从字符串中提取 5 位邮政编码

Question

我让用户输入他们的地址，我需要从中提取邮政编码。

我发现这个正则表达式应该可以工作：d{5}([ -]d{4})?但是我很难让它在 Swift 上工作。

这就是我所在的地方：

private func sanatizeZipCodeString() -> String {
        let retVal = self.drugNameTextField.text
        let regEx = try! NSRegularExpression(pattern: "", options: .CaseInsensitive)
        let match = regEx.matchesInString(retVal!, options: [], range: NSMakeRange(0, (retVal?.characters.count)!))
        for zip in match {
            let matchRange = zip.range
        }
    }

我不明白为什么我不能把第一根匹配的绳子拉出来！

Answer 1

你可以试试这个

func match() {
    do {
        let regex = try NSRegularExpression(pattern: "\b\d{5}(?:[ -]\d{4})?\b", options: [])
        let retVal = "75463 72639823764 gfejwfh56873 89765"
        let str = retVal as NSString
        let postcodes = regex.matchesInString(retVal,
        options: [], range: NSMakeRange(0, retVal.characters.count))
        let postcodesArr =  postcodes.map { str.substringWithRange($0.range)}
        // postcodesArr[0] will give you first postcode
    } catch let error as NSError {
    }
}   

Answer 2

您可以使用

"\b\d{5}(?:[ -]\d{4})?\b"

单词

边界确保仅匹配整个单词 ZIP。

反斜杠必须加倍。

字符类末尾的连字符不必转义。

要使用它：

func regMatchGroup(regex: String, text: String) -> [[String]] {
do {
    var resultsFinal = [[String]]()
    let regex = try NSRegularExpression(pattern: regex, options: [])
    let nsString = text as NSString
    let results = regex.matchesInString(text,
        options: [], range: NSMakeRange(0, nsString.length))
    for result in results {
        var internalString = [String]()
        for var i = 0; i < result.numberOfRanges; ++i{
            internalString.append(nsString.substringWithRange(result.rangeAtIndex(i)))
        }
        resultsFinal.append(internalString)
    }
    return resultsFinal
   } catch let error as NSError {
       print("invalid regex: (error.localizedDescription)")
       return [[]]
   }
}
let input = "75463 72639823764 gfejwfh56873 89765"
let matches = regMatchGroup("\b\d{5}(?:[ -]\d{4})?\b", text: input)
if (matches.count > 0) 
{ 
    print(matches[0][0]) // Print the first one
}