xcode 如何快速将字符串转换为 URL
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How to convert string to URL in swift
提问by Akshay Phulare
I want to convert below String to URL
我想将下面的字符串转换为 URL
www.mydomain.com/key=?????
www.mydomain.com/key=?????
I tried let urlToSend = URL(string: "www.mydomain.com/key=?????")!but it returns nil.
我试过了，let urlToSend = URL(string: "www.mydomain.com/key=?????")!但它返回零。
I'm getting that '?????' keyword from textField, it could be in any local language.
我得到那个'?????' textField 中的关键字，它可以是任何本地语言。
It works fine with English but not working with local language.
它适用于英语，但不适用于当地语言。
I used let url = URL(string: "www.mydomain.com")?.appendingPathComponent("key=?????")it gives www.mydomain.com/key=%E0%A4%85%E0%A4%95%E0%A5%8D%E0%A4%B7%E0%A4%AF
我用let url = URL(string: "www.mydomain.com")?.appendingPathComponent("key=?????")它给 www.mydomain.com/key=%E0%A4%85%E0%A4%95%E0%A5%8D%E0%A4%B7%E0%A4%AF
but now I want to
www.mydomain.com/key=%E0%A4%85%E0%A4%95%E0%A5%8D%E0%A4%B7%E0%A4%AFto www.mydomain.com/key=?????
但现在我要
www.mydomain.com/key=%E0%A4%85%E0%A4%95%E0%A5%8D%E0%A4%B7%E0%A4%AF到www.mydomain.com/key=?????
回答by Ahmad F
When working with URLs as strings, you should encode them to be valid as a URL; One of the most popular examples of encoding the URL is that the " " (space) would be encoded as "%20".
将 URL 作为字符串处理时，您应该将它们编码为有效的 URL；最流行的 URL 编码示例之一是“”（空格）将被编码为“%20”。
So in your case the encoded value of your url should be:
所以在你的情况下，你的 url 的编码值应该是：
www.mydomain.com%2Fkey=%E0%A4%85%E0%A4%95%E0%A5%8D%E0%A4%B7%E0%A4%AF
As you noticed the value of the keyis changed
正如您注意到的那样，值key已更改
from: "?????"
从： ”？？？？？”
to:"%E0%A4%85%E0%A4%95%E0%A5%8D%E0%A4%B7%E0%A4%AF"
到：“%E0%A4%85%E0%A4%95%E0%A5%8D%E0%A4%B7%E0%A4%AF”
which will let the URL to be valid.
这将使 URL 有效。
How to:
如何：
you could get the above result like this:
你可以像这样得到上面的结果：
let string = "www.mydomain.com/key=?????"
if let encodedString = string.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed), let url = URL(string: encodedString) {
print(url) // www.mydomain.com%2Fkey=%E0%A4%85%E0%A4%95%E0%A5%8D%E0%A4%B7%E0%A4%AF
}
Note that there is optional bindingfor both the encoded string and the url for the purpose of being safe.
请注意，为了安全起见，编码字符串和 url 都有可选绑定。
Decoding the URL:
解码网址：
You could also returns to the original unencoded url (decoding it) like this:
您还可以像这样返回原始未编码的 url（对其进行解码）：
let decodedString = "www.mydomain.com%2Fkey=%E0%A4%85%E0%A4%95%E0%A5%8D%E0%A4%B7%E0%A4%AF"
if let unwrappedDecodedString = decodedString.removingPercentEncoding {
print(unwrappedDecodedString) // www.mydomain.com/key=?????
}
Again, optional binding to be safe.
再次，可选绑定是安全的。
回答by Oliver Kulpakko
You need to encode the URL.
您需要对 URL 进行编码。
let urlString = "www.mydomain.com/key=?????".addingPercentEncoding(withAllowedCharacters: .urlPathAllowed)
let url = URL(string: urlString!)
let decodedUrl = urlString?.removingPercentEncoding
Keep in mind that you shouldn't force unwrap URL's and strings, use if let or guard statements.
请记住，您不应强制解开 URL 和字符串，而应使用 if let 或 guard 语句。
回答by Tejas Ardeshna
Try this thing:
试试这个：
let strURL = "www.mydomain.com/key=?????".addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)
let url = URL(string: strURL!)
Or you can use:
或者你可以使用：
let strURL = "www.mydomain.com/key=?????".addingPercentEncoding(withAllowedCharacters: .urlPathAllowed)
let url = URL(string: strURL!)
回答by Gereon
Instead of dealing with percent encodings explicitly, you can also build the URL piece by piece, using appendingPathComponent:
除了显式处理百分比编码之外，您还可以使用appendingPathComponent以下方法逐个构建 URL ：
let url = URL(string: "www.mydomain.com")?.appendingPathComponent("key=?????")
// www.mydomain.com/key=%E0%A4%85%E0%A4%95%E0%A5%8D%E0%A4%B7%E0%A4%AF
回答by Kathiresan Murugan
Swift3.0
斯威夫特3.0
let baseUrl = "www.mydomain.com/key=?????" // base url
let encodedUrl : String! = baseUrl.addingPercentEncoding(withAllowedCharacters: CharacterSet.urlQueryAllowed) // remove the spaces in the url string
let typeUrl = URL(string: encodedUrl)! // convert the string into url
print(typeUrl) // www.mydomain.com/key=%E0%A4%85%E0%A4%95%E0%A5%8D%E0%A4%B7%E0%A4%AF
For Checking purpose
用于检查目的
if let unwrappedDecodedString = encodedUrl.removingPercentEncoding {
print(decodedString) // www.mydomain.com/key=?????
}
