Question

我必须解析这些类型的字符串：

   public void showMessage(string message)
    {
        setMessage(message);
        Debug.Log("start_fadeIn");
        StartCoroutine(coroutine__fadeIn(delegate
        {
            Debug.Log("start_fadeOut");
            StartCoroutine(coroutine__fadeOut(delegate
            {
                Debug.Log("done");
            }));
        }));
    }


    private IEnumerator coroutine__fadeIn(Action completion)
    {
        CanvasGroup canvasGroup = GetComponent<CanvasGroup>();
        for (float f = 0f; f <= 1; f += 0.01f)
        {
            canvasGroup.alpha = f;
            yield return null;
        }

        completion();
    }

    private IEnumerator coroutine__fadeOut(Action completion)
    {
        CanvasGroup canvasGroup = GetComponent<CanvasGroup>();
        for (float f = 1f; f >= 0; f -= 0.01f)
        {
            canvasGroup.alpha = f;
            yield return null;
        }

        completion();
    }

我需要以某种方式提取let result1 = "t=20171017T1201&s=349.00&fn=8712000100030779&i=21456&fp=124519970&n=1" let result2 = "t=20171016T180757&s=92.00&fn=8710000101140572&i=17711&fp=4191934636&n=1"和t参数并将它们保存到另一个字符串：

fn

我设法通过逐符号检查整个字符串来获取这些参数，但必须有更好的方法。

Answer 1

您需要使用components(separatedBy: _)将字符串拆分为可解析对。

let result1 = "t=20171017T1201&s=349.00&fn=8712000100030779&i=21456&fp=124519970&n=1"
let result2 = "t=20171016T180757&s=92.00&fn=8710000101140572&i=17711&fp=4191934636&n=1"

func value(for name: String, in query: String) -> String? {
    for params in query.components(separatedBy: "&") {
        let pair = params.components(separatedBy: "=")
        if (pair.first == name) { 
            return pair.last
        }
    }
    return nil
}

print(value(for: "t", in: result1))
print(value(for: "fn", in: result1))
print(value(for: "t", in: result2))
print(value(for: "fn", in: result2))

Answer 2

基于这些字符串基本上是URL的查询参数的事实，您可以使用以下hack来轻松获取它：

let result1 = "t=20171017T1201&s=349.00&fn=8712000100030779&i=21456&fp=124519970&n=1"
let result2 = "t=20171016T180757&s=92.00&fn=8710000101140572&i=17711&fp=4191934636&n=1"

let params1 = queryParameters(from: result1)
let params2 = queryParameters(from: result2)

if let t1 = params1["t"], let t2 = params2["t"] {
    print(">>> t1 - \(t1) t2 - \(t2)")
}


public func queryParameters(from parameters: String) -> [String: String] {
    var components = URLComponents()
    components.query = parameters
    guard
        let queryItems = components.queryItems else {
            return [:]
    }

    var parameters = [String: String]()
    for item in queryItems {
        parameters[item.name] = item.value
    }

    return parameters
}

Answer 3

它会在这里创建一个小的解决方法，URLComponents在这种情况下可以帮到你，因为你的输入似乎是我的标准查询字符串（参见RFC-3986）。

let result1: String = "t=20171017T1201&s=349.00&fn=8712000100030779&i=21456&fp=124519970&n=1"

所以我会利用它并期望使用URLQueryItem数组，所以这个扩展已被放在一个数组上（可能没有效果，但很简单）。

extension Array where Iterator.Element == URLQueryItem {

    subscript(key: String) -> String? {
        get {
            return self.filter { $0.name == key }.first?.value
        }
    }

}

然后我会以随机顺序从随机生成但语法上有效的URL中读取组件：

if let url = URL(string: "u://?\(result1)"),
    let urlComponents = URLComponents(url: url, resolvingAgainstBaseURL: false) {

    // ...

    let t: String? = urlComponents.queryItems?["t"] // "20171017T1201"
    let s: String? = urlComponents.queryItems?["s"] // "349.00"
    let fp: String? = urlComponents.queryItems?["fp"] // "124519970"
    let random: String? = urlComponents.queryItems?["random"] // nil

    // ...
}

注意： 此解决方案基于输入字符串（查询字符串）的性质，如果输入格式将来发生变化，您可能需要直接使用正则表达式来提取信息。

Answer 4

你可以像：

那样实现它

if let t1 = result1.components(separatedBy: "t=")[1].components(separatedBy: "&").first {
    print(t1) // 20171017T1201
}

if let fn1 = result1.components(separatedBy: "fn=")[1].components(separatedBy: "&").first {
    print(fn1) // 8712000100030779
}

注意如果您非常确定您的字符串是网址参数，我建议您关注@ Milan Nosáľ answer。

将字符串的一部分复制到另一个字符串

4 个答案: