将 url 字符串编码为字典在我的例子中：一些值在里面有 =

Question

我正在尝试将下面的 url 编码字符串解码为字典。下面给出了这样做的正常方法。就我而言，它不起作用。我还需要删除 \u{05}

之类的任何字符

let params = str.components(separatedBy: "&").map({
    [=11=].components(separatedBy: "=")
}).reduce(into: [String:String]()) { dict, pair in
    if pair.count == 2 {
        dict[pair[0]] = pair[1]
    }
}

我的url编码字符串是

"id=sfghsgh=sbfsfhj&name=awsjdk_fs\u{05}"

我期待结果

{ 
"id" = "sfghsgh=sbfsfhj",
"name" = "awsjdk_fs"
}

如何实现？

Answer 1

搭载在URLComponents上：

var components = URLComponents()
components.query = "id=sfghsgh=sbfsfhj&name=awsjdk_fs"

components.queryItems
// => Optional([id=sfghsgh=sbfsfhj, name=awsjdk_fs])

let list = components.queryItems?.map { ([=10=].name, [=10=].value) } ?? []
// [("id", Optional("sfghsgh=sbfsfhj")), ("name", Optional("awsjdk_fs"))]

let dict = Dictionary(list, uniquingKeysWith: { a, b in b })
// ["name": Optional("awsjdk_fs"), "id": Optional("sfghsgh=sbfsfhj")]

如果您需要 [String: String] 而不是 [String: String?]：

let list = components.queryItems?.compactMap { ([=11=].name, [=11=].value) as? (String, String) } ?? []
// [("id", "sfghsgh=sbfsfhj"), ("name", "awsjdk_fs")]

let dict = Dictionary(list, uniquingKeysWith: { a, b in b })
// ["name": "awsjdk_fs", "id": "sfghsgh=sbfsfhj"]

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