Java 如何检测输入是否为';没有信吗?我正在写一个游戏,它需要打印;无效输入“;
我正在写一个游戏,如果输入的不是字母,我想警告玩家?。例如,如果输入是“(”或“6”,它需要打印“无效输入”。我还需要“!”作为结束游戏的有效输入Java 如何检测输入是否为';没有信吗?我正在写一个游戏,它需要打印;无效输入“;,java,Java,我正在写一个游戏,如果输入的不是字母,我想警告玩家?。例如,如果输入是“(”或“6”,它需要打印“无效输入”。我还需要“!”作为结束游戏的有效输入 +-----+-----+-----+ | | | | | | | | | | | | | | | | +-----+-----+-----+ Secret Phrase: --- Guessed Letters: Enter a letter t
+-----+-----+-----+
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+-----+-----+-----+
Secret Phrase: ---
Guessed Letters:
Enter a letter to guess or "!" to end the game
(
No, ( is not in the phrase
下面是这个游戏的演示
+-----+-----+-----+
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+-----+-----+-----+
Secret Phrase: ---
Guessed Letters:
Enter a letter to guess or "!" to end the game
(
No, ( is not in the phrase
以下是编码部分
while (gameEnd == false) {
System.out.println(drawScoreboard(totalGuesses, wrong));
System.out.println("Enter a letter to guess or \"!\" to end the game");
letterz = scnr.next();
for (int i = 0; i < secretPhrase.length(); i++) {
for (int hk = 0; hk < alreadyGuessed.length(); hk++) {
if (Character.toString(letterz.charAt(0)).equals(Character.toString(alreadyGuessed.charAt(hk)))) {
break;
}
}
if (i >= letterz.length()) {
break;
} else if ((Character.toString(letterz.charAt(i))).equalsIgnoreCase(Character.toString(secretPhrase.charAt(i)))) {
System.out.print("\n");
inc++;
} else if (letterz.equals("!")) {
gameEnd = true;
} else {
System.out.println("No, " + letterz + " is not in the phrase\n");
wrong++;
}
}
}
while(gameEnd==false){
System.out.println(drawScoreboard(totalGuesses,错误));
System.out.println(“输入一个字母进行猜测或\“!\”结束游戏”);
letterz=序列号next();
for(int i=0;i=letterz.length()){
打破
}else if((Character.toString(letterz.charAt(i))).equalSignalCase(Character.toString(secretPhrase.charAt(i))){
系统输出打印(“\n”);
inc++;
}else if(字母Z等于(!”){
gameEnd=true;
}否则{
System.out.println(“否,+letterz+”不在短语\n中”);
错误++;
}
}
}
检查字符串是否与特定字符模式匹配的最简单方法是使用方法string.matches
。例如:
guess.matches("[A-Z!]+")
当且仅当
guess
仅包含大写字母或!
字符时,才会返回true。检查字符串是否与特定字符模式匹配的最简单方法是使用方法string.matches
。例如:
guess.matches("[A-Z!]+")
当且仅当guess
仅包含大写字母或!
字符时,才会返回true。请参阅