【问题标题】:How to get the value of url parameter using swift 4.1如何使用swift 4.1获取url参数的值
【发布时间】:2018-05-30 07:09:37
【问题描述】:

我正在尝试获取 URL 的参数,并且我尝试仅获取 URL 的日期而不是时间。

http://aa.no-ip.biz:8001/hf_tracker/api/history.php?accesskey=12345&Vehilce=1618&FromDate=2018-05-10 13:11&ToDate=2018-05-14 12:11

代码:

extension URL {
    func valueOf(_ queryParameterName: String) -> String? { 
        guard let url = URLComponents(string: self.absoluteString) else { 
              return nil 
        } 

        return url.queryItems?.first(where: { $0.name == queryParameterName})?.value 
    } 
}
    
    let newURL = URL(string: "assetlinkasia.no-ip.biz:8001/hf_tracker/api/…)! 
    newURL.valueOf("toDate") 
    newURL.valueOf("fromDate") 

我怎样才能只得到日期而不是时间?

【问题讨论】:

  • 向我们展示你的代码,你尝试了什么
  • 贴出相关代码方便理解@Rehan Meo
  • 扩展 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 } } let newURL = URL(string: " assetlinkasia.no-ip.biz:8001/hf_tracker/api/…)!newURL.valueOf("toDate") newURL.valueOf("fromDate")
  • 虽然问题是三年前的问题,但即使在今天,检查URLComponents(string: self.absoluteString) 也毫无意义,永远不会失败。

标签: ios json swift object swift4.1


【解决方案1】:

你可以这样做,

您的网址扩展名,来自here

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
    }
}

你的代码放在这里,

let string = "http://aa.no-ip.biz:8001/hf_tracker/api/history.php?accesskey=12345&Vehilce=1618&FromDate=2018-05-10 13:11&ToDate=2018-05-14 12:11"
let test = string.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)
let url = URL(string: test!)!
let fromDate = url.valueOf("FromDate")
let toDate = url.valueOf("ToDate")

let date1 = fromDate?.components(separatedBy: " ").first
print(date1)

let date2 = toDate?.components(separatedBy: " ").first
print(date2 )

输出将在下方,

2018-05-10
2018-05-14

【讨论】:

    【解决方案2】:

    在我的应用程序中,使用解析 url 并返回可选字典的辅助函数:

    func parameters(for url: URL) -> [String: String]? {
        guard let urlQuery = url.query else { return nil }
    
        // Create all parameters dictionary
        let queryArray = urlQuery.split { $0 == "&" }.map(String.init)
        var parametersDict: [String: String] = [:]
        for queryParameter in queryArray {
            // split the queryParam into key / value
            let keyValueArray = queryParameter.split{ $0 == "=" }.map(String.init)
            let key = keyValueArray.first!
            let value = keyValueArray.last!.removingPercentEncoding!
            parametersDict.updateValue(value, forKey: key)
        }
    
        return parametersDict
    }
    

    并像这样使用它:

    let params = parameters(for: URL(string: "https://someurl.com?name=john")!)
    if let name = params?["name"] {
        print(name)
    }
    

    或者您可以将 var 添加到 URL 扩展:

    extension URL {
        var parameters: [String: String]? {
            guard let urlQuery = self.query else { return nil }
    
            // Create all parameters dictionary
            let queryArray = urlQuery.split { $0 == "&" }.map(String.init)
            var parametersDict: [String: String] = [:]
            for queryParameter in queryArray {
                // split the queryParam into key / value
                let keyValueArray = queryParameter.split{ $0 == "=" }.map(String.init)
                let key = keyValueArray.first!
                let value = keyValueArray.last!.removingPercentEncoding!
                parametersDict.updateValue(value, forKey: key)
            }
    
            return parametersDict
        }
    }
    

    并获取参数:

    let params = URL(string: "https://someurl.com?name=john")!.parameters
    if let name = params?["name"] {
        print(name)
    }
    

    【讨论】:

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