Java 为什么equals函数不起作用?
当我编译下面的代码时,它显示(字符串)username和myList.get(0)等于,但是equals函数返回false为什么密码也会发生同样的情况Java 为什么equals函数不起作用?,java,equals,Java,Equals,当我编译下面的代码时,它显示(字符串)username和myList.get(0)等于,但是equals函数返回false为什么密码也会发生同样的情况 btnLogIn.addActionListener(new ActionListener() { public void actionPerformed(ActionEvent e) { String Username=(String) textField.getText(); //fatch the user name
btnLogIn.addActionListener(new ActionListener()
{
public void actionPerformed(ActionEvent e) {
String Username=(String) textField.getText(); //fatch the user name from text field
String Password=(String) textField_1.getText(); //fatch password frof text field
databaseconnection connect = new databaseconnection(); // databaseconnection class object to connect to data base
ArrayList myList = connect.search(Username,Password); //serch the username and password in data base
System.out.println((String)myList.get(0)); //for testing
System.out.println((String)myList.get(1)); //for testing
System.out.println(Username); //for testing
System.out.println(Password); //for testing
System.out.println(Username.equals(myList.get(0))); //for testing
System.out.println(Password.equals((String)myList.get(1))); //for testing
if(Username.equals(myList.get(0))&&Password.equals((String)myList.get(1))){
System.out.println("Hello"+Username);
}
}
});
这是我的数据库连接类
import java.sql.*;
import java.util.ArrayList;
public class databaseconnection{
Statement stmt ;
ResultSet rs ;
Connection conn;
ArrayList<String> temp = new ArrayList<String>();
public void getconnection(){
try{
Class.forName("sun.jdbc.odbc.JdbcOdbcDriver");
Connection conn = DriverManager.getConnection("jdbc:odbc:Database1","","");
stmt = conn.createStatement();
}
catch(Exception e){
System.out.println("connection error");
}
}
public ArrayList search(String Username,String Password){
getconnection();
try{
rs = stmt.executeQuery("select username,password from login where username = \'"+Username+"\'");
if(rs.next()){
String tempString=rs.getString("username");
temp.add(tempString);
tempString= rs.getString("password");
temp.add(tempString);
}
}
catch(Exception e){
System.out.println("search error");
}
return temp;
}
}
import java.sql.*;
导入java.util.ArrayList;
公共类数据库连接{
报表stmt;
结果集rs;
连接接头;
ArrayList temp=新的ArrayList();
公共连接()无效{
试一试{
forName(“sun.jdbc.odbc.JdbcOdbcDriver”);
Connection conn=DriverManager.getConnection(“jdbc:odbc:Database1”,“”,“”);
stmt=conn.createStatement();
}
捕获(例外e){
System.out.println(“连接错误”);
}
}
公共ArrayList搜索(字符串用户名、字符串密码){
getconnection();
试一试{
rs=stmt.executeQuery(“从登录名中选择用户名、密码,其中用户名=\'”+username+“\”);
如果(rs.next()){
String tempString=rs.getString(“用户名”);
临时添加(临时字符串);
tempString=rs.getString(“密码”);
临时添加(临时字符串);
}
}
捕获(例外e){
System.out.println(“搜索错误”);
}
返回温度;
}
}
将myList定义为typeArrayList
我将您的代码改写为:
String Username = "user";
String Password = "pass";
List<String> myList = new ArrayList<>();
myList.add("user");
myList.add("pass");
System.out.println(Username);
System.out.println(Password);
System.out.println(myList.get(0));
System.out.println(myList.get(1));
System.out.println(Username.equals(myList.get(0)));
System.out.println(Password.equals(myList.get(1)));
if (Username.equals(myList.get(0)) && Password.equals(myList.get(1))) {
System.out.println("Hello, " + Username);
}
它仍然不是完美的,因为search()
方法应该返回类型为UserCredentials
的实例或List
的实例,具体取决于它应该做什么<代码>用户凭据将如下所示:
public void actionPerformed(ActionEvent e) {
String username = "user";
String password = "pass";
List<String> dtbSearchResults;
try (DatabaseConnection connection = new DatabaseConnection()) {
dtbSearchResults = connection.search(username, password);
} catch (SomeExceptionYouReallyShouldHandle e) {
// seriously, handle it here
}
System.out.println(username);
System.out.println(password);
System.out.println(dtbSearchResults.get(0));
System.out.println(dtbSearchResults.get(1));
System.out.println(username.equals(dtbSearchResults.get(0)));
System.out.println(password.equals(dtbSearchResults.get(1)));
if (username.equals(dtbSearchResults.get(0)) && password.equals(dtbSearchResults.get(1))) {
System.out.println("Hello, " + username);
}
}
public class UserCredentials {
private final String username;
private final String password;
public UserCredentials(String username, String password) {
// maybe some validity checks
this.username = username;
this.password = password;
}
public String getUsername() {
return username;
}
public String getPassword() {
return password;
}
}
实现该类后,代码将如下所示(假设search()
只返回一个结果):
如果您还有其他问题,请询问。这里或那里可能有演员,但您是卡斯图拉伯爵。打破这个习惯…我试过了,但结果是萨梅特不会改变结果。为什么列表没有参数化?@Rob+1代表“Count Castula”,我在那个(有效的)技术点上大声笑了出来。试着用引号括住打印输出,以确保没有尾随空格,例如:System.out.println(“\”“+Username+“\”“”)(顺便说一句,变量的名称很糟糕)。如果存在无法打印的字符,您也可以打印长度。@RavikumArmity,当您在声明列表中只能放置
字符串类型的对象后添加
。使用此方法意味着您不必到处使用强制类型转换。您也不必担心将所有的equals比较作为目标。您的搜索方法返回的类型是什么?我建议您看看java泛型教程好吗?@ravikumarmitry查找connect.search()
方法,并将其返回类型更改为ArrayListUsername是字符串。对原始类型列表的get调用返回一个对象
。因此,将一个对象和一个字符串兼容,这将是不相等的。按照PoiXen的说法设置列表的类型将有助于在字符串类型的临时变量中输入值<代码>字符串tmpUser=myList.get(0.toString();tmpUser.equals(用户名)代码>@RavikumArmitry尝试解决您的新问题一段时间,如果您不成功,则开始新的问题。编译器将显示未选中或不安全的警告operation@RaviKumarMistry在哪一条线上,在哪里?我打赌问题在于search()
方法的声明。将其返回类型更改为List
,并将其周围的代码更改为支持该类型。如果您无法做到这一点,请编辑您的问题并添加您的search()
方法。@ravikumanistry正是我所说的。更改为public ArrayList search(字符串用户名、字符串密码){}
到public List search(字符串用户名、字符串密码){}
。
public void actionPerformed(ActionEvent e) {
String username = "user";
String password = "pass";
UserCredentials user;
try (DatabaseConnection connection = new DatabaseConnection()) {
user = connection.search(username, password);
} catch (SomeExceptionYouReallyShouldHandle e) {
// seriously, handle it here
}
System.out.println(username);
System.out.println(password);
System.out.println(user.getUsername());
System.out.println(user.getPassword());
System.out.println(username.equals(user.getUsername()));
System.out.println(password.equals(user.getPassword()));
if (username.equals(user.getUsername())
&& password.equals(user.getPassword())) {
System.out.println("Hello, " + username);
}
}