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Java 错误:参数列表长度不同_Java_Arrays - Fatal编程技术网

Java 错误:参数列表长度不同

Java 错误:参数列表长度不同,java,arrays,Java,Arrays,错误: private class ButtonSelect implements ActionListener { @Override public void actionPerformed(ActionEvent e) { Object Ps = e.getSource(); String selectedLanguage = (String) languageList.getSelectedItem(); JOptionPan

错误:

private class ButtonSelect implements ActionListener {
    @Override
    public void actionPerformed(ActionEvent e) {
        Object Ps = e.getSource();
        String selectedLanguage = (String) languageList.getSelectedItem();
        JOptionPane.showMessageDialog(StudentRecordsGUI.this,
                "You selected the language: " + selectedLanguage);
        if (selectedLanguage != "English") {
        JOptionPane.showMessageDialog(StudentRecordsGUI.this,
            "I'm sorry. The language you have selected is currently not available. Please try again.");
        }
        else
            setVisible(false);
            new Record (student, this, "Student Records");
    }
} //end class ButtonSelect
我需要使用:

java:52: error: constructor Record in class StudentRecordsGUI.Record cannot be applied to given types;
                new Record (student, this, "Student Records");
                ^
  required: Student
  found: ArrayList<Student>,StudentRecordsGUI.ButtonSelect,String
  reason: actual and formal argument lists differ in length

从学生课堂上获取信息,但不需要。我如何编辑这一行,使它能够发布您的记录类?错误信息非常清楚。你至少试过阅读和理解它吗?还有一个你可能没有意识到的问题。因为在'else'子句中没有使用大括号,所以它只包含setVisible行。缩进的新记录。。。未在else子句中运行。这不会解决您在那里对构造函数的错误调用,但请将字符串与equals方法进行比较,而不是与==。否则你的if将永远不会评估为真。因此,在打了所有试图帮助你的人的耳光之后,你可以继续lakshman所说的:发布你的记录课程。但不管怎样:记录构造函数期望一个学生作为它的参数。把它交给那个学生,它就会起作用。
ArrayList<Student> student = new ArrayList<Student>();