【发布时间】:2018-09-29 18:28:27
【问题描述】:
我正在尝试获取用户输入的信用卡号并将其放入 Luhn 算法。当我运行代码时,我得到一个 conversion='i' 错误。拨打“检查”时,我也会收到一个两位数的号码。
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
System.out.print("Enter a credit card number (enter a blank line to quit): ");
String ccNumber=keyboard.nextLine();
long length2 = String.valueOf(ccNumber).length();
int length = toIntExact(length2);
if (length == 0) {
System.out.print("Goodbye!");
System.exit(0);
}
if (length != 16) {
System.out.println("ERROR! Number MUST have exactly 16 digits");
System.exit(0);
}
int check = ccNumber.charAt(15);
int sum = 0;
boolean alternate = false;
int i = length-1;
while (i >= 0)
{
int n = Integer.parseInt(ccNumber.substring(i, i + 1));
if (alternate)
{
n *= 2;
if (n > 9)
{
n = (n % 10) + 1;
}
}
sum += n;
alternate = !alternate;
i--;
}
if (sum % 10 == 0) {
System.out.printf("Check number should be: %i",check);
System.out.printf("Number is valid\n");
}
else {
System.out.printf("Check number should be: %s", check+"\n");
System.out.printf("Number is NOT valid\n");
}
}
}
【问题讨论】:
-
请注意,
String.charAt()返回一个字符。如果需要值,则需要将 ASCII 转换为 int(例如,char1的十进制值是 49,而不是 1)。 -
我收到一个 conversion='i' 错误:这是什么意思?什么是完整和准确的错误? 调用'check'时我也得到一个两位数。:您发布的代码中没有检查方法。
-
线程“main”中的异常 java.util.UnknownFormatConversionException: Conversion = 'i' at java.util.Formatter$FormatSpecifier.conversion(Unknown Source) at java.util.Formatter$FormatSpecifier.
(Unknown Source) at java.util.Formatter.parse(Unknown Source) at java.util.Formatter.format(Unknown Source) at java.io.PrintStream.format(Unknown Source) at java.io.PrintStream.printf( Unknown Source) at osu.cse1223.Project06.main(Project06.java:44) 这是完整的错误