在給定索引處從NSString中提取字子字符串

Question

我想從給定索引處的NSString中提取子字符串。例如：

NSString = @"Hello, welcome to the jungle"; 
int index = 9; 

指數點位「9」是一個詞「歡迎」的中間，我想是能夠提取詞「歡迎」作爲一個子字符串。任何人都可以告訴我怎麼做到這一點？用正則表達式？

Answer 1

這裏有一個解決方案作爲NSString的一個類別：

- (NSString *) wordAtIndex:(NSInteger) index { 
    __block NSString *result = nil; 
    [self enumerateSubstringsInRange:NSMakeRange(0, self.length) 
          options:NSStringEnumerationByWords 
          usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) { 
           if (NSLocationInRange(index, enclosingRange)) { 
            result = substring; 
            *stop = YES; 
           } 
          }]; 
    return result; 
} 

而另一位，這是比較複雜，但可以讓您精確指定單詞字符你想要的：

- (NSString *) wordAtIndex:(NSInteger) index { 
    if (index < 0 || index >= self.length) 
     [NSException raise:NSInvalidArgumentException 
        format:@"Index out of range"]; 

    // This definition considers all punctuation as word characters, but you 
    // can define the set exactly how you like 
    NSCharacterSet *wordCharacterSet = 
    [[NSCharacterSet whitespaceAndNewlineCharacterSet] invertedSet]; 

    // 1. If [self characterAtIndex:index] is not a word character, find 
    // the previous word. If there is no previous word, find the next word. 
    // If there are no words at all, return nil. 
    NSInteger adjustedIndex = index; 
    while (adjustedIndex < self.length && 
      ![wordCharacterSet characterIsMember: 
      [self characterAtIndex:adjustedIndex]]) 
     ++adjustedIndex; 
    if (adjustedIndex == self.length) { 
     do 
      --adjustedIndex; 
     while (adjustedIndex >= 0 && 
       ![wordCharacterSet characterIsMember: 
       [self characterAtIndex:adjustedIndex]]); 
     if (adjustedIndex == -1) 
      return nil; 
    } 

    // 2. Starting at adjustedIndex which is a word character, find the 
    // beginning and end of the word 
    NSInteger beforeBeginning = adjustedIndex; 
    while (beforeBeginning >= 0 && 
      [wordCharacterSet characterIsMember: 
      [self characterAtIndex:beforeBeginning]]) 
     --beforeBeginning; 

    NSInteger afterEnd = adjustedIndex; 
    while (afterEnd < self.length && 
      [wordCharacterSet characterIsMember: 
      [self characterAtIndex:afterEnd]]) 
     ++afterEnd; 

    NSRange range = NSMakeRange(beforeBeginning + 1, 
           afterEnd - beforeBeginning - 1); 
    return [self substringWithRange:range]; 
} 

的第二個版本，也有長串更高效，假設話則短。

Answer 2

這裏是做一個相當hackish的方式，但它的工作：

的NSString有一個方法：

- (NSArray *)componentsSeparatedByString:(NSString *)separator; 

，所以你可以這樣做：

NSString *myString = @"Blah blah blah"; 
NSString *output = @""; 
int index = 9; 
NSArray* myArray = [myString componentsSeparatedByString:@" "]; // <-- note the space in the parenthesis 

for(NSString *str in myArray) { 
    if(index > [str length]) index -= [str length] + 1; // don't forget the space that *was* there 
    else output = str; 
}