Java - StringIndexOutOfBoundsException
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Java - StringIndexOutOfBoundsException
提问by G?khan Nas
First of all, here is the problematic part of my code; these are very basic classes
首先，这是我的代码有问题的部分；这些是非常基础的课程
public Passenger(String Name, String adress, String number, String password){
count++;
accId+=count;
this.Name=Name;
this.adress=adress;
this.number=number;
if(checkPw(password)==true){
this.password=password;
}
}
private boolean checkPw(String password){
int length;
length = password.length();
if(length != 6){
return false;
}
else if(password.charAt(0)==0){
return false;
}
else {
for (int i = 0; i < password.length();i++){
if((password.charAt(i))==(password.charAt(i+1))){
return false;
}
}
}
return true;
}
testClass:
测试类：
public static void main(String[] args){
Passenger gokhan=new Passenger("Gokhan","Istanbul","xxx","254651");
System.out.println(gokhan.password);
}
So, I think the problem is in the Passenger class. Its my first time that class in class things (I meant the if(checkPw(password)==true) part). In the test class, it looks very clear and I never thought that this error will appear. How can I avoid this message?
所以，我认为问题出在Passenger 类上。这是我第一次在课堂上上课（我的意思是 if(checkPw(password)==true) 部分）。在测试类中，看起来很清晰，没想到会出现这个错误。如何避免此消息？
Full error:
完整错误：
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 6
at java.lang.String.charAt(String.java:658)
at project1.Passenger.checkPw(Passenger.java:45)
at project1.Passenger.<init>(Passenger.java:27)
at project1.testClass.main(testClass.java:11)
Java Result: 1
Java 结果：1
采纳答案by Alexis C.
The problem is here :
问题在这里：
for (int i = 0; i < password.length();i++){
if((password.charAt(i))==(password.charAt(i+1))){
return false;
}
}
When you're in the last iteration, you're trying to access the charin the stringat the position i+1which doesn't exists.
当你在最后一次迭代的时候，你要访问char在string该位置i+1不存在。
text
^
|
when i = 3 charAt(i) will return t and charAt(i+1) will throw the exception
回答by rgettman
This line appears to be the problem:
这一行似乎是问题所在：
if((password.charAt(i))==(password.charAt(i+1))){
When on the last iteration of the forloop, iis 5and i+1, or 6, goes off the end of the string, because indexes range from 0to length() - 1. The solution here is to stop the forloop iteration after the second-to-last character instead of the last character. Change
在for循环的最后一次迭代中，iis 5andi+1或 or6离开字符串的末尾，因为索引范围从0到length() - 1。这里的解决方案是在for倒数第二个字符而不是最后一个字符之后停止循环迭代。改变
for (int i = 0; i < password.length();i++){
to
到
for (int i = 0; i < password.length() - 1; i++){
So that the maximum value ihas in the forloop is 4, so i+1or 5isn't off the end of the string.
这样的最大值i已在for环路4，所以i+1还是5不脱字符串的结尾。
