更改属性字符串的文本并保留 Swift 中的属性

Question

对于数据库程序中的输出，我有一些文本，我插入了标记以指示粗体或斜体，以及一些替换图像的文本。例如：

"%Important% ^All employees to the breakroom^" 最终输出应为：

重要 全体员工到休息室

我编写了代码来查找周围带有“%”符号和“^”符号的文本，但我现在遇到的问题是文本输出如下：

%重要%^全体员工到休息室^

我想删除这些 % 和 ^，同时保留字符串的格式。

这是我一直用到它失效的代码：

func processText(inString string: String) -> NSAttributedString {

let pattern = ["(?<=\^).*?(?=\^)","(?<=\%).*?(?=\%)","\^", "\%"]
let italicsRegex = NSRegularExpression(pattern: pattern[0], options: .allZeros, error: nil)
let range = NSMakeRange(0, count(string))
let italicsMatches = italicsRegex?.matchesInString(string, options: .allZeros, range: range) as? [NSTextCheckingResult]

var attributedText = NSMutableAttributedString(string: string)

for match in italicsMatches! {
    attributedText.addAttribute(NSFontAttributeName, value: UIFont(name: "Helvetica-Oblique", size: 14.0)!, range: match.range)
}

let boldRegex = NSRegularExpression(pattern: pattern[1], options: .allZeros, error: nil)
let boldMatches = boldRegex?.matchesInString(string, options: .allZeros, range: range) as? [NSTextCheckingResult]

for match in boldMatches!   {
    attributedText.addAttribute(NSFontAttributeName, value: UIFont(name: "Helvetica-Bold", size: 14.0)!, range: match.range)
}

let removeItalicsMarksRegex = NSRegularExpression(pattern: pattern[2], options: .allZeros, error: nil)
let removeItalicsMarksMatches = removeItalicsMarksRegex?.matchesInString(string, options: .allZeros, range: range) as? [NSTextCheckingResult]

var numberOfLoops = 0
for match in removeItalicsMarksMatches! {

    attributedText.replaceCharactersInRange(match.range, withString: "")

}

return attributedText.copy() as! NSAttributedString
}

这适用于 % 匹配（但仅适用于第一个字符）并立即导致 ^ 字符崩溃。

对于解决此问题的任何帮助或建议，我们将不胜感激。谢谢

Answer 1

这是一个可能的解决方案，本质上是对 how to catch multiple instances special indicated **characters** in an NSString and bold them in between? 从 Objective-C 到 Swift.

想法是在一个循环中添加属性和删除定界符。 shift 在删除第一个分隔符后，需要变量来调整匹配范围。

为简单起见，仅显示“^...^”处理。

func processText(inString string: String) -> NSAttributedString {

    let pattern = "(\^)(.*?)(\^)"

    let regex = NSRegularExpression(pattern: pattern, options: nil, error: nil)!
    var shift = 0 // number of characters removed so far
    let attributedText = NSMutableAttributedString(string: string)
    regex.enumerateMatchesInString(string, options: nil, range: NSMakeRange(0, count(string.utf16))) {
        (result, _, _) -> Void in
        var r1 = result.rangeAtIndex(1) // Location of the leading delimiter
        var r2 = result.rangeAtIndex(2) // Location of the string between the delimiters
        var r3 = result.rangeAtIndex(3) // Location of the trailing delimiter
        // Adjust locations according to the string modifications:
        r1.location -= shift
        r2.location -= shift
        r3.location -= shift
        // Set attribute for string between delimiters:
        attributedText.addAttribute(NSFontAttributeName, value: UIFont(name: "Helvetica-Oblique", size: 14.0)!, range: r2)
        // Remove leading and trailing delimiters:
        attributedText.mutableString.deleteCharactersInRange(r3)
        attributedText.mutableString.deleteCharactersInRange(r1)
        // Update offset:
        shift += r1.length + r3.length
    }

    return attributedText.copy() as! NSAttributedString
}

请注意 enumerateMatchesInString() 需要一个 NSRange，因此您必须计算 UTF-16 字符的数量而不是 Swift 个字符的数量。

示例：

let text = "aaa ^bbb^ eee"
let attrText = processText(inString: text)
println(attrText)

输出：

aaa {
}bbb{
    NSFont = " font-family: \"Helvetica-Oblique\"; font-weight: normal; font-style: italic; font-size: 14.00pt";
} eee{
}

Answer 2

马丁，

我最终使用了非常相似的东西，但我决定更改正则表达式以包含 ^ 标记。在这样做的过程中，我能够使用 "replaceCharactersInRange" 方法剪辑包含的属性子字符串的第一个和最后一个字符。到目前为止，这对我的目的来说效果更好一些，因为它是从属性字符串开始工作的，所以它不会搞砸或删除它的任何属性。

我附上了正则表达式和处理斜体的代码部分以供任何人将来参考（再次感谢！）：

func processText(inString string: String) -> NSAttributedString {

let pattern = ["\^.*?\^"] //Presented as an array here because in the full code there are a lot of patterns that are run.




let italicsRegex = NSRegularExpression(pattern: pattern[0], options: .allZeros, error: nil) 

//In addition to building the match for this first regular expression, I also gather build the regular expressions and gather matches for all other matching patterns on the initial string ("string") before I start doing any processing.

    let range = NSMakeRange(0, count(string.utf16))

let italicsMatches = italicsRegex?.matchesInString(string, options: .allZeros, range: range) as? [NSTextCheckingResult]

var attributedText = NSMutableAttributedString(string: string)

var charactersRemovedFromString = 0

for match in italicsMatches! {

    let newRange = NSMakeRange(match.range.location - charactersRemovedFromString, match.range.length) // Take the updated range for when this loop iterates, otherwise this crashes.
    attributedText.addAttribute(NSFontAttributeName, value: UIFont(name: "Helvetica-Oblique", size: 12.0)!, range: newRange)

    let rangeOfFirstCharacter = NSMakeRange(match.range.location - charactersRemovedFromString, 1)

    attributedText.replaceCharactersInRange(rangeOfFirstCharacter, withString: "")

    charactersRemovedFromString += 2

    let rangeOfLastCharacter = NSMakeRange(match.range.location + match.range.length - charactersRemovedFromString, 1)

    attributedText.replaceCharactersInRange(rangeOfLastCharacter, withString: "")
    }

return attributedText
}

Answer 3

这对我有用！

extension UILabel {
    func updateAttributedText(_ text: String) {
        if let attributedText = attributedText {
            let mutableAttributedText = NSMutableAttributedString(attributedString: attributedText)
            mutableAttributedText.mutableString.setString(text)
            self.attributedText = mutableAttributedText
        }
    }
}

更改属性字符串的文本并保留 Swift 中的属性

Change text of an attributed string and retain attributes in Swift

string

string-substitution

nsmutablestring

swift