Question

有没有办法实现JAVA函数 int indexOf（int ch，int fromIndex）。

让我解释一下：

NSString *steSample = @"This is sample test string";

现在我希望获得 i 的索引，但是在第二个索引之后。我怎样才能做到这一点。

提前致谢

Answer 1

正则表达式对我来说太复杂了：）

我们不能修剪它然后找它吗？

那是3行...

包装在一个类别中：

#import <Foundation/Foundation.h>

@interface NSString (Extend)
-(NSUInteger)indexOfSubstring:(NSString*)needle afterIndex:(NSUInteger)index;
@end
@implementation NSString (Extend) 
-(NSUInteger)indexOfSubstring:(NSString*)needle afterIndex:(NSUInteger)index {
    id str = [self substringFromIndex:index];

    NSUInteger i = [str rangeOfString:needle].location;
    return i==NSNotFound ? i : i+index;
}
@end

演示用法：

int main(int argc, char *argv[]) {
    @autoreleasepool {
        id str = @"@asd@asd";
        NSUInteger index = [str indexOfSubstring:@"@" afterIndex:2];
        NSLog(@"index of @ is: %d", index);
    }
}

Answer 2

我会做这样的事情：

NSString *_sample = @"This is sample test string";
NSError *_error;
NSRegularExpression *_regExp = [NSRegularExpression regularExpressionWithPattern:@"i" options:NSRegularExpressionCaseInsensitive error:&_error];
NSArray *_matches = [_regExp matchesInString:_sample options:NSMatchingReportCompletion range:NSMakeRange(0, _sample.length)];
[_matches enumerateObjectsUsingBlock:^(NSTextCheckingResult * result, NSUInteger idx, BOOL *stop) {
    if (idx == 0) {
        NSLog(@"ignoring first occurance...");
    } else {
        NSLog(@"occurance's index : %d, character's index in string : %d", idx, result.range.location); // that line is simplified for your problem
    }
}];

注意：您可以重新排列实际的if语句，它当前“跳过”第一次出现并打印其余内容 - 但它可以根据您的进一步愿望进行自定义。

我的控制台显示如下内容：

ignoring first occurance...
occurance's index : 1, character's index in string : 5
occurance's index : 2, character's index in string : 23

Answer 3

NSString *steSample = @"This is sample test string";
NSUInteger count = 0, length = [steSample length];
NSRange range = NSMakeRange(0, length);
while(range.location != NSNotFound)
{
    range = [steSample rangeOfString: @"i" options:0 range:range];
    if(range.location != NSNotFound)
    {
        range = NSMakeRange(range.location + range.length, length - (range.location + range.length));

        count++;
        if (count == 2)
        {
            NSLog(@"%d", range.location); // print 6 which is location of second 'i'
        }
    }
}

特定索引后出现字符

3 个答案: