Question

我正在尝试调用showArrayList方法但它不断给我的ContactList无法转换为java.util.ArrayList<Contact>。我究竟做错了什么？到目前为止，我有这个：

public class ControlPanel extends JPanel implements ActionListener
{
    // Fields (instance variables/attributes)
    private JLabel      phoneNumberLabel;  // a reference to the phoneNumberLabel that is displayed in the PhoneNumberDisplayPanel.

    private Contact     nameAndNumber;    // Holds the name and number for a phone Contact.
    // Contact lists;
    private ContactList contacts;          // Encapsulates a list of saved contacts, ie.e phone Contacts.
    private ContactList callsMade;         // Encapsulates a list of calls made, i.e. phone Contacts.

private void findNumber( )
    {
        if(nameLabel.getText(  ).equals(contacts.searchByName(nameLabel.getText())))
            showArrayList(contacts,nameLabel.getText());
        else
            showArrayList(contacts, "");
    }

showArrayList方法是：

public void showArrayList( ArrayList<Contact> list, String title ) 
    {
        int x = 410;
        int y = 445;
        int width  = 350;
        int height = 200;
        boolean disableCloseButton = false;
        Window newWindow  = new Window( title, x, y, width, height, Color.RED, disableCloseButton );
        {
            newWindow.println( "No contact information saved" );
        }
        else
        {
            for ( Contact c: list )
            {
                newWindow.println( c.toString() );
            }
        }

现在我做了这个并编译：

private void findNumber( )
    {
        ArrayList<Contact> list = new ArrayList<Contact>();
        if(nameLabel.getText().equals(contacts.searchByName(nameLabel.getText())))
            showArrayList(list,nameLabel.getText());
        else
            showArrayList(list, "");
    }

我的ContactList类：

public class ContactList
{
    private ArrayList<Contact> list;

    /**
     * Constructor for a ContactList, to initialize the list from file.
     * 
     * @param  fileName   a reference to an existing file
     * @param  window     a reference to an existing window to display the list contacts
     */
    public ContactList( String fileName, Window window )
    {
        list = new ArrayList<Contact>( );

        File file = new File( fileName );

        // Remark. A file needs to have been written, before it can be read.
        if (file.exists())
        {
            try
            {
                Scanner fileReader = new Scanner( file ); 
                readFile( fileReader );
                window.print( this.toString () );
            }
            catch( FileNotFoundException error )   // could not find file
            {
                System.out.println( "File not found " );
            }
        }
    }

    /********************************************************************
     * Searches the list for a particular contact, 
     * comparing this name and number (in lower case) 
     * with the contact name and number (in lower case).
     * 
     * @param   contact
     * @return  returns true if the contact is found, otherwise false
     */
    public boolean found( Contact contact )
    {  
        for(Contact c: list){
            if(c == contact){
                return true;
            }   
        }
        return false;

    }

    /******************************************************************
     * Adds one contact to the list
     * 
     * @param   contact
     */
    public void add( Contact contact )
    {
        list.add(contact);


    }

    /******************************************************************
     * searches the list for each name that contains the specified substring
     * 
     * @param   substring
     * @return  returns the result as an ArrayList
     */
    public ArrayList<Contact> searchByName(String substring) 
    {
        ArrayList<Contact> l = new ArrayList<Contact>( );
        for(Contact c: list){
            if(c.getName() == substring){
                l.add(c);   
            }
        }
        return l;   
    }

Answer 1

ContactList不会延伸ArrayList<Contact>，因此您无法将其作为参数传递给期望ArrayList<Contact>的方法。也许您可以在getter中使用ContactList来展示list，以便将其传递给showArrayList方法，或者您可以实施showArrayList加入ContactList已键入argument的方法。

Answer 2

private void findNumber( )
    {
        if(nameLabel.getText().equals(contacts.searchByName(nameLabel.getText())))
            showArrayList(contacts.searchByName(""),nameLabel.getText());
        else
            showArrayList(contacts.searchByName(""), "");
    }

试图调用showArrayList方法

2 个答案: