Open
Description
Return the above exception
when using this code
import com.twilio.Twilio;
import com.twilio.base.ResourceSet;
import com.twilio.rest.api.v2010.account.IncomingPhoneNumber;
public class Example {
// Find your Account Sid and Token at twilio.com/console
// DANGER! This is insecure. See http://twil.io/secure
public static final String ACCOUNT_SID = "ACXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX";
public static final String AUTH_TOKEN = "your_auth_token";
public static void main(String[] args) {
Twilio.init(ACCOUNT_SID, AUTH_TOKEN);
ResourceSet<IncomingPhoneNumber> incomingPhoneNumbers =
IncomingPhoneNumber.reader()
.limit(20)
.read();
for(IncomingPhoneNumber record : incomingPhoneNumbers) {
System.out.println(record.getSid());
}
}
}
Error details:
com.twilio.exception.ApiConnectionException: Unable to deserialize response: Can not deserialize value of type java.net.URI from String
"voice_url": "http://sb360.com/forward?PhoneNumber=208-209-3062&FailUrl=http://sb360.com/voicemail?Email=chris@jisekihealth.com&Message=Thanks for calling Jiseki. Please leave a message and%we will get back to you shortly.&Transcribe=true&", is not a valid textual representation ,problem: Malformed escape pair at index 199```