【发布时间】:2013-12-07 06:01:23
【问题描述】:
我正在处理我的刽子手程序的错误。如果用户输入数字(int/double)而不是字母(char/string),我想打印一条错误消息。我该怎么做?
这是引擎类的代码:
//hangman viewer stuff
//////////////////////////////////////////////////////////////
JFrame frame = new JFrame();
frame.setSize(200,375); //invoked the method setSize on the implicit parameter frame
frame.setTitle("Hangman");
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
HangmanComponent g = new HangmanComponent();
frame.add(g);
frame.setVisible(true);
///////////////////////////////////////////////////////////////
String wordd = JOptionPane.showInputDialog("Type in a word.");
int length = wordd.length();
String blank = "_ ";
String word2 = new String("");
int guesscount = 10;
ArrayList<String>answers=new ArrayList<String>(); //creates reference to empty structure that will contain references
char blanks[]=new char[wordd.length()]; //creates an array with the same number of terms as the length of the word
for (int i=0; i<length; i++)//fills the array with blanks corresponding to the length of the word
{
blanks[i] = '_';
}
HangmanComponent y = new HangmanComponent();
while (true)
{
String letter = JOptionPane.showInputDialog("Guess a letter! You have "+guesscount+" guesses."+"\n"+answers+"\n"+Arrays.toString(blanks).replace(",", " ").replace("[","").replace("]","")); //Prints a space
char letterchar = letter.charAt(0); //converts string letter to char letterchar
int idx = 0;
boolean found = false;
answers.add(letter); //adds the string to the arraylist answers
while (idx >= 0 && idx < length) //idx is greater than or equal to 0 but less than the length of the word
{
//System.out.println("idx = " + idx);
idx = wordd.indexOf(letter, idx); //idx is the index of "letter" in "wordd" and finds all instances of the letter
//System.out.println("idx = " + idx + ", guesscount = " + guesscount);
if (idx != -1) //if idx is not -1 (the letter exists in the word)
{
found = true;
blanks[idx] = letterchar; //sets the term in the array equal to the letter
idx += 1; //idx=idx+1
}
else
{
guesscount=guesscount-1;
y.nextStage();
y.printStage();
frame.add(y);
frame.setVisible(true);
break;
}
}
if (found)
{
JOptionPane.showMessageDialog(null, Arrays.toString(blanks).replace(",", " ").replace("[","").replace("]","")+"\n"+"You found a letter!"+"\n"+answers);
}
else
{
JOptionPane.showMessageDialog(null, Arrays.toString(blanks).replace(",", " ").replace("[","").replace("]","")+"\n"+"That letter is not in the word! Guess again!"+"\n"+answers);
if (guesscount == 0)
{
JOptionPane.showMessageDialog(null, "Sorry, you're all out of guesses. The answer was '"+wordd+".' Thanks for playing!");
break;
}
}
char [] lettersArray = wordd.toCharArray(); //converts word to array of chars
if (Arrays.equals(blanks, lettersArray))//compares array of blanks to array of letters
{
JOptionPane.showMessageDialog(null, "You guessed the word! Thanks for playing!");
break;
}
}
}
【问题讨论】:
-
好奇这都是写在main函数里的吗?
-
您可以创建一个可接受的字符数组,以及一个对其进行排序并检查变量(无论输入是什么)是否等于其中任何一个的方法;如果不是,则抛出错误并停止程序或再次向他们询问正确的字符(据说使用 while 循环)。这就是我的做法,尽管我只使用 Ruby 做过,从未在 Java 中尝试过。自动检查可能有很多更好的方法,但我不知道。
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@theGreenCabbage 是的,这一切都在主函数中
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作为为我的 Java 初学者课程完成这个项目的人,你真的应该学习面向对象编程。它让生活变得如此轻松。
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一些“A3”允许吗?
标签: java error-handling