我必须检测字符串是否包含任何特殊字符。我怎样才能检查它？ Swift 支持正则表达式吗？
var characterSet:NSCharacterSet = NSCharacterSet(charactersInString: "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789")
if (searchTerm!.rangeOfCharacterFromSet(characterSet).location == NSNotFound){
println("Could not handle special characters")
}
我尝试了上面的代码，但只有当我输入第一个字符作为特殊字符时它才会匹配。
您的代码检查字符串中是否没有来自给定集合的字符。 你想要的是检查给定集合中的any字符是否not：
if (searchTerm!.rangeOfCharacterFromSet(characterSet.invertedSet).location != NSNotFound){
println("Could not handle special characters")
}
您还可以使用正则表达式来实现此目的：
let regex = NSRegularExpression(pattern: ".*[^A-Za-z0-9].*", options: nil, error: nil)!
if regex.firstMatchInString(searchTerm!, options: nil, range: NSMakeRange(0, searchTerm!.length)) != nil {
println("could not handle special characters")
}
模式
[^A-Za-z0-9]Swift 2 更新：
let searchTerm = "a+b"
let characterset = NSCharacterSet(charactersInString: "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789")
if searchTerm.rangeOfCharacterFromSet(characterset.invertedSet) != nil {
print("string contains special characters")
}
Swift 3 更新：
let characterset = CharacterSet(charactersIn: "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789")
if searchTerm.rangeOfCharacter(from: characterset.inverted) != nil {
print("string contains special characters")
}
使用 Swift 5 你就可以做到
if let hasSpecialCharacters = "your string".range(of: ".*[^A-Za-z0-9].*", options: .regularExpression) != nil {}
这个答案可能会对使用 Swift 4.1 的人有所帮助
func hasSpecialCharacters() -> Bool {
do {
let regex = try NSRegularExpression(pattern: ".*[^A-Za-z0-9].*", options: .caseInsensitive)
if let _ = regex.firstMatch(in: self, options: NSRegularExpression.MatchingOptions.reportCompletion, range: NSMakeRange(0, self.count)) {
return true
}
} catch {
debugPrint(error.localizedDescription)
return false
}
return false
}
参考@Martin R的回答。
反转你的字符集将会起作用，因为在你的字符集中你拥有所有有效的字符：
var characterSet:NSCharacterSet = NSCharacterSet(charactersInString: "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789")
if (searchTerm!.rangeOfCharacterFromSet(characterSet.invertedSet).location == NSNotFound){
println("No special characters")
}
希望这有帮助..:)
密码验证如下：-（密码至少八个字符长，一个特殊字符，一个大写字母，一个小写字母和一个数字）
var isValidateSecialPassword : Bool {
if(self.count>=8 && self.count<=20){
}else{
return false
}
let nonUpperCase = CharacterSet(charactersIn: "ABCDEFGHIJKLMNOPQRSTUVWXYZ").inverted
let letters = self.components(separatedBy: nonUpperCase)
let strUpper: String = letters.joined()
let smallLetterRegEx = ".*[a-z]+.*"
let samlltest = NSPredicate(format:"SELF MATCHES %@", smallLetterRegEx)
let smallresult = samlltest.evaluate(with: self)
let numberRegEx = ".*[0-9]+.*"
let numbertest = NSPredicate(format:"SELF MATCHES %@", numberRegEx)
let numberresult = numbertest.evaluate(with: self)
let regex = try! NSRegularExpression(pattern: ".*[^A-Za-z0-9].*", options: NSRegularExpression.Options())
var isSpecial :Bool = false
if regex.firstMatch(in: self, options: NSRegularExpression.MatchingOptions(), range:NSMakeRange(0, self.count)) != nil {
print("could not handle special characters")
isSpecial = true
}else{
isSpecial = false
}
return (strUpper.count >= 1) && smallresult && numberresult && isSpecial
}
@Martin R 的回答很棒，我只是想将其（第二部分）更新到 Swift 2.1 版本
let regex = try! NSRegularExpression(pattern: ".*[^A-Za-z0-9].*", options: NSRegularExpressionOptions())
if regex.firstMatchInString(searchTerm!, options: NSMatchingOptions(), range:NSMakeRange(0, searchTerm!.characters.count)) != nil {
print("could not handle special characters")
}
我使用了
try!在 Swift 5 及更高版本中你可以这样做
func validateGenericString(_ string: String) -> Bool {
return string.range(of: ".*[^A-Za-z0-9].*", options: .regularExpression) == nil
}
检查字符串是否有特殊字符
出于文件名清理的目的，我更喜欢检测无效字符，而不是提供允许的字符集。毕竟，许多非英语用户需要重音字符。以下函数的灵感来自于这个要点：
func checkForIllegalCharacters(string: String) -> Bool {
let invalidCharacters = CharacterSet(charactersIn: "\\/:*?\"<>|")
.union(.newlines)
.union(.illegalCharacters)
.union(.controlCharacters)
if string.rangeOfCharacter(from: invalidCharacters) != nil {
print ("Illegal characters detected in file name")
// Raise an alert here
return true
} else {
return false
}
根据特殊字符的定义，您可以使用：
let chars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
chars.canBeConvertedToEncoding(NSASCIIStringEncoding)
两种解决方案：
1)
extension String {
var stripped: String {
let okayChars = Set("abcdefghijklmnopqrstuvwxyz ABCDEFGHIJKLMNOPQRSTUVWXYZ")
return self.filter {okayChars.contains($0) }
}
}
2)
class TrimDictionary {
static func trimmedWord(wordString: String) -> String {
var selectedString = wordString
let strFirst = selectedString.first
let strLast = selectedString.last
let characterset = CharacterSet(charactersIn: "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ")
if strFirst?.description.rangeOfCharacter(from: characterset.inverted) != nil {
selectedString = String(selectedString.dropFirst())
}
if strLast?.description.rangeOfCharacter(from: characterset.inverted) != nil {
selectedString = String(selectedString.dropLast())
}
return selectedString
}
}
Mahendra 的答案可以通过在正则表达式子句中使用反转（^）来简化一点。此外，在使用 caseInsensitive 选项时，您不需要 A-Z 和 a-z，因为 Swift 会为您解决这种可能性：
extension String {
func containsSpecialCharacters(string: String) -> Bool {
do {
let regex = try NSRegularExpression(pattern: "[^a-z0-9 ]", options: .caseInsensitive)
if let _ = regex.firstMatch(in: string, options: [], range: NSMakeRange(0, string.count)) {
return true
} else {
return false
}
} catch {
debugPrint(error.localizedDescription)
return true
}
}我是韩国人。 （不是英文）
/// 다음 문자를 포함하는지 여부
/// ! @ # $ & ( ) - ‘ . / + , “ ^ 공백
var hasSpecialChar: Bool {
if self.isEmpty {
return false
}
do {
let pattern = "[*\\^\\s!@#$&()\\-`.+,/\"]+"
let regex = try NSRegularExpression(pattern: pattern, options: .caseInsensitive)
return regex.matches(in: self, options: [], range: NSRange(location: 0, length: self.count)).count > 0
} catch {
return false
}
}