如果未检测到“ @”符号，则会在屏幕上显示错误消息

Question

如果用户输入的电子邮件地址中不包含“ @”符号，则必须使用if / else语句退出该程序。 这是我这部分的代码：

//prompt user for their email address

System.out.print("\nPlease enter your email address: ");

//read user's input
emailAddress = keyboard.next();

//create username
String username = emailAddress.substring(0, emailAddress.indexOf('@'));

 //create username
atSign = emailAddress.lastIndexOf("@");
if(atSign >= 0){
    username = emailAddress.substring(0, atSign);
}
else{
    System.out.print("You've entered an invalid email address!");
    System.out.println("Goodbye!");
}

错误：

Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: -1
        at java.lang.String.substring(String.java:1911)
        at Riccio_Lesson6.main(Riccio_Lesson6.java:51)

Answer 1

问题在于您的电话：

String username = emailAddress.substring(0,emailAddress.indexOf('@'));

如果电子邮件地址不包含“ @”，则emailAddress.indexOf('@')返回-1 ，表示您正在尝试调用emailAddress.substring(0, -1) 。

-1不是有效的索引，它导致StringIndexOutOfBoundsException 。

Answer 2

String username = emailAddress.substring(0, emailAddress.indexOf('@'));//this will throw exception

该怎么办？

在之前创建username ，并将其设置为null或空字符串。 然后检查该符号是否存在。

 String username = null;
 // or String username = "";

     //create username 
 atSign = emailAddress.lastIndexOf("@"); 
  if(atSign >= 0){ 
    username = emailAddress.substring(0, atSign);
 } 
 else{ 
     System.out.print("You've entered an invalid email address!");
     System.out.println("Goodbye!");


   // use return to stop code here if in a method(better choice) or
   // if in the main() uncomment the code below
   // System.exit(0);
 }