NSString 到 NSArray

Question

I want to split an NSString into an NSArray . For example, given:

NSString *myString=@"ABCDEF";

I want an NSArray like:

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

How to do this with Objective-C and Cocoa?

Answer 1

NSMutableArray *letterArray = [NSMutableArray array];
NSString *letters = @"ABCDEF𝍱क्";
[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);
}

results in

A
B
C
D
E
F
𝍱
क्

enumerateSubstringsInRange:options:usingBlock: _{available for iOS 4+} can enumerate a string with different styles. One is called NSStringEnumerationByComposedCharacterSequences , what will enumerate letter by letter but is sensitive to surrogate pairs, base characters plus combining marks, Hangul jamo, and Indic consonant clusters , all referred as Composed Character

Note, that the accepted answer "swallows" and breaks क् into क and ् .

Answer 2

Conversion

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

Logging

NSLog(@"%@", array);

---

This is what the console returned

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 contains examples to split a string into a array based on sub strings and also based on strings in a character set. I hope that post may help you.

here is the code snip

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

Without loop you can use this:

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);
}

This solution append space in front of each character with the help of regular expression and uses componentsSeparatedByString with <space> to return an array

Answer 6

Swift 4.2:

String to Array

let list = "Karin, Carrie, David"

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

Output: ["Karin", "Carrie", "David"]

Array to String

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

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

Output: "Karin, Carrie, David"

Answer 7

In Swift, this becomes very simple.

Swift 3:

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

Swift 4:

myString.map { String($0) }