我正在制作一个用户输入他/她的名字和姓氏的游戏。程序从名称中提取第一个字母，然后从字符串数组中输出名称。
我认为这个问题出现在代码的最后部分，我将Strings firstletter和lastLetter与数组进行比较。但我不确定。我花了一些时间研究，而且我被卡住了。
欢迎任何评论。你不会伤害我的感情。
import java.util.Arrays;
import javax.swing.*;
import java.awt.*;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.util.Scanner;
public abstract class ChristmasName extends JFrame implements ActionListener {
public static void main(String[] args) {
JFrame frame = new JFrame("What is your Christmas Name?");
frame.setVisible(true);
frame.setSize(400,300);
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
JPanel panel = new JPanel();
frame.add(panel);
JLabel firstName = new JLabel("First name:");
JTextField first = new JTextField(15);
JLabel lastName = new JLabel("Last name:");
JTextField last = new JTextField(15);
panel.add(firstName);
panel.add(first);
panel.add(lastName);
panel.add(last);
JButton submit = new JButton("Submit");
panel.add(submit);
submit.addActionListener(new ActionListener() {
@Override
public actionPerformed(ActionEvent e) {
String[] first_name = {"Apple","Eggnogg","Candy","Jingle","Holly","Goldie","Ho Ho","Frosty","Joyous","Mittens","Tinsel","Turkey","Tiny","Cranberry","Bloated","Angel","Bauble","Bulb","Ginger","Blitzen","Eve","Faith","Fruitcake","Goose","Glitter","Grinch"};
String[] last_name = {"Tidings","Swan","Jolly","Claus","Mistletoe","Punch","Chimney","Coal","Igloo","Jumper","Myrhh","Pudding","Reindeer","Rejoice","Icicle","Midnight","Shepherd","Surprise","Gift","Magi","Train","Tree","White","Donkey","Wreath","Stuffing"};
String firstLetter = first.getText();
firstLetter = firstLetter.substring(0,1);
String lastLetter = last.getText();
lastLetter = lastLetter.substring(0,1);
if (firstLetter == "A") {
firstLetter = first_name[0];
}
JOptionPane.showMessageDialog(null, firstLetter + " " + lastLetter);
System.exit(0);
}
});
}
}
答案 0 :(得分：1)
不确定您要问的是什么，但我在您的代码中看到的第一个问题是：
if（firstLetter ==＆＃34; A＆＃34;）{ firstLetter = first_name [0];
}
请参阅以下有关如何检查两个字符串是否具有相同值的信息： How do I compare strings in Java?
答案 1 :(得分：1)
因为您只需要一个字符，所以应使用charAt()代替substring。虽然substring是可能的，但我总是忘记哪个参数是包含性的还是排他性的。我想你也一样。
你应该声明2个字符：
char firstChar = first.getText().charAt(0);
char lastChar = last.getText ().charAt (0);
然后你可以查看它们：
if (firstChar == 'A') { //Remember to use single quotes!
答案 2 :(得分：1)
这是我解决问题的方法：
String[] first_names = {"Apple","Eggnogg","Candy","Jingle","Holly","Goldie","Ho Ho","Frosty","Joyous","Mittens","Tinsel","Turkey","Tiny","Cranberry","Bloated","Angel","Bauble","Bulb","Ginger","Blitzen","Eve","Faith","Fruitcake","Goose","Glitter","Grinch"};
String[] last_names = {"Tidings","Swan","Jolly","Claus","Mistletoe","Punch","Chimney","Coal","Igloo","Jumper","Myrhh","Pudding","Reindeer","Rejoice","Icicle","Midnight","Shepherd","Surprise","Gift","Magi","Train","Tree","White","Donkey","Wreath","Stuffing"};
// User Input:
// Note: converted to lower case so the chars can be compared easily.
