我正在尝试创建一个使用文本字段将字符串添加到链接列表的Java applet。我无法使用搜索按钮。我试图在文本字段中获取用户指定的字符串,然后搜索列表并打印该单词被查找的次数。
import javax.swing.*;
import java.awt.*;
import java.awt.event.*;
import java.util.LinkedList;
import java.util.ListIterator;
/**
* Created by joshuaogunnote on 31/10/2015.
*/
public class Applet2 extends JApplet {
private String text;
private int text1;
JTextField value1, value2;
LinkedList<String> list = new LinkedList<String>();
public JLabel jLabel;
public int count = 0;
我创建了这个search_count变量来计算找到该单词的次数。
public int search_count = 0;
public void init() {
JLabel prompt = new JLabel("Please enter a word");
JLabel prompt1 = new JLabel("Please enter a certain letter");
value1 = new JTextField(10);
value2 = new JTextField(10);
JPanel textPanel = new JPanel();
textPanel.add(prompt);
textPanel.add(value1);
add(textPanel, BorderLayout.NORTH);
textPanel.add(prompt1);
textPanel.add(value2);
JPanel centrePanel = new JPanel();
text = "";
jLabel = new JLabel(text);
centrePanel.add(jLabel);
add(centrePanel, BorderLayout.CENTER);
JButton but = new JButton("Add word");
JButton but1 = new JButton("Clear");
JButton but2 = new JButton("Remove first occurrence");
JButton but3 = new JButton("Remove all occurrences");
JButton but4 = new JButton("Display all words begging with certain letter");
JButton but5 = new JButton("Search");
JPanel butPanel = new JPanel();
butPanel.add(but);
butPanel.add(but1);
butPanel.add(but5);
butPanel.add(but2);
butPanel.add(but3);
butPanel.add(but4);
add(butPanel, BorderLayout.SOUTH);
but.addActionListener(new ButtonHandler(this));
but1.addActionListener(new ButtonHandler1(this));
but5.addActionListener(new ButtonHandler2(this));
}
class ButtonHandler implements ActionListener {
private Applet2 theApplet;
public ButtonHandler(Applet2 app) {
theApplet = app;
}
public void actionPerformed(ActionEvent e) {
text = theApplet.value1.getText();
try {
text1 = Integer.parseInt(text);
jLabel.setText("ERROR - The string " + "'" + text1 + "'" + " is not a valid word");
} catch (NumberFormatException e1) {
if (text.length() != 0) {
jLabel.setText("Word " + "'" + text + "'" + " has been added to the list");
count = count + 1;
} else {
jLabel.setText("ERROR - Please enter a word");
}
}
}
}
class ButtonHandler1 implements ActionListener {
private Applet2 theApplet;
public ButtonHandler1(Applet2 app) {
theApplet = app;
}
public void actionPerformed(ActionEvent e) {
list.clear();
jLabel.setText("List has been cleared");
count = 0;
}
}
class ButtonHandler2 implements ActionListener {
private Applet2 theApplet;
public ButtonHandler2(Applet2 app) {
theApplet = app;
}
public void actionPerformed(ActionEvent e) {
String text = theApplet.value1.getText();
这里我试图使用for循环迭代列表中的所有字符串,如果找到匹配则增加search_count。然而,它并没有产生正确的答案。当用户尝试搜索不在列表中的单词时,我也尝试生成ERROR消息。如何获取search_count变量以及如何在正确的时间显示ERROR消息?
for (int i = 0; i < list.size(); i++) {
if(text.equals(list.get(i))){
search_count = search_count + 1;
} else {
jLabel.setText("ERROR - word is not in the list")
}
jLabel.setText("Word " + "'" + text + "'" + " was found " + search_count + " time(s) in the list");
if (text.length() == 0) {
jLabel.setText("Please enter a word - The total number of words in the list are: " + count);
}
}
}
}
答案 0 :(得分:0)
在循环中使用它之前,似乎没有将count声明为变量。如果输出不是您想要/期望的,那么循环的条件会为您提供其他内容,而不是您的想法。
int count = 0;