如何验证一个电子邮件地址在swift?

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。

当前回答

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
    }
}

2016-08-09 21:26:31

其他回答

我唯一添加到响应列表的是，对于Linux, NSRegularExpression不存在，它实际上是RegularExpression

    func isEmail() -> Bool {

    let patternNormal = "[A-Z0-9a-z._%+-]+@[A-Za-z0-9.-]+\\.[A-Za-z]{2,6}"

    #if os(Linux)
        let regex = try? RegularExpression(pattern: patternNormal, options: .caseInsensitive)
    #else
        let regex = try? NSRegularExpression(pattern: patternNormal, options: .caseInsensitive)
    #endif

    return regex?.firstMatch(in: self, options: [], range: NSMakeRange(0, self.characters.count)) != nil

在macOS和Ubuntu上都能成功编译。

2016-10-24 20:15:07

这是@Fattie的“THE REASONABLE SOLUTION”的新版本，在Swift 4.1的一个名为String+Email.swift的新文件中测试:

import Foundation

extension String {
    private static let __firstpart = "[A-Z0-9a-z]([A-Z0-9a-z._%+-]{0,30}[A-Z0-9a-z])?"
    private static let __serverpart = "([A-Z0-9a-z]([A-Z0-9a-z-]{0,30}[A-Z0-9a-z])?\\.){1,5}"
    private static let __emailRegex = __firstpart + "@" + __serverpart + "[A-Za-z]{2,6}"

    public var isEmail: Bool {
        let predicate = NSPredicate(format: "SELF MATCHES %@", type(of:self).__emailRegex)
        return predicate.evaluate(with: self)
    }
}

所以它的用法很简单:

let str = "mail@domain.com"
if str.isEmail {
    print("\(str) is a valid e-mail address")
} else {
    print("\(str) is not a valid e-mail address")
}

我只是不喜欢向String对象添加func，因为作为电子邮件地址是它们固有的(或不是)。根据我的理解，Bool属性比func更适合。

2018-05-15 17:52:07

这是一个最新的游乐场兼容版本，使用标准库，所以你不需要维护一个正则表达式:

import Foundation

func isValid(email: String) -> Bool {
  do {
    let detector = try NSDataDetector(types: NSTextCheckingResult.CheckingType.link.rawValue)
    let range = NSRange(location: 0, length: email.count)
    let matches = detector.matches(in: email, options: .anchored, range: range)
    guard matches.count == 1 else { return false }
    return matches[0].url?.scheme == "mailto"
  } catch {
    return false
  }
}

extension String {
  var isValidEmail: Bool {
    isValid(email: self)
  }
}

let email = "test@mail.com"
isValid(email: email) // prints 'true'
email.isValidEmail // prints 'true'

2021-11-17 11:47:09

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
    }
}

2016-08-09 21:26:31

对于Swift 3:

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?.firstMatch(in: self, options: [], range: NSMakeRange(0, self.characters.count)) != nil
    }
}

2016-10-14 14:06:07

如何验证一个电子邮件地址在swift?

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