我这样做是一个项目。其重点是创建一个密码，其中包含您在所选长度中选择的字符类型。但是当我使用选项D时，它不会返回任何内容以及选项B和C，无论我输入的长度如何，我总是得到一个随机数字的字符。如果你可以运行它，并指出我正确的方向，这将是非常感谢。
import java.util.Scanner;
public class Password
{
public static void main(String [] args){
String password = "";
String temp = "";
int randLetter = 0;
int randNumber = 0;
int randPunct = 0;
int charSelection = 0;
int counter = 0;
String lowerCase = "abcdefghijklmnopqrstuvwxyz";
String upperCase = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
String number = "0123456789";
String punctuation = "!?";
Scanner in = new Scanner(System.in);
System.out.println("Please choose what characters you would like to use.");
System.out.println("[A] Lowercase Letters");
System.out.println("[B] Lowercase & Uppercase Letters");
System.out.println("[C] Lowercase, Uppercase, Numbers");
System.out.println("[D] Lowercase, Uppercase, Numbers, Punctuation");
System.out.print("Selection: ");
String selection = in.next();
System.out.println();
System.out.print("Select a length (1 - 14): ");
int length = in.nextInt();
if(selection.equalsIgnoreCase("A")){
for (int i = 0; i < length; i++){
randLetter = 1 + (int)(Math.random() * 26);
temp = lowerCase.substring(randLetter - 1 , randLetter);
password += temp;
}
}
else if(selection.equalsIgnoreCase("B")){
while (counter < length){
for (int i = 0; i < length; i++){
charSelection = 1 + (int)(Math.random() * 10);
if (charSelection < 5){
randLetter = 1 + (int)(Math.random() * 26);
temp = lowerCase.substring(randLetter - 1 , randLetter);
counter++;
password += temp;
}
else if (charSelection > 5 && charSelection < 10){
randLetter = 1 + (int)(Math.random() * 26);
temp = upperCase.substring(randLetter - 1 , randLetter);
counter++;
password += temp;
}
}
}
}
else if(selection.equalsIgnoreCase("C")){
while (counter < length){
for (int i = 0; i < length; i++){
charSelection = 1 + (int)(Math.random() * 17);
randNumber = 1 + (int)(Math.random() * 9);
if (charSelection < 5){
randLetter = 1 + (int)(Math.random() * 26);
temp = lowerCase.substring(randLetter - 1 , randLetter);
counter++;
password += temp;
}
else if (charSelection > 5 && charSelection < 10){
randLetter = 1 + (int)(Math.random() * 26);
temp = upperCase.substring(randLetter - 1 , randLetter);
counter++;
password += temp;
}
else if (charSelection > 15 && charSelection < 17){
randLetter = 1 + (int)(Math.random() * 26);
temp = number.substring(randNumber - 1 , randNumber);
counter++;
password += temp;
}
}
}
}
else if(selection.equalsIgnoreCase("D")){
while (counter < length){
for (int i = 1; i < 0; i++){
charSelection = 1 + (int)(Math.random() * 20);
randNumber = 1 + (int)(Math.random() * 9);
randPunct = 1 + (int)(Math.random() * 2);
if (charSelection < 5){
randLetter = 1 + (int)(Math.random() * 26);
temp = lowerCase.substring(randLetter - 1 , randLetter);
counter++;
password += temp;
}
else if (charSelection > 5 && charSelection < 10){
randLetter = 1 + (int)(Math.random() * 26);
temp = upperCase.substring(randLetter - 1 , randLetter);
counter++;
password += temp;
}
else if (charSelection > 15 && charSelection < 17){
randLetter = 1 + (int)(Math.random() * 26);
temp = number.substring(randNumber - 1 , randNumber);
counter++;
password += temp;
}
else if (charSelection > 17){
randLetter = 1 + (int)(Math.random() * 26);
temp = punctuation.substring(randPunct - 1 , randPunct);
counter++;
password += temp;
}
}
}
}
System.out.println(password);
}
}
答案 0 :(得分：1)
for (int i = 1; i < 0; i++){
这种情况永远不会成立，所以循环体永远不会执行。
目前还不清楚你真正想要的是什么，所以很难建议正确的条件 - 你的意思是length而不是0，就像你在其他循环中那样吗？
答案 1 :(得分：0)
模式“B”和“C”：
此循环：while (counter < length){将再次运行密码生成，如果此行中至少生成一个值：
charSelection = 1 + (int)(Math.random() * 10);
等于5（模式“B”），或者等于5或者在（包括）10和15之间（模式“C”）。所以基本上你尝试length次来生成一个新值，然后重复这个过程，直到你生成了比counter更多的字符（length），这很可能会产生超过length的字符。 {1}}字符。我建议您删除未使用的值并删除while - 循环以提供一致的行为。
模式“D”：
for (int i = 1; i < 0; i++){
此循环将终止生成，而不会生成单个字符，因为i < 0未在循环开始处保持（i初始化为1）。
并建议： 我建议你将代码分成几个部分，因为有大量可重用的代码：
Random r = new Random();
public char nextChar(String valid_chars){
return valid_chars.charAt(r.nextInt(valid_chars.length()));
}
public String generate(String chars , int len){
String tmp = "";
for(int i = 0 ; i < len ; i++)
tmp += nextChar(chars);
return tmp;
}
public static void main(String[] args){
...
switch(selection.toLowerCase()){
case "a":
{
String password = generate("abcdefghijklmnop..." , length);
System.out.println(password);
}
break;
...
}
...
}
这有助于保持代码的可读性，可维护性，并允许简单扩展到其他模式。