Question

我是新手，需要帮助。我需要从下面的QueryString创建一个baseString。这个baseString最终会看起来像这样：

＆amp; ap =裕廊坊心邻坊＆amp; oq = c #nunit mac＆amp; q = c #nunit mac

QueryString：

Foo

问题：

如何从上面的QueryString中获取KeyValue

1）通过＆＃34;＆amp;＆＃34;将所有组件分开。如下所示

- 查询字符串中的键值：

q = c％23 + nunit + mac＆amp; oq
oq = c％23 + nunit + mac
AP =裕坊％20邻坊

HTTPS://www.sky.com/api/v1/rest/level2/in-in/?q=c%23+nunit+mac&oq=c%23+nunit+mac&ap=裕坊%20邻坊

感谢。非常感谢你的帮助。

Answer 1

您可以在URLQueryItem类的帮助下从URL的查询项中获取Key-Value字典，如下所示

let urlString = "https://www.sky.com/api/v1/rest/level2/in-in/?q=c%23+nunit+mac&oq=c%23+nunit+mac&ap=裕坊%20邻坊"
let encodedUrlString = urlString.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)!

let items = URLComponents(string: encodedUrlString)?.queryItems
var keyValues: [String: String] = [:]
items?.forEach{
    item in
    keyValues[item.name] = item.value
}
print(keyValues)
//["q": "c%23+nunit+mac", "ap": "裕坊%20邻坊", "oq": "c%23+nunit+mac"]

希望这会对你有所帮助。如果需要从url字符串中获取Query子字符串，则需要从中创建URL并获取查询。

guard let url = URL(string: encodedUrlString) else {
    fatalError()
}
let queryString = url.query!
print(queryString.removingPercentEncoding)
//q=c%23+nunit+mac&oq=c%23+nunit+mac&ap=裕坊%20邻坊

如果您需要在查询中添加新组件，

var components = URLComponents(string: encodedUrlString)
let item = URLQueryItem(name: "NEWVKEY", value: "NEWVALUE")
components?.queryItems?.append(item)

let url = components?.url
let resultString = url?.absoluteString
//or
let resultString2 = url?.absoluteString.removingPercentEncoding

这个想法是使用Swift标准库的URL处理能力。请查看URL，URLComponents，URLQueryItem结构的文档。不要编写字符串处理代码，而是操纵URL。

https://developer.apple.com/documentation/foundation/url https://developer.apple.com/documentation/foundation/urlcomponents https://developer.apple.com/documentation/foundation/urlqueryitem

如何从QueryString获取KeyValue并以特定的字符串格式返回字符串

1 个答案: