格式化Swift中的电话号码

Question

一旦用户开始在这种格式类型中输入电话号码,我就会格式化我的文本文本,0 (555) 444 66 77并且它工作正常但是一旦我从服务器获得号码,我就这样得到它05554446677所以请你告诉我如何编辑它一旦我从服务器上获取相同的格式？

我开始输入后的代码:

func textField(textField: UITextField, shouldChangeCharactersInRange range: NSRange, replacementString string: String) -> Bool {

    if textField == phoneNumberTextField{
        var newString = (textField.text as NSString).stringByReplacingCharactersInRange(range, withString: string)
        var components = newString.componentsSeparatedByCharactersInSet(NSCharacterSet.decimalDigitCharacterSet().invertedSet)

        var decimalString = "".join(components) as NSString
        var length = decimalString.length
        var hasLeadingOne = length > 0 && decimalString.characterAtIndex(0) == (1 as unichar)

        if length == 0 || (length > 11 && !hasLeadingOne) || length > 12{
            var newLength = (textField.text as NSString).length + (string as NSString).length - range.length as Int

            return (newLength > 11) ? false : true
        }
        var index = 0 as Int
        var formattedString = NSMutableString()

        if hasLeadingOne{
            formattedString.appendString("1 ")
            index += 1
        }

        if (length - index) > 1{
            var zeroNumber = decimalString.substringWithRange(NSMakeRange(index, 1))
            formattedString.appendFormat("%@ ", zeroNumber)
            index += 1
        }
        if (length - index) > 3{
            var areaCode = decimalString.substringWithRange(NSMakeRange(index, 3))
            formattedString.appendFormat("(%@) ", areaCode)
            index += 3
        }
        if (length - index) > 3{
            var prefix = decimalString.substringWithRange(NSMakeRange(index, 3))
            formattedString.appendFormat("%@ ", prefix)
            index += 3
        }
        if (length - index) > 3{
            var prefix = decimalString.substringWithRange(NSMakeRange(index, 2))
            formattedString.appendFormat("%@ ", prefix)
            index += 2
        }

        var remainder = decimalString.substringFromIndex(index)
        formattedString.appendString(remainder)
        textField.text = formattedString as String
        return false
    }else{
        return true
    }
}

Answer 1

斯威夫特4

func formattedNumber(number: String) -> String {
    let cleanPhoneNumber = number.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()
    let mask = "+X (XXX) XXX-XXXX"

    var result = ""
    var index = cleanPhoneNumber.startIndex
    for ch in mask where index < cleanPhoneNumber.endIndex {
        if ch == "X" {
            result.append(cleanPhoneNumber[index])
            index = cleanPhoneNumber.index(after: index)
        } else {
            result.append(ch)
        }
    }
    return result
}

但是如果字符串字符大于或等于11,则它不起作用

蒙面数字打字

func textField(_ textField: UITextField, shouldChangeCharactersIn range: NSRange, replacementString string: String) -> Bool {
    guard let text = textField.text else { return false }
    let newString = (text as NSString).replacingCharacters(in: range, with: string)
    textField.text = formattedNumber(number: newString)
    return false
}

从UITextField委托方法调用上面的函数:

"" => ""
"0" => "+0"
"412" => "+4 (12"
"12345678901" => "+1 (234) 567-8901"
"a1_b2-c3=d4 e5&f6|g7?h8" => "+1 (234) 567-8"

所以,这样做会更好.

func formattedNumber(number: String) -> String {
    let cleanPhoneNumber = number.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()
    let mask = "+X (XXX) XXX-XXXX"

    var result = ""
    var index = cleanPhoneNumber.startIndex
    for ch in mask where index < cleanPhoneNumber.endIndex {
        if ch == "X" {
            result.append(cleanPhoneNumber[index])
            index = cleanPhoneNumber.index(after: index)
        } else {
            result.append(ch)
        }
    }
    return result
}

Answer 2

斯威夫特3和4

此解决方案在应用格式之前删除所有非数字字符.nil如果无法根据假设格式化源电话号码,则返回.

