根据if语句显示GUI中的不同按钮

时间:2016-04-30 17:19:32

标签: java user-interface if-statement javafx

我正在尝试创建一个有两个选项的游戏:学生可以通过引入答案或单击包含正确答案的按钮来回答问题。我为两种类型的问题创建了texfield和按钮,并且如果要显示按钮或文本字段但是它不起作用,则根据问题的类型(类型是预定的)制作if语句(它不是进入if)。 我的代码如下所示:

    SanswerTextField = new TextField();
    SanswerTextField.setPrefWidth(300.0);
    SanswerTextField.setMinWidth(300.0);
    SanswerTextField.setMaxWidth(300.0);
    SanswerTextField.setPrefHeight(50.0);
    SanswerTextField.setStyle("-fx-font: 30 timesnewroman; -fx-base: #AE3522");



    SanswerButton1 = new Button();
    SanswerButton1.setPrefWidth(50.0);
    SanswerButton1.setMinWidth(50.0);
    SanswerButton1.setMaxWidth(50.0);
    SanswerButton1.setPrefHeight(50.0);
    SanswerButton1.setStyle("-fx-font: 30 timesnewroman; -fx-base: #AE3522");


    SanswerButton2 = new Button();
    SanswerButton2.setPrefWidth(50.0);
    SanswerButton2.setMinWidth(50.0);
    SanswerButton2.setMaxWidth(50.0);
    SanswerButton2.setPrefHeight(50.0);
    SanswerButton2.setStyle("-fx-font: 30 timesnewroman; -fx-base: #AE3522");



    SanswerButton3 = new Button();
    SanswerButton3.setPrefWidth(50.0);
    SanswerButton3.setMinWidth(50.0);
    SanswerButton3.setMaxWidth(50.0);
    SanswerButton3.setPrefHeight(50.0);
    SanswerButton3.setStyle("-fx-font: 30 timesnewroman; -fx-base: #AE3522");



    if(saveTypeQ.getText().equals("Choose Answer"))
    {
        SanswerButton1.setText(saveAnswer.getText());
        System.out.println("answer in button: " + SanswerButton1.getText());
        SanswerButton2.setText(saveAnswer.getText());
        SanswerButton3.setText(saveAnswer.getText());

        mathTestButtonPane.add(SanswerButton1,1,4);
        mathTestButtonPane.add(SanswerButton2,2,4);
        mathTestButtonPane.add(SanswerButton3,3,4);

        System.out.println("I am a question to choose an answer");
    }
    else if(saveTypeQ.getText().equals("Introduce Answer"))
    {
        mathTestButtonPane.add(SanswerTextField,1,4);
        System.out.println("I am a question to introduce an answer");
    }
    else
    {
        System.out.println("Not working");

    }

它总是访问其他人,我不知道为什么。我真的很感激帮助。提前谢谢。

0 个答案:

没有答案