如何使用正则表达式查找以三个字符前缀开头的单词

Question

我的目标是计算以多个字母的指定前缀开头的单词数(以字符串形式).案例是以"非"开头的单词.所以在这个例子中......

NSString * theFullTestString = @"nonsense non-issue anonymous controlWord";

......我想,但不是"匿名"或"控制字""胡说"和"不是问题的问题"命中.我的点击总数应为2.

所以这是我的测试代码似乎很接近,但我试过的正则表达式形式都没有正常工作.此代码捕获"废话"(正确)和"匿名"(错误)但不"非问题"(错误).它的数量是2,但出于错误的原因.

NSUInteger countOfNons = 0;
NSString * theFullTestString = @"nonsense non-issue anonymous controlWord";
NSError *error = nil;

NSRegularExpression *regex = [NSRegularExpression regularExpressionWithPattern:@"non(\\w+)" options:0 error:&error];

NSArray *matches = [regex matchesInString:theFullTestString options:0 range:NSMakeRange(0, theFullTestString.length)];

for (NSTextCheckingResult *match in matches) {
    NSRange wordRange = [match rangeAtIndex:1];
    NSString* word = [theFullTestString substringWithRange:wordRange];
    ++countOfNons;
    NSLog(@"Found word:%@  countOfNons:%d", word, countOfNons);
}

我很难过.

Answer 1

正则表达式\bnon[\w-]*应该做的伎俩

\bnon[\w-]*
^ (\b) Start of word
  ^ (non) Begins with non
     ^ ([\w-]) A alphanumeric char, or hyphen
          ^ (*) The character after 'non' zero or more times

所以,在你的情况下:

NSUInteger countOfNons = 0;
NSString * theFullTestString = @"nonsense non-issue anonymous controlWord";
NSError *error = nil;

NSRegularExpression *regex = [NSRegularExpression regularExpressionWithPattern:@"(\\bnon[\\w-]*)" options:0 error:&error];

NSArray *matches = [regex matchesInString:theFullTestString options:0 range:NSMakeRange(0, theFullTestString.length)];

for (NSTextCheckingResult *match in matches) {
    NSRange wordRange = [match rangeAtIndex:1];
    NSString* word = [theFullTestString substringWithRange:wordRange];
    ++countOfNons;
    NSLog(@"Found word:%@  countOfNons:%d", word, countOfNons);
}