斯威夫特4

Swift 4解决方案考虑了CharacterView的弃用和Sting成为CharacterView的字符集合.

import Foundation

func format(phoneNumber sourcePhoneNumber: String) -> String? {
    // Remove any character that is not a number
    let numbersOnly = sourcePhoneNumber.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()
    let length = numbersOnly.count
    let hasLeadingOne = numbersOnly.hasPrefix("1")

    // Check for supported phone number length
    guard length == 7 || (length == 10 && !hasLeadingOne) || (length == 11 && hasLeadingOne) else {
        return nil
    }

    let hasAreaCode = (length >= 10)
    var sourceIndex = 0

    // Leading 1
    var leadingOne = ""
    if hasLeadingOne {
        leadingOne = "1 "
        sourceIndex += 1
    }

    // Area code
    var areaCode = ""
    if hasAreaCode {
        let areaCodeLength = 3
        guard let areaCodeSubstring = numbersOnly.substring(start: sourceIndex, offsetBy: areaCodeLength) else {
            return nil
        }
        areaCode = String(format: "(%@) ", areaCodeSubstring)
        sourceIndex += areaCodeLength
    }

    // Prefix, 3 characters
    let prefixLength = 3
    guard let prefix = numbersOnly.substring(start: sourceIndex, offsetBy: prefixLength) else {
        return nil
    }
    sourceIndex += prefixLength

    // Suffix, 4 characters
    let suffixLength = 4
    guard let suffix = numbersOnly.substring(start: sourceIndex, offsetBy: suffixLength) else {
        return nil
    }

    return leadingOne + areaCode + prefix + "-" + suffix
}

extension String {
    /// This method makes it easier extract a substring by character index where a character is viewed as a human-readable character (grapheme cluster).
    internal func substring(start: Int, offsetBy: Int) -> String? {
        guard let substringStartIndex = self.index(startIndex, offsetBy: start, limitedBy: endIndex) else {
            return nil
        }

        guard let substringEndIndex = self.index(startIndex, offsetBy: start + offsetBy, limitedBy: endIndex) else {
            return nil
        }

        return String(self[substringStartIndex ..< substringEndIndex])
    }
}

斯威夫特3

import Foundation

func format(phoneNumber sourcePhoneNumber: String) -> String? {

    // Remove any character that is not a number
    let numbersOnly = sourcePhoneNumber.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()
    let length = numbersOnly.characters.count
    let hasLeadingOne = numbersOnly.hasPrefix("1")

    // Check for supported phone number length
    guard length == 7 || (length == 10 && !hasLeadingOne) || (length == 11 && hasLeadingOne) else {
        return nil
    }

    let hasAreaCode = (length >= 10)
    var sourceIndex = 0

    // Leading 1
    var leadingOne = ""
    if hasLeadingOne {
        leadingOne = "1 "
        sourceIndex += 1
    }

    // Area code
    var areaCode = ""
    if hasAreaCode {
        let areaCodeLength = 3
        guard let areaCodeSubstring = numbersOnly.characters.substring(start: sourceIndex, offsetBy: areaCodeLength) else {
            return nil
        }
        areaCode = String(format: "(%@) ", areaCodeSubstring)
        sourceIndex += areaCodeLength
    }

    // Prefix, 3 characters
    let prefixLength = 3
    guard let prefix = numbersOnly.characters.substring(start: sourceIndex, offsetBy: prefixLength) else {
        return nil
    }
    sourceIndex += prefixLength

    // Suffix, 4 characters
    let suffixLength = 4
    guard let suffix = numbersOnly.characters.substring(start: sourceIndex, offsetBy: suffixLength) else {
        return nil
    }

    return leadingOne + areaCode + prefix + "-" + suffix
}

extension String.CharacterView {
    /// This method makes it easier extract a substring by character index where a character is viewed as a human-readable character (grapheme cluster).
    internal func substring(start: Int, offsetBy: Int) -> String? {
        guard let substringStartIndex = self.index(startIndex, offsetBy: start, limitedBy: endIndex) else {
            return nil
        }

        guard let substringEndIndex = self.index(startIndex, offsetBy: start + offsetBy, limitedBy: endIndex) else {
            return nil
        }

        return String(self[substringStartIndex ..< substringEndIndex])
    }
}

例

func testFormat(sourcePhoneNumber: String) -> String {
    if let formattedPhoneNumber = format(phoneNumber: sourcePhoneNumber) {
        return "'\(sourcePhoneNumber)' => '\(formattedPhoneNumber)'"
    }
    else {
        return "'\(sourcePhoneNumber)' => nil"
    }
}

print(testFormat(sourcePhoneNumber: "1 800 222 3333"))
print(testFormat(sourcePhoneNumber: "18002223333"))
print(testFormat(sourcePhoneNumber: "8002223333"))
print(testFormat(sourcePhoneNumber: "2223333"))
print(testFormat(sourcePhoneNumber: "18002223333444"))
print(testFormat(sourcePhoneNumber: "Letters8002223333"))
print(testFormat(sourcePhoneNumber: "1112223333"))

