ios Swift 正则表达式：字符串是否与模式匹配？

Swift regex: does a string match a pattern?

提问by joelparkerhenderson

In Swift, what is a simple way to see if a string matches a pattern?

在 Swift 中，查看字符串是否与模式匹配的简单方法是什么？

Pseudocode examples:

伪代码示例：

if string matches pattern ...

if string =~ pattern ...

(I have read the Swift docs and haven't seen a regex capability. I've read about adding a new =~operator which is a good idea yet more complex than I'd like because this is for a teaching project. I have tried rangeOfStringbut get the error: 'String' does not have a member 'rangeOfString'. I am looking for a Swift solution, i.e. not typing NSRegularExpression. I do not need to do anything with the match result data.)

（我已经阅读了 Swift 文档，但还没有看到正则表达式功能。我读过关于添加一个新的=~运算符，这是一个好主意，但比我想要的更复杂，因为这是一个教学项目。我试过了，rangeOfString但是得到错误：'String' 没有成员 'rangeOfString'。我正在寻找一个 Swift 解决方案，即不输入 NSRegularExpression。我不需要对匹配结果数据做任何事情。）

回答by joelparkerhenderson

Swift version 3 solution:

Swift 版本 3 解决方案：

if string.range(of: regex, options: .regularExpression, range: nil, locale: nil) != nil ...

Swift version 2 solution:

Swift 版本 2 解决方案：

if string.rangeOfString(pattern, options: .RegularExpressionSearch) != nil ...

Example -- does this string contain two letter "o" characters?

示例 -- 此字符串是否包含两个字母“o”字符？

"hello world".rangeOfString("o.*o", options: .RegularExpressionSearch) != nil

Note: If you get the error message 'String' does not have a member 'rangeOfString', then add this before: import Foundation. This is because Foundation provides the NSString methods that are automatically bridged to the Swift String class.

注意：如果您收到错误消息'String' does not have a member 'rangeOfString'，请在此之前添加：import Foundation。这是因为 Foundation 提供了自动桥接到 Swift String 类的 NSString 方法。

import Foundation

Thanks to Onno Eberhard for the Swift 3 update.

感谢 Onno Eberhard 为 Swift 3 更新。

回答by Onno Eberhard

The solutions mentioned above didn't work for me anymore, so I'm posting my solution here (I used an extension for String):

上面提到的解决方案不再适合我，所以我在这里发布我的解决方案（我使用了 String 的扩展名）：

extension String {
    func matches(_ regex: String) -> Bool {
        return self.range(of: regex, options: .regularExpression, range: nil, locale: nil) != nil
    }
}

Example:

例子：

if str.matches("^[a-zA-Z0-9._-]{1,30}$") {
    //...
}

回答by David Berry

To get the syntax you actually ask about, you can easily define a new operator which wraps the bridged NSStringfunctionality:

要获得您实际询问的语法，您可以轻松定义一个包含桥接NSString功能的新运算符：

infix operator =~ {}
func =~(string:String, regex:String) -> Bool {
    return string.rangeOfString(regex, options: .RegularExpressionSearch) != nil
}

"abcd" =~ "ab*cd"
"abcd" =~ "abcde+"

回答by P J Szekszardi

In Swift 4.2the following doesn't return nil even if the string does not match the pattern:

在Swift 4.2 中，即使字符串与模式不匹配，以下内容也不会返回 nil：

string.range(of: regex, options: .regularExpression, range: nil, locale: nil)

As Xcode suggests:

正如 Xcode 所建议的：

Comparing non-optional value of type 'NSRange' (aka '_NSRange') to nil always returns true

将“NSRange”（又名“_NSRange”）类型的非可选值与 nil 进行比较总是返回 true

Therefore what I ended up using is:

因此，我最终使用的是：

if string.range(of: regex, options: .regularExpression)
        .location != NSNotFound {
    // do stuff if string matches regexp
}

回答by Gary Davies

This will return whether a string matches a pattern

这将返回字符串是否与模式匹配

eg. matchesPattern(text: "sd56Yz", pattern: "[a-zA-Z0-9]+") true

例如。匹配模式（文本：“sd56Yz”，模式：“[a-zA-Z0-9]+”）真

eg. matchesPattern(text: "sd56$Yz", pattern: "[a-zA-Z0-9]+") false

例如。匹配模式（文本：“sd56$Yz”，模式：“[a-zA-Z0-9]+”）假

func matchesPattern(text: String?, pattern: String?) -> Bool {
    guard hasValue(text) && hasValue(pattern) else {
        return false
    }

    let range = text!.rangeOfString(pattern!, options: .RegularExpressionSearch)

    guard range != nil else {
        return false
    }

    let rangeString = text?.substringWithRange(range!)

    return areEqual(text, string2: rangeString)
}