在属性字符串中分隔符之间的快速粗体和斜体

Question

我需要将带有分隔符的字符串转换为属性字符串。 我的分隔符是$$斜体$$和$$$大胆$$$。实际上，只有当我没有使用斜体和粗体时，我的代码才是函数。我的麻烦是$$$$$大胆和斜体$$$$$。只设置了斜体...

这是我在UIViewcontroller类中的代码：

func pikipiki2AttributedString(text: String) -> NSAttributedString {
    let attributedText = NSMutableAttributedString(string: text)
    let size = [NSFontAttributeName : UIFont.systemFontOfSize(15.0)]
    attributedText.addAttributes(size, range: NSRange(location: 0, length: attributedText.length))

    let attributeBold = [NSFontAttributeName: UIFont.boldSystemFontOfSize(15)]
    try! attributedText.addAttributes(attributeBold, delimiter: "$$$")

    let attributeItalic = [NSFontAttributeName: UIFont.italicSystemFontOfSize(15)]
    try! attributedText.addAttributes(attributeItalic, delimiter: "$$")

    return attributedText
}

创建了here的扩展程序：

public extension NSMutableAttributedString {
    func addAttributes(attrs: [String : AnyObject], delimiter: String) throws {
        let escaped = NSRegularExpression.escapedPatternForString(delimiter)
        let regex = try NSRegularExpression(pattern:"\(escaped)(.*?)\(escaped)", options: [])

        var offset = 0
        regex.enumerateMatchesInString(string, options: [], range: NSRange(location: 0, length: string.characters.count)) { (result, flags, stop) -> Void in
            guard let result = result else {
                return
            }

            let range = NSRange(location: result.range.location + offset, length: result.range.length)enter code here
            self.addAttributes(attrs, range: range)
            let replacement = regex.replacementStringForResult(result, inString: self.string, offset: offset, template: "$1")
            self.replaceCharactersInRange(range, withString: replacement)
            offset -= (2 * delimiter.characters.count)
        }
    }
}