因此，我目前遇到的问题是，在执行完所有步骤之后，“输入您的命令（反向，首先替换，最后替换，全部删除，删除）”这条语句打印了两次。
我认为正在发生的是循环执行了两次，但我不知道为什么。任何帮助将不胜感激解决这个问题。如果我的代码格式不好，请提前抱歉，仍然在学习如何正确设置格式。
import java.util.Scanner;
public class StringChangerenter {
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
// Output Variables
String userInput = "";
// Variables
String removeChar = "", removeAllChar = "";
int removeIndex = 0;
// First Output
System.out.println("Enter the string to be manipulated");
userInput = keyboard.nextLine();
String command = "";
// While loop
while (!command.equalsIgnoreCase("quit")) {
// Output
System.out.println("Enter your command (reverse, replace first, replace last, remove all, remove)");
command = keyboard.nextLine();
if (command.equalsIgnoreCase("remove")) {
System.out.println("Enter the character to remove");
removeChar = keyboard.nextLine();
int totalCount = 0;
for (int j = 0; j < userInput.length(); j++) {
if (userInput.charAt(j) == removeChar.charAt(0)) {
totalCount = totalCount + 1;
}
}
System.out.println("Enter the " + removeChar
+ " you would like to remove (Not the index - 1 = 1st, 2 = 2nd, etc.):");
removeIndex = keyboard.nextInt();
int currentIndex = 1;
if (removeIndex <= totalCount) {
for (int i = 0; i < userInput.length(); i++) {
if (userInput.charAt(i) == removeChar.charAt(0)) {
if (currentIndex == removeIndex) {
String firstpartOfString = userInput.substring(0, i);
String secondpartOfString = userInput.substring(i + 1, userInput.length());
System.out.println("The new sentence is " + firstpartOfString + secondpartOfString);
userInput = firstpartOfString + secondpartOfString;
break;
} else {
currentIndex = currentIndex + 1;
}
}
}
} else {
System.out.println("Can't find " + removeChar + " occuring at " + removeIndex + " int the string.");
}
// Remove All Code
} else if (command.equalsIgnoreCase("remove all")) {
System.out.println("Enter the character to remove");
removeAllChar = keyboard.next();
String newString = "";
for (int i = 0; i < userInput.length(); i++) {
if (userInput.charAt(i) != removeAllChar.charAt(0)) {
newString = newString + userInput.charAt(i);
}
}
userInput = newString;
System.out.println("The new sentence is " + userInput);
}
// Bracket for while loop
}
}
}
答案 0 :(得分：1)
发生的事情是，循环为您提供的条件
while (!command.equalsIgnoreCase("quit"))
用英语表示，只要命令不等于“ quit”，就运行此循环。
在循环内部，命令实际上从未设置为“退出”。例如，如果我将输入字符串指定为“ abcde”并要求删除位置1处的“ c” 然后您的逻辑将命令设置为在此处“删除”
command = keyboard.nextLine();
，然后将最终值打印为“ abde”。现在，当循环结束时，命令仍然处于“删除”状态，因此循环再次执行。
可能的解决方案是使用do while循环明确询问用户是否要重试。也只是一个提示，我看到您已经使用了nextInt。建议在下一个int之后立即使用nextLine。看到它的原因如下：Java Scanner doesn't wait for user input
如果您要运行更多命令，则明确征得用户的同意，这就是您的代码：
public static void main (String[] args) throws java.lang.Exception
{
Scanner keyboard = new Scanner(System.in);
// Output Variables
String userInput = "";
// Variables
String removeChar = "", removeAllChar = "";
int removeIndex = 0;
// First Output
System.out.println("Enter the string to be manipulated");
userInput = keyboard.nextLine();
String command = "";
String retry = "";
// While loop
do {
// Output
System.out.println("Enter your command (reverse, replace first, replace last, remove all, remove)");
command = keyboard.nextLine();
if (command.equalsIgnoreCase("remove")) {
System.out.println("Enter the character to remove");
removeChar = keyboard.nextLine();
int totalCount = 0;
for (int j = 0; j < userInput.length(); j++) {
if (userInput.charAt(j) == removeChar.charAt(0)) {
totalCount = totalCount + 1;
}
}
System.out.println("Enter the " + removeChar
+ " you would like to remove (Not the index - 1 = 1st, 2 = 2nd, etc.):");
removeIndex = keyboard.nextInt();
keyboard.nextLine();
int currentIndex = 1;
if (removeIndex <= totalCount) {
for (int i = 0; i < userInput.length(); i++) {
if (userInput.charAt(i) == removeChar.charAt(0)) {
if (currentIndex == removeIndex) {
String firstpartOfString = userInput.substring(0, i);
String secondpartOfString = userInput.substring(i + 1, userInput.length());
System.out.println("The new sentence is " + firstpartOfString + secondpartOfString);
userInput = firstpartOfString + secondpartOfString;
break;
} else {
currentIndex = currentIndex + 1;
}
}
}
} else {
System.out.println("Can't find " + removeChar + " occuring at " + removeIndex + " int the string.");
}
// Remove All Code
} else if (command.equalsIgnoreCase("remove all")) {
System.out.println("Enter the character to remove");
removeAllChar = keyboard.next();
String newString = "";
for (int i = 0; i < userInput.length(); i++) {
if (userInput.charAt(i) != removeAllChar.charAt(0)) {
newString = newString + userInput.charAt(i);
}
}
userInput = newString;
System.out.println("The new sentence is " + userInput);
}
System.out.println("Do you want to go again?");
retry = keyboard.nextLine();
// Bracket for while loop
}while("yes".equalsIgnoreCase(retry));
}
答案 1 :(得分：1)
处理完一个字符后得到两个条目的原因是，您没有完全读取包含该字符的行。
具体来说，您在上部分支中使用keyboard.nextInt();，在下部分支中使用keyboard.next();。虽然它们分别读取下一个整数和字符，但它们不处理行尾标记。
然后，当您到达循环的顶部时，调用keyboard.nextLine()，它处理从int之后（或字符，在全部删除的情况下）到行标记结束之前出现的所有字符。使用预期的用户输入，那只是一个空字符串。
要解决此问题，在仅读取整数或单个字符的情况下，需要确保从头到尾keyboard.nextLine()都进行读取。