如何在iOS中删除NSURL查询中的参数？

Question

我使用这个字符串来构建NSURL

https://123.com?p1=AA&p2=BB&p3=CC

并且我想要移除p1&p2参数，只留下p3:

https://123.com?p3=CC

有没有比字符串compare&删去

Answer 1

guard let urlFromString = URL(string: "https://123.com?p1=AA&p2=BB&p3=CC") else {
return
}
var url = urlFromString

获取没有查询参数的url：

var components = URLComponents(url: url, resolvingAgainstBaseURL: true)
components?.query = nil // remove the query
url = components.url

创建QueryParams字典：

let stringDictionary = [
"p3": "CC"
]

创建最终url:

guard let newURl = url.append(queryParameters: stringDictionary) else {
return
}