度辅音簇敏感，所有这些都称为组合字符

请注意，accepted answer“吞噬”? 并将क् 分解为क 和्。

Answer 2

转化率

NSString * string = @"A B C D E F";
NSArray * array = [string componentsSeparatedByString:@" "];
//Notice that in this case I separated the objects by a space because that's the way they are separated in the string

记录

NSLog(@"%@", array);

---

这是控制台返回的内容

Answer 3

NSMutableArray *chars = [[NSMutableArray alloc] initWithCapacity:[theString length]];
for (int i=0; i < [theString length]; i++) {
    NSString *ichar  = [NSString stringWithFormat:@"%C", [theString characterAtIndex:i]];
    [chars addObject:ichar];
}

Answer 4

This link 包含将字符串拆分为基于子字符串和字符集中字符串的数组的示例。希望这篇文章对你有所帮助。

这里是代码片段

NSMutableArray *characters = [[NSMutableArray alloc] initWithCapacity:[myString length]];
for (int i=0; i < [myString length]; i++) {
    NSString *ichar  = [NSString stringWithFormat:@"%c", [myString characterAtIndex:i]];
    [characters addObject:ichar];
}

Answer 5

没有循环你可以使用这个：

NSString *myString = @"ABCDEF";
NSMutableString *tempStr =[[NSMutableString alloc] initWithString:myString];

if([myString length] != 0)
{
    NSError  *error  = NULL;

    // declare regular expression object
    NSRegularExpression *regex =[NSRegularExpression regularExpressionWithPattern:@"(.)" options:NSMatchingReportCompletion error:&error];

    // replace each match with matches character + <space> e.g. 'A' with 'A '
    [regex replaceMatchesInString:tempStr options:NSMatchingReportCompletion range:NSMakeRange(0,[myString length]) withTemplate:@"$0 "];

    // trim last <space> character
    [tempStr replaceCharactersInRange:NSMakeRange([tempStr length] - 1, 1) withString:@""];

    // split into array
    NSArray * arr = [tempStr componentsSeparatedByString:@" "];

    // print
    NSLog(@"%@",arr);
}

此解决方案借助正则表达式在每个字符前面附加空格，并使用componentsSeparatedByString 和<space> 返回一个数组

Answer 6

Swift 4.2：

字符串到数组

let list = "Karin, Carrie, David"

let listItems = list.components(separatedBy: ", ")

输出：[“Karin”、“Carrie”、“David”]

数组转字符串

let list = ["Karin", "Carrie", "David"]

let listStr = list.joined(separator: ", ")

输出：“卡琳、嘉莉、大卫”

Answer 7

在 Swift 中，这变得非常简单。

斯威夫特 3：

myString.characters.map { String($0) }

斯威夫特 4：

myString.map { String($0) }