Question

我需要在两个＆＃34;％＆＃34;之间提取字符串。字符，多次出现可以出现在查询字符串中。现在使用以下正则表达式，有人可以帮助获得确切的Regax格式。

let query =  "Hello %test% ho do you do %test1%"
let regex = try! NSRegularExpression(pattern:"%(.*?)%", options: [])

  if let results = regex?.matchesInString(query, options: .Anchored,  range: NSMakeRange(0,query.characters.count)){
    for match in results{
         }
      }

Answer 1

您的模式很好，但您的代码无法编译。试试这个：

Swift 4

let query = "Hello %test% how do you do %test1%"
let regex = try! NSRegularExpression(pattern:"%(.*?)%", options: [])
var results = [String]()

regex.enumerateMatches(in: query, options: [], range: NSMakeRange(0, query.utf16.count)) { result, flags, stop in
    if let r = result?.range(at: 1), let range = Range(r, in: query) {
        results.append(String(query[range]))
    }
}

print(results) // ["test", "test1"]

NSString使用UTF-16编码，因此使用UTF-16代码单元的数量调用NSMakeRange。

Swift 2

let query = "Hello %test% how do you do %test1%"
let regex = try! NSRegularExpression(pattern:"%(.*?)%", options: [])
let tmp = query as NSString
var results = [String]()

regex.enumerateMatchesInString(query, options: [], range: NSMakeRange(0, tmp.length)) { result, flags, stop in
    if let range = result?.rangeAtIndex(1) {
        results.append(tmp.substringWithRange(range))
    }
}

print(results) // ["test", "test1"]

从Swift的原生String类型中获取子字符串有点麻烦。这就是我将query投放到NSString

的原因

Answer 2

我写了一个常规快递的方法。你的正则表达没问题。您可以测试正则表达式here。方法是：

 func regexInText(regex: String!, text: String!) -> [String] {

        do {
            let regex = try NSRegularExpression(pattern: regex, options: [])
            let nsString = text as NSString
            let results = regex.matchesInString(text,
                options: [], range: NSMakeRange(0, nsString.length))
            return results.map { nsString.substringWithRange($0.range)}
        } catch let error as NSError {
            print("invalid regex: \(error.localizedDescription)")
            return []
        }
    }

您可以随时调用它。

正则表达式获取两个％字符之间的字符串

2 个答案:

Swift 4

Swift 2