Question

我有一个字符串说例如“a，b，c，d，e，f，g，h”，现在我想要替换从索引4＆amp;以指数6结束。

因此，在示例中，结果字符串将是“a，b，c，d，f，e，g，h”。

仅限FYI动态中的所有内容，包括要替换的索引..

我不知道如何实现这一点..感谢任何帮助!!

Answer 1

在这种情况下，如果你NSMutableString会更好。请参阅以下示例：

int a = 6; // Assign your start index.
int b = 9; // Assign your end index.

NSMutableString *abcd = [NSMutableString stringWithString:@"abcdefghijklmano"]; // Init the NSMutableString with your string.
[abcd deleteCharactersInRange:NSMakeRange(a, b)]; //Now remove the charachter in your range.

[abcd insertString:@"new" atIndex:a]; //Insert your new string at your start index.

Answer 2

从您的示例中可以看出，您要替换字符串中的组件（即索引4是第四个分隔的字母 - “e”）。如果是这种情况，那么解决方案就在于NSString componentsSeparatedByString：和componentsJoinedByString：

// string is a comma-separated set of characters.  replace the chars in string starting at index
// with chars in the passed array

- (NSString *)stringByReplacingDelimitedLettersInString:(NSString *)string withCharsInArray:(NSArray *)chars startingAtIndex:(NSInteger)index {

    NSMutableArray *components = [[string componentsSeparatedByString:@","] mutableCopy];

    // make sure we start at a valid position
    index = MIN(index, components.count-1);

    for (int i=0; i<chars.count; i++) {
        if (index+i < components.count)
            [components replaceObjectAtIndex:index+i withObject:chars[i]];
        else
            [components addObject:chars[i]];
    }
    return [components componentsJoinedByString:@","];
}

- (void)test {
    NSString *start = @"a,b,c,d,e,f,g";
    NSArray *newChars = [NSArray arrayWithObjects:@"x", @"y", @"y", nil];
    NSString *finish = [self stringByReplacingDelimitedLettersInString:start withCharsInArray:newChars startingAtIndex:3];
    NSLog(@"%@", finish);  // logs @"a,b,c,x,y,z,g"

    finish = [self stringByReplacingDelimitedLettersInString:start withCharsInArray:newChars startingAtIndex:7];
    NSLog(@"%@", finish);  // logs @"a,b,c,d,e,f,x,y,z"
}

用其他内容替换NSString的中间内容

2 个答案: