2

所以我需要做的是使用 JTextField 以字符串的形式获取用户输入，虽然我可以做这部分，但我无法从方法中获取字符串以在程序的其他地方进行分析，这就是我到目前为止：

 import javax.swing.* ;
 import java.awt.event.* ;
 import java.util.* ;
 class Games extends JFrame implements ActionListener
{
  JPanel pnl = new JPanel() ;
  JTextField input = new JTextField(38) ;
  JTextArea output = new JTextArea(6, 37) ;
  JScrollPane pane = new JScrollPane(output) ;
  String inputString ;

public Games()
{
    super("Text Based RPG") ;
    setSize(600,200) ;
    setDefaultCloseOperation(EXIT_ON_CLOSE) ;
    output.setLineWrap(true) ;
    output.setWrapStyleWord(true) ;
    pane.setVerticalScrollBarPolicy(JScrollPane.VERTICAL_SCROLLBAR_ALWAYS) ;
    output.setEditable(false) ;
    pane.getVerticalScrollBar().setValue(100); //scrollbar always at bottom
    input.requestFocus() ;

    /*Edit: Should I call for the input string here, e.g. output.append(inputString)
     i get the compilation error message:
    Games.java:29: error: cannot find symbol
    output.append(inputString) ;
                  ^
    symbol:   variable inputString
    location: class Games
    */

    add(pnl) ;
    pnl.add(pane) ;
    pnl.add(input) ;

    input.addActionListener(this) ;

    setVisible(true) ;
}

public void actionPerformed(ActionEvent event) 
{
    inputString = input.getText() ;
    input.setText("") ;
}         
/*The string i need to analyse is inputString, but I cannot, get it out of this method
  any help you guys can give me would be greatly appreciated. */

public static void main(String[] args) 
{
    Games gui = new Games() ;
}
}

**********************************
    try
    {
        a = reader.readLine() ;
    }
    catch (IOException e ) 
    {
        System.out.println("An Input Error Has Occured") ;
    }
    if (a.indexOf("hide") != -1 ) switchvariable = 'b' ;
    if  ((a.indexOf("exit") != -1 ) || (a.indexOf("leave") != -1) || (a.indexOf("out") != -1)) switchvariable = 'a' ;
    if ((a.indexOf("exit") == -1) && (a.indexOf("hide") == -1) && (a.indexOf("leave") == -1) && (a.indexOf("out") == -1)) switchvariable= 'c' ;
    String outcome = "" ;
    switch (switchvariable) 
    {
    case 'a' : outcome = "You exit the tavern to be discovered by\na group of bandits!\n" ; i = -1 ; break ;
    case 'b' : outcome = "You hide behind the bar for 10 minutes until\nyou can no longer hear the group." ; i = -1 ; break ;
    case 'c' : outcome = "You must choose between the two given options, exit the tavern, or hide." ; break ; // the variable i will still be greater
    //than 1 so the loop will continue until you reach an agreeable option, which in this case is either hiding, or addressing the group of people
    }
    System.out.println(outcome) ;

理想情况下我想做的是使用文本字段进行用户输入，所以我需要做的是将文本放在输出文本区域中，然后这应该根据用户输入进行更改，我想使用的代码示例显示在星号下方

4

1 回答 1

3

如果我理解正确，当用户在文本字段中按 Enter 时，您希望将在文本字段中输入的文本附加到输出文本区域。

当按下 Enter 时，文本字段会触发一个动作事件。你在课堂上听这个事件。因此，此时，您只需要将值附加到文本区域即可。

Swing 是基于事件的：

public void actionPerformed(ActionEvent event) {
    String inputString = input.getText() ;
    output.append(inputString);
    input.setText("");
}

构造函数中的代码仅在执行时执行。所以获取文本字段的值当然会返回一个空字符串，因为在构建时，文本字段甚至还不可见。

编辑：做你想做的事，只需要以下内容：

public void actionPerformed(ActionEvent event) {
    String inputString = input.getText() ;
    printOutcome(inputString);
    input.setText("");
}

private void printOutcome(String input) {
    char switchvariable = ' ';
    if (input.indexOf("hide") != -1) {
        switchvariable = 'b' ;
    }
    if ((input.indexOf("exit") != -1 ) || (input.indexOf("leave") != -1) || (input.indexOf("out") != -1)) {
        switchvariable = 'a';
    }
    if ((input.indexOf("exit") == -1) && (input.indexOf("hide") == -1) && (input.indexOf("leave") == -1) && (input.indexOf("out") == -1)) {
        switchvariable= 'c';
    }
    String outcome = "" ;
    switch (switchvariable) {
        case 'a' : 
            outcome = "You exit the tavern to be discovered by\na group of bandits!\n"; 
            break;
        case 'b' : 
            outcome = "You hide behind the bar for 10 minutes until\nyou can no longer hear the group." ; 
            break ;
        case 'c' : 
            outcome = "You must choose between the two given options, exit the tavern, or hide."; 
            break;
    }
    System.out.println(outcome);
}

不过，这种方法太复杂了。使用 if 子句设置 char 变量，然后切换此 char 变量是没有意义的。你的 if 子句也应该清理干净。

于 2012-12-18T20:00:50.540 回答