快速在特定号码上打开 Whatsapp

Question

我正在尝试在 whatsapp 中打开特定的联系人聊天，但没有得到任何解决方案。请帮助我完全陷入困境。我试过这个：

let whatsAppURL: NSURL = NSURL(string: "whatsapp://send?abid=\(primary)&;text=lOL;")!
        if UIApplication.sharedApplication().canOpenURL(whatsAppURL){
            UIApplication.sharedApplication().openURL(whatsAppURL)
        }

Answer 1

您可以向特定用户发送消息。

直接应用聊天网址打开

let urlWhats = "whatsapp://send?phone=+919789384445&abid=12354&text=Hello"
    if let urlString = urlWhats.addingPercentEncoding(withAllowedCharacters: NSCharacterSet.urlQueryAllowed) {
        if let whatsappURL = URL(string: urlString) {
            if UIApplication.shared.canOpenURL(whatsappURL!) {
                UIApplication.shared.openURL(whatsappURL!)
            } else {
                print("Install Whatsapp")
            }
        }
    }

注意：国家代码（例如：+91）是打开手机号码聊天的必填项

WebUrl 链接打开聊天

 let whatsappURL = URL(string: "https://api.whatsapp.com/send?phone=9512347895&text=Invitation")
    if UIApplication.shared.canOpenURL(whatsappURL) {
        UIApplication.shared.open(whatsappURL, options: [:], completionHandler: nil)
    }

查看以下链接，

https://www.whatsapp.com/faq/en/general/26000030

注意：在 info.plist 中添加 url scheme

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

Answer 2

适用于 Swift 4.2 / Swift 5

func openWhatsapp(){
    let urlWhats = "whatsapp://send?phone=(mobile number with country code)"
    if let urlString = urlWhats.addingPercentEncoding(withAllowedCharacters: CharacterSet.urlQueryAllowed){
        if let whatsappURL = URL(string: urlString) {
            if UIApplication.shared.canOpenURL(whatsappURL){
                if #available(iOS 10.0, *) {
                    UIApplication.shared.open(whatsappURL, options: [:], completionHandler: nil)
                } else {
                    UIApplication.shared.openURL(whatsappURL)
                }
            }
            else {
                print("Install Whatsapp")
            }
        }
    }
}

对于 Swift 2.0

let urlWhats = "whatsapp://send?phone=(mobile number with country code)"
        if let urlString = urlWhats.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.URLQueryAllowedCharacterSet()){
            if let whatsappURL = NSURL(string: urlString) {
                if UIApplication.sharedApplication().canOpenURL(whatsappURL){
                    UIApplication.sharedApplication().openURL(whatsappURL)
                }
                else {
                    print("Install Whatsapp")
                }
            }
        }

注意：在 info.plist 中添加 url scheme

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

Answer 3

这对我很有效

    if let url = URL(string: "https://wa.me/91xxxxxxxxxx?text=Hello"),
            UIApplication.shared.canOpenURL(url) {
                UIApplication.shared.open(url, options: [:])
    }

Answer 4

1. 向Info.plist 源代码文件添加属性，如下所示：

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

2. 在您的应用中自定义点击时使用以下功能：

   func openWhatsapp(){
       let urlWhats = "https://wa.me/numberWithCountryCode"
       if let urlString = urlWhats.addingPercentEncoding(withAllowedCharacters: CharacterSet.urlQueryAllowed){
           if let whatsappURL = URL(string: urlString) {
               if UIApplication.shared.canOpenURL(whatsappURL){
                   if #available(iOS 10.0, *) {
                       UIApplication.shared.open(whatsappURL, options: [:], completionHandler: nil)
                   } else {
                       UIApplication.shared.openURL(whatsappURL)
                   }
               } else {
                   Alert(withMessage: "You do not have WhatsApp installed! \nPlease install first.").show(andCloseAfter: 2.0)
               }
           }
       }
   }
   

Answer 5

根据whatsapp forum link，您无法向特定用户发送消息，这在 whatsapp URL 方案中不可用。

您只需设置预定义的消息，然后使用 URL 方案，您就可以打开 whatsapp 最近的控制器。

Answer 6

这个问题真的很有帮助

只需输入不带“+”的人员编号 示例：60161234567

Answer 7

这是不可能的，你可以用 URL 方案打开 WhatsApp。