当链接中包含“&”时，WhatsApp 分享无法正常工作

Question

示例代码：

let sampleURL = "http://Hello/site/link?id=MTk=&fid=MTA="
let urlWhats = "whatsapp://send?text=\(sampleURL)"
if let urlString = urlWhats.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed) {
   if let whatsappURL = NSURL(string: urlString) {
      if UIApplication.shared.canOpenURL(whatsappURL as URL) {
           UIApplication.shared.open(whatsappURL as URL)
      }
      else {
           // Open App Store.
      }
   }
}

添加到.plist文件中

<key>LSApplicationQueriesSchemes</key>
<array>
    <string>whatsapp</string>
</array>

面临的问题是与 & 共享链接时文本不在 &

之后

对于上面的网址文本，就像 http://Hello/site/link?id=MTk=

提前致谢

Answer 1

URL的不同部分有不同的编码规则。

addingPercentEncoding

正确使用非常棘手。通常，您不应尝试对自己的 URL 进行编码。让系统使用 URLComponents 为您完成这件事：

//For the constant part of the link, it's fine to just use a string
var sample = URLComponents(string: "whatsapp://send")!

// Then add a query item
sample.queryItems = [URLQueryItem(name: "text", value: "http://Hello/site/link?id=MTk=&fid=MTA=")]

// Extract the URL, which will have the correct encoding
print(sample.url!)
// whatsapp://send?text=http://Hello/site/link?id%3DMTk%3D%26fid%3DMTA%3D

Answer 2

尝试直接初始化为 URL 而不是 NSURL

let whatsappURL = URL(string: urlString)

也许可以使用完成处理程序来进一步调试

UIApplication.shared.open(whatsappURL, options: [:], completionHandler: nil)

Answer 3

经过一些研究和测试，我发现在 .urlQueryAllowed

 中使用 

AllowedCharacters 时，将允许 URL 并且不会更改 URL，这就是发生此问题的原因。

只需在不包含在该编码中的文本上使用

.init()

如下所示

whatsapp://send?text=

像下面这样

let sampleURL = "http://Hello/site/link?id=MTk=&fid=MTA="
if let urlString = sampleURL.addingPercentEncoding(withAllowedCharacters: .init()) {
    let urlWhats = "whatsapp://send?text=\(urlString)"
    if let whatsappURL = NSURL(string: urlWhats) {
       if UIApplication.shared.canOpenURL(whatsappURL as URL) {
            UIApplication.shared.open(whatsappURL as URL)
       }
    }
}

Answer 4

对于那些在 flutter web 上的 Whatsapp 中遇到相同问题的人可以使用以下代码片段。

非工作方法❌：

final String _siteLink = "http://clientsite/products?id=2&vc=premium";
String _whatsappLink = "https://api.whatsapp.com/send?text=${_siteLink}";
await launchUrl(_whatsappLink);
OUTPUT:
http://clientsite/products?id=2 //&vc=premium is removed when opening whatsapp web page

工作方法✅：

    final Uri _whatsappURI = Uri(
      scheme: 'https',
      host: 'api.whatsapp.com',
      path: 'send',
      queryParameters: <String, String>{
        'text': _siteLink,
      },
    );
    await launchUrl(_whatsappURI.toString());

OUTPUT:
http://clientsite/products?id=2&vc=premium