示例输出

'1 800 222 3333' => '1 (800) 222-3333'

'18002223333' => '1 (800) 222-3333'

'8002223333' => '(800) 222-3333'

'2223333' => '222-3333'

'18002223333444' => nil

'Letters8002223333' => '(800) 222-3333'

'1112223333' => nil

Answer 3

真正简单的解决方案：

extension String {
    func applyPatternOnNumbers(pattern: String, replacmentCharacter: Character) -> String {
        var pureNumber = self.replacingOccurrences( of: "[^0-9]", with: "", options: .regularExpression)
        for index in 0 ..< pattern.count {
            guard index < pureNumber.count else { return pureNumber }
            let stringIndex = String.Index(encodedOffset: index)
            let patternCharacter = pattern[stringIndex]
            guard patternCharacter != replacmentCharacter else { continue }
            pureNumber.insert(patternCharacter, at: stringIndex)
        }
        return pureNumber
    }
}

用法：

guard let text = textField.text else { return }
textField.text = text.applyPatternOnNumbers(pattern: "+# (###) ###-####", replacmentCharacter: "#")

Answer 4

在String中使用字符进行操作并不是非常简单.你需要以下:

斯威夫特2.1

let s = "05554446677"
let s2 = String(format: "%@ (%@) %@ %@ %@", s.substringToIndex(s.startIndex.advancedBy(1)),
    s.substringWithRange(s.startIndex.advancedBy(1) ... s.startIndex.advancedBy(3)),
    s.substringWithRange(s.startIndex.advancedBy(4) ... s.startIndex.advancedBy(6)),
    s.substringWithRange(s.startIndex.advancedBy(7) ... s.startIndex.advancedBy(8)),
    s.substringWithRange(s.startIndex.advancedBy(9) ... s.startIndex.advancedBy(10))
)

Swift 2.0

let s = "05554446677"
let s2 = String(format: "%@ (%@) %@ %@ %@", s.substringToIndex(advance(s.startIndex, 1)),
    s.substringWithRange(advance(s.startIndex, 1) ... advance(s.startIndex, 3)),
    s.substringWithRange(advance(s.startIndex, 4) ... advance(s.startIndex, 6)),
    s.substringWithRange(advance(s.startIndex, 7) ... advance(s.startIndex, 8)),
    s.substringWithRange(advance(s.startIndex, 9) ... advance(s.startIndex, 10))
)

代码将打印 0 (555) 444 66 77

Answer 5

你可以使用这个库https://github.com/luximetr/AnyFormatKit

例子

let phoneFormatter = DefaultTextFormatter(textPattern: "### (###) ###-##-##")
phoneFormatter.format("+123456789012") // +12 (345) 678-90-12

使用起来非常简单。

Answer 6

斯威夫特 4

创建此函数并调用文本字段事件 Editing Changed

private func formatPhone(_ number: String) -> String {
    let cleanNumber = number.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()
    let format: [Character] = ["X", "X", "X", "-", "X", "X", "X", "-", "X", "X", "X", "X"]

    var result = ""
    var index = cleanNumber.startIndex
    for ch in format {
        if index == cleanNumber.endIndex {
            break
        }
        if ch == "X" {
            result.append(cleanNumber[index])
            index = cleanNumber.index(after: index)
        } else {
            result.append(ch)
        }
    }
    return result
}

Answer 7

Swift 5.1 更新 ????? ？？？？？？？？？很好的解决方案

extension String {

    func applyPatternOnNumbers(pattern: String, replacmentCharacter: Character) -> String {
        var pureNumber = self.replacingOccurrences( of: "[^0-9]", with: "", options: .regularExpression)
        for index in 0 ..< pattern.count {
            guard index < pureNumber.count else { return pureNumber }
            let stringIndex = String.Index(utf16Offset: index, in: self)
            let patternCharacter = pattern[stringIndex]
            guard patternCharacter != replacmentCharacter else { continue }
            pureNumber.insert(patternCharacter, at: stringIndex)
        }
        return pureNumber
    }
}

用法：

let formattedText = text.applyPatternOnNumbers(pattern: "+# (###) ###-####", replacmentCharacter: "#")