Because I was the keyboard input number into a String type first, and then use the substring to obtain bits ten with ten digits automatically generated judgment

 import Java. Util. Scanner; 
The class lotterygame {
Public static void main (String [] args) {
Scanner variable=new Scanner (System. In); 
Int bonus=(int) (Math. The random () * 90 + 10);//random in the production of 10-99 digital 
String bonus1=bonus + ""//into type String 
String bonusone=bonus1. Substring (0, 1);//get the generated digital bits 
String bonustwo=bonus1. Substring (1, 2);//get the generated number 10 

Scanner input=new Scanner (System. In); 
Int inputl=input. NextInt ();//get input number 
String input1=inputl + ""//into type String 
String inputone=input1. Substring (0, 1);//get the generated digital bits 
String inputtwo=input1. Substring (1, 0).//get the generated number 10 

If (bonus1. Equals (input1)) {//I heard==judgment is not good, use the equals judgment 
System. The out. Println (" 10000 "); 
} else if (bonusone equals (inputtwo) & amp; & Bonustwo. Equals (inputone)) {//see the generated bits with the input of ten 

If, on the other hand,, 
System. The out. Println (" 3000 "); 
} else if (inputone equals (bonusone) | | inputtwo. Equals (bonustwo)) {
System. The out. Println (" 1000 "); 
} else if (inputone equals (bonusone) | | inputone. Equals (bonustwo) | | 

Inputtwo. Equals (bonusone) | | inputtwo. Equals (bonustwo)) {
System. The out. Println (" 500 "); 
} else {
System. The out. Println (" no money "); 
} 
System. The out. Println (" the winning Numbers: "+ bonus1); 
} 
} 

Error message:
The Exception in the thread "main" Java. Lang. StringIndexOut
The ex out of range: 1
The at Java. Lang. String. The substring (String. Java: 1
The at lotterygame. Main (lotterygame. Java: 14)


Enter the Numbers, then an error

CodePudding user response:

Has been solved, type is caused by inconsistent problem, as long as the types of the won't error!