String firstName = first.getText().toLowerCase();
String lastName = last.getText().toLowerCase();
// Vars for output:
String sChristmasFirstName = null;
String sChristmasLastName = null;
// Do first name (if entered)
if(firstName.length() > 0){
// Loop all names to find the match:
for(String name : first_names){
if(name.toLower().charAt(0) == firstName.charAt(0)){
// We found a Christmas first name for the user
sChristmasFirstName = name;
// we can now exit the loop
break;
}
}
} // else, the user left this field blank
// Do same thing for last name
if(firstName.length() > 0){
// Loop all names to find the match:
for(String name : last_names){
if(name.toLower().charAt(0) == lastName.charAt(0)){
// We found a Christmas last name for the user
sChristmasLastName = name;
// we can now exit the loop
break;
}
}
} // else, the user left this field blank
// Prepare output string:
String output = "";
String outputErrorPortion = "";
if(sChristmasFirstName != null){
output += sChristmasFirstName;
}else{
if(firstName.length() == 0){
outputErrorPortion += "It looks like you didn't enter a first name.";
}else{
// Could not find an applicable first name
output += firstName;
ouputErrorPortion += "It looks like we couldn't find a Christmas first name for you :-(";
}
}
if(sChristmasLastName != null){
output += " " + sChristmasLastName;
}else{
if(lastName.length() == 0){
outputErrorPortion += " It looks like you didn't enter a last name.";
}else{
// Could not find an applicable last name
output += " " + lastName;
ouputErrorPortion += " It looks like we couldn't find a Christmas last name for you :-(";
}
}
// trim leading and trailing spaces if there are any:
output = output.trim();
outputErrorPortion = outputErrorPortion.trim();
// Variable 'output' now contains the Christmas first name, last name, both, or neither.
// Error message now contains a descriptive error about what happened (if anything)
用于选择圣诞节名称的字符串比较发生在每个循环中，循环遍历first_names和last_names字符串数组以查找以与用户相同的字母开头的第一个匹配＆＃39 ;输入名字和姓氏。然后圣诞节名称最后连接到output变量，如果在阵列中找不到相关条目，则用户输入的第一个和/或最后一个名称代替圣诞节等价物圣诞节的名字。如果在处理名称时发生任何错误，则outputErrorPortion变量中也会构造错误消息。
答案 3 :(得分：0)
以下是您需要仔细阅读的代码：
String[] firstNames = { "Apple", "Eggnogg", "Candy", "Jingle", "Holly", "Goldie", "Ho Ho", "Frosty","Joyous", "Mittens", "Tinsel", "Turkey", "Tiny", "Cranberry", "Bloated", "Angel", "Bauble","Bulb", "Ginger", "Blitzen", "Eve", "Faith", "Fruitcake", "Goose", "Glitter", "Grinch" };
String[] lastNames = { "Tidings", "Swan", "Jolly", "Claus", "Mistletoe", "Punch", "Chimney", "Coal","Igloo", "Jumper", "Myrhh", "Pudding", "Reindeer", "Rejoice", "Icicle", "Midnight", "Shepherd","Surprise", "Gift", "Magi", "Train", "Tree", "White", "Donkey", "Wreath", "Stuffing" };
// ArrayLists will contain the matching items
ArrayList<String> firstNamesMatching = new ArrayList<String>();
ArrayList<String> lastNamesMatching = new ArrayList<String>();
// Check which names from firstNames matching the firstLetter
String firstLetter = first.getText().substring(0, 1).toUpperCase();
for (String s : firstNames) {
if (s.startsWith(firstLetter))
firstNamesMatching.add(s);
}
// Check which names from lastNames matching the lastLetter
String lastLetter = last.getText().substring(0, 1).toUpperCase();
for (String s : lastNames) {
if (s.startsWith(lastLetter))
lastNamesMatching.add(s);
}
JOptionPane.showMessageDialog(null, firstNamesMatching.toArray() + " " + lastNamesMatching);
此外：
虽然数组firstNames和lastNames等必须在actionListener之外。你应该记住程序使用的内存，而不是一次又一次地初始化相同的内容。