Question

嘿大家有人可能会问过这个但是我试图从用户回答中实现一个ArrayList我已经做了一些研究而且它不会这样做。这是我的代码＆lt;部分内容

public class Breed
{

public static void cross_breed()
    {
        // Creates 3 strings named male_breed, female_breed, and cross_breed. 
        String male_breed;   
        String female_breed;
        String cross_breed;
        Cross_breed crossbreed;
        List<String> clist = new ArrayList<String>();

        while (true)  // While the male_breed is a real breed continue on with the program
        {
            male_breed = JOptionPane.showInputDialog("What is the breed of the male?");
            if (male_breed.length() == 0) break; // Makes it so if the male dog breed wasn't entered the program will exit
            if (List_Breeds.blist.containsKey(male_breed))  //If the given input of the users is on the list of breeds continue with program
            {
                while (true) // Loops the female breed
                {
                    female_breed = JOptionPane.showInputDialog("What is the breed of the Female");
                    if (List_Breeds.blist.containsKey(female_breed)) //
                    {
                        if (male_breed.equals(female_breed)) // if both breeds are the same
                        {
                            clist.add(JOptionPane.showMessageDialog(null,  "Your dog is a Purebread " + male_breed));
                            break; //Ends the program
                        } 
                        else  
                        {

                            cross_breed = male_breed.substring(0, 3) + "a" + female_breed.substring(0, 3);     //If both breeds arent the same take the           
                            clist.add(JOptionPane.showMessageDialog(null, "Your dog breed is a " + cross_breed));         // first three letters of the male_breed and inserts
                            break;  // Ends the program                                                                     // an 'A' at the end.
                        }

Eclipse给我一个错误说明两行

clist.add(JOptionPane.showMessageDialog(null,  "Your dog is a Purebread " + male_breed));

和

clist.add(JOptionPane.showMessageDialog(null, "Your dog breed is a " + cross_breed));

表示“类型列表中的方法添加不适用于参数（void）

如果我做错了请帮助......我要做的是将品种（或品种）名称添加到arraylist中，其中将显示在第二个菜单中

Answer 1

“类型列表中的方法添加不适用于参数（void）”表示您正在调用的方法 - JOptionPane.showMessageDialog(JComponent, String)不会返回任何内容。

这是因为showMessageDialog用于显示消息，而不是捕获输入。

您需要查看JOptionPane.showInputDialog()的各种风格，以便从用户那里获得输入。

将JOptionPane中的某些输入添加到Array中

1 个答案: