Open Whatsapp on a particular number in swift

问题

I am trying to open a particular contact chat in whatsapp but not getting any solution. Please help i am totally stuck. I have tried this:

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

回答1:

As per this whatsapp forum link, there is no way you can send message to specific user, this is not available within whatsapp URL scheme.

You just set predefined message and then with URL scheme you are able to open whatsapp recent controller.

回答2:

Its possible You can send messages to Specfic user.

Direct app chat url open

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")
            }
        }
    }

Note:Country code (Ex:+91) is mandatory to open mobile number Chat

WebUrl Link open Chat

 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)
    }

Check below link,

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

Note: Add url scheme in info.plist

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

回答3:

For 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")
            }
        }
    }
}

For 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")
                }
            }
        }

Note: Add url scheme in info.plist

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

回答4:

This is not possible you can just open WhatsApp with URL scheme.

来源：https://stackoverflow.com/questions/40535309/open-whatsapp-on-a-particular-number-in-swift

标签

ios