获取URL参数的值


80

我正在尝试使用Swift从URL获取参数。假设我有以下网址:

http://mysite3994.com?test1=blah&test2=blahblah

如何获取test1和test2的值?

Answers:


181

您可以使用以下代码获取参数

func getQueryStringParameter(url: String, param: String) -> String? {
  guard let url = URLComponents(string: url) else { return nil }
  return url.queryItems?.first(where: { $0.name == param })?.value
}

像这样调用方法 let test1 = getQueryStringParameter(url, param: "test1")

其他扩展方法:

extension URL {
    public var queryParameters: [String: String]? {
        guard
            let components = URLComponents(url: self, resolvingAgainstBaseURL: true),
            let queryItems = components.queryItems else { return nil }
        return queryItems.reduce(into: [String: String]()) { (result, item) in
            result[item.name] = item.value
        }
    }
}

为什么要投向NSURLQueryItem?一些不必要的
亚历山大-恢复莫妮卡

3
另外,最好使用first(where:)代替filter(_:)then first。一旦找到第一个匹配项,它将立即停止,因此速度更快。另外,它还节省了阵列分配
亚历山大-莫妮卡(Monica)恢复

1
@现在user2423476检查我在操场上进行测试与SWIFT 3
PARTH Adroja

1
@ Matian2049有一个名为“ EZSwiftExtensions”的库可以使用。
Parth Adroja

1
正确的调用方法是:---- let test1 = getQueryStringParameter(url:url,param:“ test1”)
pw2

71

第1步:创建URL扩展

extension URL {
    func valueOf(_ queryParamaterName: String) -> String? {
        guard let url = URLComponents(string: self.absoluteString) else { return nil }
        return url.queryItems?.first(where: { $0.name == queryParamaterName })?.value
    }
}

步骤2:如何使用扩充功能

let newURL = URL(string: "http://mysite3994.com?test1=blah&test2=blahblah")!

newURL.valueOf("test1") // Output i.e "blah"
newURL.valueOf("test2") // Output i.e "blahblah"

26

我还做了一个URL扩展,但是将查询参数查找放入下标中。

extension URL {
    subscript(queryParam:String) -> String? {
        guard let url = URLComponents(string: self.absoluteString) else { return nil }
        return url.queryItems?.first(where: { $0.name == queryParam })?.value
    }
}

用法:

let url = URL(string: "http://some-website.com/documents/127/?referrer=147&mode=open")!

let referrer = url["referrer"]  // "147"
let mode     = url["mode"]      // "open"

您可以这样使用,让urlString = URL(string:url.absoluteString)!警卫让userName = urlString.valueOf(“ userName”),让代码= urlString.valueOf(“ code”),让userId = urlString.valueOf(“ userName”)否则{return false} print(“ Details:(urlString)” )print(“ userName:(userName)”)print(“ code:(code)”)print(“ userId:(userId)”)
Mehul

@Mehul为什么您要从另一个URL创建URLlet urlString = URL(string: url.absoluteString)!是没有意义的。顺便说一句URL没有valueOf方法。
Leo Dabus

9

链接指向在Angular上创建的网站时,似乎没有现有答案可用。这是因为Angular的路径通常#在所有链接中都包含一个(哈希)符号,这将导致url.queryItems始终返回nil。

如果链接看起来像这样: http://example.com/path/#/morepath/aaa?test1=blah&test2=blahblah

然后只能从中获取参数url.fragment。在@Matt的扩展中添加了一些附加的解析逻辑后,通用代码将如下所示:

extension URL {
    subscript(queryParam: String) -> String? {
        guard let url = URLComponents(string: self.absoluteString) else { return nil }
        if let parameters = url.queryItems {
            return parameters.first(where: { $0.name == queryParam })?.value
        } else if let paramPairs = url.fragment?.components(separatedBy: "?").last?.components(separatedBy: "&") {
            for pair in paramPairs where pair.contains(queryParam) {
                return pair.components(separatedBy: "=").last
            }
            return nil
        } else {
            return nil
        }
    }
}

用法保持不变:

let url = URL(string: "http://example.com/path/#/morepath/aaa?test1=blah&test2=blahblah")!

let referrer = url["test1"]  // "blah"
let mode     = url["test2"]  // "blahblah"

您通过提及url.fragment节省了我的时间。非常感谢。
Keyhan Kamangar

2

这样做的另一种方法是在URL上创建扩展名以返回组件,然后在[URLQueryItem]上创建扩展名以从queryItems中检索值。

extension URL {
    var components: URLComponents? {
        return URLComponents(url: self, resolvingAgainstBaseURL: false)
    }
}

extension Array where Iterator.Element == URLQueryItem {
    subscript(_ key: String) -> String? {
        return first(where: { $0.name == key })?.value
    }
}

这是如何使用它的一个示例:

if let urlComponents = URL(string: "http://mysite3994.com?test1=blah&test2=blahblah")?.components,
    let test1Value = urlComponents.queryItems?["test1"] {
    print(test1Value)
}
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