ios - NSRegularExpression 搜索字符串 "*string*"

Question

我有一个字符串列表，我想搜索像 A 这样的字符串，这意味着它只出现在中间，而不是第一个或最后一个。我的代码在这里:当我运行所有 3 个案例时，但只有案例“う上あ”为真

- (void)viewDidLoad {
    [super viewDidLoad];
    // Do any additional setup after loading the view from its nib.
    NSArray *array = @[@"上あう", @"う上あ", @"あう上"];
    NSString *stringNeedToFind = @"*上*";
    for (NSString *string in array) {
        if ([[self regularExpressionFromValue: stringNeedToFind] numberOfMatchesInString:string options: 0 range: NSMakeRange(0, string.length)]) {
            NSLog(@"String containt.");
        }
    }
}

- (NSRegularExpression *) regularExpressionFromValue: (NSString *) value
{
    NSString *pattern = [NSRegularExpression escapedPatternForString: value];
    pattern = [pattern stringByReplacingOccurrencesOfString: @"\\*" withString: @".*"];
    pattern = [pattern stringByReplacingOccurrencesOfString: @"\\?" withString: @"."];
    pattern = [pattern stringByReplacingOccurrencesOfString: @".*.*.*" withString: @"\\*"];
    pattern = [pattern stringByReplacingOccurrencesOfString: @"..." withString: @"\\?\\?\\?"];
    pattern = [NSString stringWithFormat: @"^%@$", pattern];
    return [NSRegularExpression regularExpressionWithPattern: pattern options: 0 error: NULL];
}

Answer 1

据我所知，星号 (*) 用作运算符来检查字符(集)是否出现 0 次或多次。我认为您应该使用加号 (+)，例如像这样:^.+上.+$，意思是开头和结尾应该有1个或多个字符。