[英]How do I extract a single digit from a string of digits?
例:
String number = "1234567890123456";
如何將倒數第二位本身存儲在int類型變量中?
詳細:
for (int i = number.length()-2; i>-1; i-=2)
int x = (Extract the number) at i;
我該怎么做呢?
這是代碼:
package creditcard;
import java.io.FileWriter;
import java.io.BufferedWriter;
import java.io.IOException;
public class Card {
//Declarations
private String cardNumber;
private boolean test = false;
private String cardType = "Unknown";
public Card(){
cardNumber = "0";
}//end Default Constructor
public Card(String input){
cardNumber = input;
}//end Constructor
//
public void typeMatcher(){
if (cardNumber.startsWith("37"))
cardType = "American Express";
if (cardNumber.startsWith("4"))
cardType = "Visa";
if (cardNumber.startsWith("5"))
cardType = "Mastercard";
if (cardNumber.startsWith("6"))
cardType = "Discover";
}//end typeMatcher
// Returns true if card number is Valid.
public boolean isValid(String cardNumber){
if (cardNumber.length() > 12 && cardNumber.length() < 17)
if((sumOfDoubleEvenPlace(cardNumber)+sumOfOddPlace(cardNumber))/10==0)
test = true;
return test;
}//end isValid
// Get result from Step 2.
public int sumOfDoubleEvenPlace(String cardNumber){
int sum = 0;
int num = 0;
for (int i = cardNumber.length()-2; i>-1; i-=2){
num = Integer.parseInt(cardNumber.substring(i,i+1));
if (num<10)
sum+=(num*2);
else
sum+=getDigit(num);
}//end for
System.out.println(sum);
return sum;
}//end SumOfDoubleEvenPlace
// Return this number if it is a single digit
// Otherwise return the sum of the two digits.
public int getDigit (int num){
int no1 = num/10;
int no2 = num%10;
return no1+no2;
}//end getDigit
// Returns sum of odd place digits in number.
public int sumOfOddPlace(String cardNumber){
int sum = 0;
for (int i = cardNumber.length()-1; i>-1; i-=2)
sum += Integer.parseInt(cardNumber.substring(i,i));
System.out.println(sum);
return sum;
}//end sumOfOddPlace
public void writeToFile(){
try{
FileWriter fw = new FileWriter("G:\\Output.txt",true);
BufferedWriter bw = new BufferedWriter(fw);
if (isValid(cardNumber)){
bw.write(cardNumber+" is a valid "+cardType+" card.");
bw.newLine();
}//end if
else{
bw.write(cardNumber+" is an invalid "+cardType+" card.");
bw.newLine();
}//end else
bw.close();
System.out.println("Success");
}//end try
catch(IOException ioe){
ioe.printStackTrace();
}//end catch
}//end outputToFile
}//end Card
目的是檢查信用卡號是否有效(學校項目)
只要這樣做:
int x = number.charAt(i) - '0'
Integer.parseInt
在這里過大。
您的方法sumOfOddPlace中的子字符串代碼有一個缺陷:
sum += Integer.parseInt(cardNumber.substring(i,i));
那應該是
sum += Integer.parseInt(cardNumber.substring(i,i+1));
這可能是格式異常的原因。
有一個名為“ charAt()”的小方法可以完成您想要的操作。
char c = someString.charAt(someString.length - 2);
然后,您需要將字符轉換為整數。
選擇@Andrew White的答案,因為這樣可以使整數轉換更加容易。 您可以通過char進行操作,但這有點棘手(對我來說不是,因為我本質上是C程序員)。
如果要倒數第二個:
if (number.length() >= 2) {
int myNumber = number.charAt(number.length() - 2) - '0';
}
或為您的循環:
for (int i = number.length()-2; i>-1; i-=2)
int x = number.charAt(i) - '0';
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