有人知道如何在Swift中验证电子邮件地址吗?我找到了这个代码:
- (BOOL) validEmail:(NSString*) emailString {
if([emailString length]==0){
return NO;
}
NSString *regExPattern = @"[A-Z0-9a-z._%+-]+@[A-Za-z0-9.-]+\\.[A-Za-z]{2,4}";
NSRegularExpression *regEx = [[NSRegularExpression alloc] initWithPattern:regExPattern options:NSRegularExpressionCaseInsensitive error:nil];
NSUInteger regExMatches = [regEx numberOfMatchesInString:emailString options:0 range:NSMakeRange(0, [emailString length])];
NSLog(@"%i", regExMatches);
if (regExMatches == 0) {
return NO;
} else {
return YES;
}
}
但我无法翻译成斯威夫特。
更新答案@Arsonik对Swift 2.2的回答,使用比其他提供的解决方案更少的冗长代码:
extension String {
func isValidEmail() -> Bool {
let regex = try? NSRegularExpression(pattern: "^[a-zA-Z0-9.!#$%&'*+/=?^_`{|}~-]+@[a-zA-Z0-9](?:[a-zA-Z0-9-]{0,61}[a-zA-Z0-9])?(?:\\.[a-zA-Z0-9](?:[a-zA-Z0-9-]{0,61}[a-zA-Z0-9])?)*$", options: .CaseInsensitive)
return regex?.firstMatchInString(self, options: [], range: NSMakeRange(0, self.characters.count)) != nil
}
}
编辑,针对Swift 3更新:
func validateEmail(enteredEmail:String) -> Bool {
let emailFormat = "[A-Z0-9a-z._%+-]+@[A-Za-z0-9.-]+\\.[A-Za-z]{2,64}"
let emailPredicate = NSPredicate(format:"SELF MATCHES %@", emailFormat)
return emailPredicate.evaluate(with: enteredEmail)
}
Swift 2的原始答案:
func validateEmail(enteredEmail:String) -> Bool {
let emailFormat = "[A-Z0-9a-z._%+-]+@[A-Za-z0-9.-]+\\.[A-Za-z]{2,64}"
let emailPredicate = NSPredicate(format:"SELF MATCHES %@", emailFormat)
return emailPredicate.evaluateWithObject(enteredEmail)
}
它工作得很好。
这是Swift 2.0 - 2.2的更新版本
var isEmail: Bool {
do {
let regex = try NSRegularExpression(pattern: "^[a-zA-Z0-9.!#$%&'*+/=?^_`{|}~-]+@[a-zA-Z0-9](?:[a-zA-Z0-9-]{0,61}[a-zA-Z0-9])?(?:\\.[a-zA-Z0-9](?:[a-zA-Z0-9-]{0,61}[a-zA-Z0-9])?)*$", options: .CaseInsensitive)
return regex.firstMatchInString(self, options: NSMatchingOptions(rawValue: 0), range: NSMakeRange(0, self.characters.count)) != nil
} catch {
return false
}
}
下面是一个基于rangeOfString的方法:
class func isValidEmail(testStr:String) -> Bool {
let emailRegEx = "[A-Z0-9a-z._%+-]+@[A-Za-z0-9.-]+\\.[A-Za-z]{2,64}"
let range = testStr.rangeOfString(emailRegEx, options:.RegularExpressionSearch)
return range != nil
}
备注:更新TLD长度。
下面是符合RFC 5322的电子邮件的最终RegEx,请注意,最好不要使用它,因为它只检查电子邮件地址的基本语法,而不检查顶级域是否存在。
(?:[a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:\.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*
| "(?:[\x01-\x08\x0b\x0c\x0e-\x1f\x21\x23-\x5b\x5d-\x7f]
| \\[\x01-\x09\x0b\x0c\x0e-\x7f])*")
@ (?:(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?\.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?
| \[(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}
(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?|[a-z0-9-]*[a-z0-9]:
(?:[\x01-\x08\x0b\x0c\x0e-\x1f\x21-\x5a\x53-\x7f]
| \\[\x01-\x09\x0b\x0c\x0e-\x7f])+)
\])
有关电子邮件regex的更完整信息,请参阅Regular-Expressions.info。
注意,在Objective-C或Swift等语言中没有转义。
@JeffersonBe的答案是接近的,但如果字符串是“包含someone@something.com一个有效的电子邮件”,这不是我们想要的。下面是String上的一个扩展,它工作得很好(并且允许测试有效的phoneNumber和其他数据检测器来引导。
/// Helper for various data detector matches.
/// Returns `true` iff the `String` matches the data detector type for the complete string.
func matchesDataDetector(type: NSTextCheckingResult.CheckingType, scheme: String? = nil) -> Bool {
let dataDetector = try? NSDataDetector(types: type.rawValue)
guard let firstMatch = dataDetector?.firstMatch(in: self, options: NSRegularExpression.MatchingOptions.reportCompletion, range: NSRange(location: 0, length: length)) else {
return false
}
return firstMatch.range.location != NSNotFound
// make sure the entire string is an email, not just contains an email
&& firstMatch.range.location == 0
&& firstMatch.range.length == length
// make sure the link type matches if link scheme
&& (type != .link || scheme == nil || firstMatch.url?.scheme == scheme)
}
/// `true` iff the `String` is an email address in the proper form.
var isEmail: Bool {
return matchesDataDetector(type: .link, scheme: "mailto")
}
/// `true` iff the `String` is a phone number in the proper form.
var isPhoneNumber: Bool {
return matchesDataDetector(type: .phoneNumber)
}
/// number of characters in the `String` (required for above).
var length: Int {
return self.characters.count
}