[获取字符串中多个相同字符的位置index]

idx.m

#pragma mark - 获取这个字符串ASting中的所有abc的所在的index
- (NSMutableArray *)getRangeStr:(NSString *)text findText:(NSString *)findText
{
    
    NSMutableArray *arrayRanges = [NSMutableArray arrayWithCapacity:3];
    
    if (findText == nil && [findText isEqualToString:@""])
    {
        
        return nil;
        
    }
    
    NSRange rang = [text rangeOfString:findText]; //获取第一次出现的range
    
    if (rang.location != NSNotFound && rang.length != 0)
    {
        
        [arrayRanges addObject:[NSNumber numberWithInteger:rang.location]];//将第一次的加入到数组中
        
        NSRange rang1 = {0,0};
        
        NSInteger location = 0;
        
        NSInteger length = 0;
        
        for (int i = 0;; i++)
        {
            
            if (0 == i)
            {
                
               //去掉这个abc字符串
                location = rang.location + rang.length;
                
                length = text.length - rang.location - rang.length;
                
                rang1 = NSMakeRange(location, length);
                
            }
            else
            {
                
                location = rang1.location + rang1.length;
                
                length = text.length - rang1.location - rang1.length;
                
                rang1 = NSMakeRange(location, length);
                
            }
            
            //在一个range范围内查找另一个字符串的range
            
            rang1 = [text rangeOfString:findText options:NSCaseInsensitiveSearch range:rang1];
            
            if (rang1.location == NSNotFound && rang1.length == 0)
            {
                
                break;
                
            }
            else//添加符合条件的location进数组
            
            [arrayRanges addObject:[NSNumber numberWithInteger:rang1.location]];
            
        }
        
        return arrayRanges;
        
    }
    
    return nil;
    
}

//第一种办法简单粗暴（思路获取第一次xxx出现的位置然后替换成名字 替换之后string中就只有一个xxx了  然后用物品替换string中仅有的一个xxx）

// 实际数据中，$$apple$$，对下划线提问，用第一种简单粗暴的办法

//第二种方法（思路 首先遍历这个字符串 然后找到所有的xxx 所在的位置的index    然后通过index将字符串进行替换）
        NSMutableArray *arrayShare = [self getRangeStr:share6 findText:@"xxx"];
        shareContent = [[_m_dataDic[@"gender"]integerValue] == 1?share5:share6 stringByReplacingCharactersInRange:NSMakeRange([arrayShare[1]integerValue], 3) withString:((!_m_dataDic[@"content"] || [_m_dataDic[@"content"] isKindOfClass:[NSNull class]])?@"顺风车":_m_dataDic[@"content"])];
        shareContent = [shareContent stringByReplacingCharactersInRange:NSMakeRange([arrayShare[0]integerValue], 3) withString:_m_dataDic[@"nickName"]];
//获取这个字符串中的所有xxx的所在的index
- (NSMutableArray *)getRangeStr:(NSString *)text findText:(NSString *)findText
{
    NSMutableArray *arrayRanges = [NSMutableArray arrayWithCapacity:20];
    if (findText == nil && [findText isEqualToString:@""]) {
        return nil;
    }
    NSRange rang = [text rangeOfString:findText]; //获取第一次出现的range
    if (rang.location != NSNotFound && rang.length != 0) {
        [arrayRanges addObject:[NSNumber numberWithInteger:rang.location]];//将第一次的加入到数组中
        NSRange rang1 = {0,0};
        NSInteger location = 0;
        NSInteger length = 0;
        for (int i = 0;; i++)
        {
            if (0 == i) {//去掉这个xxx
                location = rang.location + rang.length;
                length = text.length - rang.location - rang.length;
                rang1 = NSMakeRange(location, length);
            }else
            {
                location = rang1.location + rang1.length;
                length = text.length - rang1.location - rang1.length;
                rang1 = NSMakeRange(location, length);
            }
            //在一个range范围内查找另一个字符串的range
            rang1 = [text rangeOfString:findText options:NSCaseInsensitiveSearch range:rang1];
            if (rang1.location == NSNotFound && rang1.length == 0) {
                break;
            }else//添加符合条件的location进数组
                [arrayRanges addObject:[NSNumber numberWithInteger:rang1.location]];
        }
        return arrayRanges;
    }
    return nil;
}