URL构造函数不适用于某些字符 [英] URL constructor doesn't work with some characters

问题描述

我正在尝试使用URLRequest从我的应用程序调用php脚本.Url路径是在字符串变量 query 中生成的，对于请求，我将其转换为这样

I'm trying to call a php-script from my app using URLRequest. The Url path is generated in the String-Variable query and for the request I convert it like this

    guard let url = URL(string: query) else {
        print("error")
        return
    }

通常有效，但是当请求包含ä，ö，ü，ß之类的字符时，将触发错误.我该如何运作?

usually it works, but when the request contains characters like ä, ö, ü, ß the error is triggered. How can I make it work?

推荐答案

URL(string:)初始化程序不需要将 String 编码为有效的URL String ，它假定 String 已被编码为仅包含在 URL 中有效的字符.因此，如果您的 String 包含无效的URL字符，则必须进行编码.您可以通过调用 String.addingPercentEncoding(withAllowedCharacters:)来实现.

The URL(string:) initializer doesn't take care of encoding the String to be a valid URL String, it assumes that the String is already encoded to only contain characters that are valid in a URL. Hence, you have to do the encoding if your String contains non-valid URL characters. You can achieve this by calling String.addingPercentEncoding(withAllowedCharacters:).

let unencodedUrlString = "áűáeqw"
guard let encodedUrlString = unencodedUrlString.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed), let url = URL(string: encodedUrlString) else { return }

您可以根据URL的哪个部分包含需要编码的字符来更改 CharacterSet ，我只是出于演示目的使用 urlQueryAllowed .

You can change the CharacterSet depending on what part of your URL contains the characters that need encoding, I just used urlQueryAllowed for presentation purposes.

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